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00:09:570Paolo Guiotto: Great weekend.
00:14:130Paolo Guiotto: Okay.
00:15:820Paolo Guiotto: So, we… Just the… We start from the statement of the differentiability TOM.
00:29:580Paolo Guiotto: So these theorems say… Sorry.
00:43:520Paolo Guiotto: Okay, so these theorem gives a sufficient condition, under which we can differentiate a function
00:52:870Paolo Guiotto: Which is defined as an integral where the function, inside the integral depends on some parameter XC.
01:01:700Paolo Guiotto: And basically, it says that we can carry the derivative inside, okay? The assumptions are, of course, that the function capital F must be defined, which is this assumption here.
01:15:780Paolo Guiotto: then, of course, that the partial derivative with respect to the parameter is the finder, and that should be controlled uniformly in the parameter by a function g of x, which is integral in the variable X.
01:34:250Paolo Guiotto: Now, we will see some applications of this, this,
01:40:110Paolo Guiotto: theorem. Let's do… the first, one is, actually a remarkable, nice, application.
01:49:550Paolo Guiotto: Which is the computation of the Fourier transform of the Gaussian distribution.
01:55:460Paolo Guiotto: Of course, we have not yet seen what is the Fourier transform, but let's take as an exercise, okay? Fourier transform of…
02:03:640Paolo Guiotto: Goshen…
02:08:970Paolo Guiotto: distribution.
02:16:690Paolo Guiotto: Okay, so, I remind you that, D.
02:22:830Paolo Guiotto: Curie transformer.
02:24:950Paolo Guiotto: of a function F,
02:27:860Paolo Guiotto: is this thing, is the function f hat of c, defined as the integral on the real line of f of x times this exponential E minus iCX dx.
02:44:190Paolo Guiotto: By the way, even if the function is real-valued, as you can see, that exponential is a complex number.
02:52:290Paolo Guiotto: So that function in the integral is a complex valued, but that's why we need also the integral be defined for complex valued functions.
03:03:100Paolo Guiotto: Now, for the Gaussian, this means that we take, this function f.
03:10:410Paolo Guiotto: the… in probability, this type of function… well, actually, to say the Gaussian distribution is a function like this, E minus X squared divided by 2 sigma squared. This is how it is written in probability.
03:27:690Paolo Guiotto: Well, now…
03:32:330Paolo Guiotto: Having this… the exponential written in this way, perhaps will carry away some calculation, but however, let's say let F of X be this function, X real.
03:46:950Paolo Guiotto: Where, sigma square stands for a positive quantity.
03:51:970Paolo Guiotto: Okay?
03:53:970Paolo Guiotto: So this function is, well-defined, and,
04:02:270Paolo Guiotto: And, okay, well, let FDB this, and, we compute, the goal is,
04:14:640Paolo Guiotto: compute the… the, F hat of Xi for this F.
04:24:420Paolo Guiotto: So, this means that we have to compute the integral on R of E minus X squared divided to sigma squared e minus iCX DX. Now, this integral can be computed in different ways.
04:42:200Paolo Guiotto: Depending on what kind of tools we have. For example, if we have tools
04:46:740Paolo Guiotto: provided by calculus with olomorphic functions. This can be computed with methods coming up from that theory, but we suppose that we do not have these tools.
04:59:480Paolo Guiotto: Also because we want to illustrate a typical feature of this kind of… this kind of operations will be in the gas.
05:11:140Paolo Guiotto: Okay, so let's check, first of all, that this function, f is well-defined. As you can see, we can call what is inside here. I know that it's a bit an abuse of notation, but I want to keep the same notations of the statements.
05:29:520Paolo Guiotto: So we will call this as F of XC.
05:33:490Paolo Guiotto: the function we have here, in such a way that you can see that our F hat of X is just the kind of object we are here, no? It's the integral on some domain E, that in this case is the real line.
05:49:60Paolo Guiotto: The measure is the Lebec measure of a function F, depending on two variables. One is the integration variable X, and the other is the parameter.
05:59:680Paolo Guiotto: Okay, so, step So, the function f of x is, well, defined… for every C in R.
06:20:750Paolo Guiotto: What does it mean, this? It means that… these… follows.
06:30:300Paolo Guiotto: Once… We prove… that.
06:38:140Paolo Guiotto: you remind what is the condition, let's say the minimal condition, that makes that function as a function of T well defined? It is this one. The function F
06:50:340Paolo Guiotto: as function of integration variable for parameter X fixed is in L1. That's what we have to…
06:58:410Paolo Guiotto: Sure, no? Because we want to show that that function is interval. So the function F, as function of X, with the parameter C fixed, is in L1 of the real line.
07:12:50Paolo Guiotto: But this is, if and only if.
07:14:350Paolo Guiotto: Well, in our case, well, let's say… this. Now, Our F… as function of variable X,
07:29:100Paolo Guiotto: see fixed them, is, first of all, when you look at that function as a function of X,
07:35:530Paolo Guiotto: You see, there is that exponentially minus X player.
07:38:930Paolo Guiotto: And then we have another exponential. As function of X, this is a continuous function.
07:44:260Paolo Guiotto: Okay, it depends nice from X. So it is continuous in the rear line, and therefore it is measurable.
07:59:770Paolo Guiotto: And, F… as a function of X,
08:04:600Paolo Guiotto: is now in L1 if and only if the integral on R of the absolute value of FXC
08:14:70Paolo Guiotto: DX is 5.
08:17:540Paolo Guiotto: But what is this?
08:19:680Paolo Guiotto: You see that this is E2 minus X squared over 2 sigma squared times E2 minus iCX.
08:30:600Paolo Guiotto: When they take the absolute value of this.
08:33:250Paolo Guiotto: The modulus of the product is the product of the modulus, then
08:37:860Paolo Guiotto: models of E minus X squared, etc, that's a positive number.
08:42:799Paolo Guiotto: So this is a positive number. So the modulus will be himself. So, E minus X squared divided 2 sigma squared, times the modulus of E minus ICX. That's a complex number.
08:57:550Paolo Guiotto: But to remind that here, our X is the integration variable, is real.
09:02:730Paolo Guiotto: Our C is the parameter, is real as well, I've not said that.
09:08:670Paolo Guiotto: Perhaps so, so… sees real.
09:15:710Paolo Guiotto: So, this means that that is something E to i theta.
09:20:820Paolo Guiotto: is a unitary complex number, no? You know that e to i theta is cosine theta plus i sine theta. So, in particular, the absolute value of EI theta is just 1.
09:37:540Paolo Guiotto: So this number is constantly equal to 1, So, this condition here…
09:45:360Paolo Guiotto: The integral of absolute value of FXXE in the xfinite means what? That integral on r of e minus X squared over 2 sigma squared times 1 dx is finite.
10:01:730Paolo Guiotto: Which is true, because that's the Gaussian. We can even say that the value of this is equal to the square root of 2 pi sigma squared.
10:12:750Paolo Guiotto: Okay, we know also, what is the value, the exact value of that integral. So, in particular, it is 5.
10:20:540Paolo Guiotto: This says that,
10:22:740Paolo Guiotto: our function f of XC verifies this condition is in L1, and this happens whatever is X, because you see, C enters only at this… in this exponential, and whatever is X, this argument worth. So.
10:41:830Paolo Guiotto: for every C real, So this means that the function f hat of X is well-defined.
10:53:770Paolo Guiotto: for every C real.
10:57:340Paolo Guiotto: So that's the step one. We know now that that quantity is defined. Step two.
11:05:80Paolo Guiotto: Now, we pass to the calculation.
11:08:340Paolo Guiotto: to compute…
11:13:360Paolo Guiotto: F at C.
11:16:300Paolo Guiotto: We… start, huh?
11:20:240Paolo Guiotto: computing.
11:25:120Paolo Guiotto: It's derivative.
11:26:840Paolo Guiotto: Respectuxi.
11:28:540Paolo Guiotto: So this is kind of,
11:32:210Paolo Guiotto: physicist argument. Let's differentiate them and see what happens.
11:37:710Paolo Guiotto: And at the end, it happens something nice, because it comes a differential equation for this quantity that we can solve and determine the quantity. This is not an unusual thing. Actually, this is, you know, who was Richard Feynman, right?
11:54:960Paolo Guiotto: I mean, I heard of this guy, he was a Nobel…
12:00:650Paolo Guiotto: love it. But it was one of the most incredible, main physicists, that…
12:10:560Paolo Guiotto: And he used it to say that this trick of differentiating integral was what made the difference between him and all the others, because he used a lot of tricks to compute very complex quantities by introducing parameters, differentiating, and then computing in this way, with this kind of tricks, integrals. So, it's a sort of techniques that
12:34:370Paolo Guiotto: Pretty nice, and…
12:35:720Paolo Guiotto: And the original. So, we want to start computing this derivative, so we want to do derivative with respect to Xi of this FXC, which is our derivative with respect to Xi of integral in R of that function FXC.
12:53:900Paolo Guiotto: DX. Of course, we would like to know if we can carry the derivative inside, and say that this is the integral of the derivative with respect to C of FXC DX.
13:07:670Paolo Guiotto: Then we will, once we will, ensure that this is possible, we will see what is this quantity. Now, to be able to do that, we apply the differentiation theorem. So, we apply…
13:24:620Paolo Guiotto: the differentiation.
13:27:70Paolo Guiotto: TRM?
13:29:560Paolo Guiotto: It says, so, what are the conditions?
13:33:240Paolo Guiotto: Once we have checked that the function FXC is in L1 in X for every value of the parameter C, we have to check, first, that there exists the partial derivative in C,
13:44:830Paolo Guiotto: for every value of C, where we expect this derivative exists, and this, almost every value of X. And number two, that we have a control of that partial derivative, which is uniform in the parameter, okay? So,
14:03:400Paolo Guiotto: Step number one, the derivative with respect to C of FXC, this is easy, because this is the derivative with respect to C of E to minus X squared over 2 sigma squared, e to minus iCX. Now, as you can see, C is only here.
14:24:100Paolo Guiotto: In this, unitar, it's exponential.
14:27:400Paolo Guiotto: So we treat x as a constant. This means that this can be carried out. We have e to minus X squared over 2 sigma squared times derivative with respect to c of that exponential. That will be D exponential, E minus x.
14:46:00Paolo Guiotto: times the derivative of the exponent with respect to C, so this is minus iX.
14:53:840Paolo Guiotto: So that's the derivative. Now, this derivative exists for every C real, there is no restriction, and for every X real.
15:03:520Paolo Guiotto: No restriction on the two. So, in particular, almost every… Almost every X.
15:10:570Paolo Guiotto: Okay, so Condition 1 has been verified. Condition 2, we have to take the absolute value of this derivative.
15:17:860Paolo Guiotto: And we have to bound this…
15:20:600Paolo Guiotto: With a function independent of the parameter.
15:24:870Paolo Guiotto: Now, if we take the absolute value here, it is written there, so we have modulus of E minus X squared over 2 sigma squared.
15:32:540Paolo Guiotto: E minus IX, and minus IX.
15:39:520Paolo Guiotto: Of course, we do the product of the models, then we have the exponential E minus X squared divided 2 sigma squared.
15:47:120Paolo Guiotto: Well, let's take the absolute value of this. The absolute value of minus i is 1.
15:52:240Paolo Guiotto: Then we have absolute value of X that we put in front of this, and then what remains is the absolute value of E minus ICX.
16:01:280Paolo Guiotto: Which we know is constant equal to 1.
16:05:50Paolo Guiotto: So the absolute value of the partial derivative with respect to parameter is equal to modulus of X E minus X squared over 2 sigma squared.
16:15:890Paolo Guiotto: which is independent of parameter C.
16:19:820Paolo Guiotto: And as a function of X, it's a nice function, integrable, because the exponential will always kill this power in front. So I can call this G ,
16:30:370Paolo Guiotto: This function is in L1R, and this happens, what, for every value of the parameter X,
16:40:440Paolo Guiotto: And of course, for every X, because there is no restriction.
16:44:540Paolo Guiotto: And it's, in particular, almost every X in R.
16:49:350Paolo Guiotto: So, as you can see, the two conditions of the differentiability theorem are verified, and this means that we are allowed to compute the derivative of the function we called f hat of x, carrying inside the derivative, inside the integral. So we have…
17:08:480Paolo Guiotto: There exists the derivative with respect to parameter X of F at.
17:14:220Paolo Guiotto: And this is integral… equal to integral on R of the derivative of F. We computed the derivative here, so I just…
17:23:730Paolo Guiotto: Copy this expression and plug into the integral.
17:28:310Paolo Guiotto: So, I have.
17:30:900Paolo Guiotto: minus IX, exponential minus X squared divided 2 sigma squared, E minus IC X, so… yes.
17:42:900Paolo Guiotto: Now, this one doesn't look to be better than the initial one, no?
17:48:640Paolo Guiotto: Because it is even more complicated, apparently, because there is this IX. The idea is now that we integrate by parts, and we return back to F, as we will see. Look at this one.
18:02:370Paolo Guiotto: Well, the i can be written in front of the integral.
18:06:760Paolo Guiotto: And now, more or less there, we have the derivative of the quadratic exponential. So if we do the derivative with respect to X of E minus X squared over 2 sigma squared, we get e to minus X squared over 2 sigma squared times the derivative of the exponent.
18:26:140Paolo Guiotto: which is a minus… derivative of X squared is 2x divided 2 sigma squared. So the 2 and 2 simplified. So you see that basically what we miss is a sigma square.
18:38:800Paolo Guiotto: So, we can do… divide by sigma square and multiply by sigma square here.
18:45:730Paolo Guiotto: And now we have, so I, sigma square integral on R,
18:52:680Paolo Guiotto: The expression here, minus X divided by sigma squared times the exponential, now becomes the derivative with respect to X of E minus X squared divided to sigma squared.
19:05:110Paolo Guiotto: And out of this derivative, there is this E2 minus iCX.
19:11:60Paolo Guiotto: This is an integration in X.
19:13:610Paolo Guiotto: Now, we integrate by parts.
19:22:410Paolo Guiotto: So I have I sigma squared. Of course, it will be… there will be an evaluation at infinity, but there is no danger here, because we will see easily that everything is fine. So we have…
19:34:400Paolo Guiotto: the evaluation of the product E minus X squared over 2 sigma squared times E minus iCX,
19:43:800Paolo Guiotto: Evaluated between, X equals minus infinity, and X equal plus infinity.
19:51:650Paolo Guiotto: provided this evaluation is a finite value. Minus…
19:56:400Paolo Guiotto: the integral on R. Now, derivative moves on the other exponential here.
20:02:200Paolo Guiotto: So we have E2 minus X squared divided 2 sigma squared.
20:06:730Paolo Guiotto: Be careful, because now it is the derivative with respect to X.
20:10:570Paolo Guiotto: So it's similar, because it is still the exponential, E minus iCX, but since we are differentiating in X, this now comes minus IC.
20:20:970Paolo Guiotto: VX.
20:25:200Paolo Guiotto: Okay, now if we look at the evaluation, here we have something like limit at as X goes to plus or minus infinity of e to minus X squared divided to sigma squared times e to minus x.
20:42:980Paolo Guiotto: It is clear that this factor does not have a limit in general, you know? The option AD, complex numbers are tooling around the origin factor, a part of a spatial values that has equal zero. But in any case, it is bounded, and this goes to zero.
21:00:980Paolo Guiotto: Because it is exponential with negative exponent going to class infinity, to minus infinity. So this goes to zero, this is bounded, and therefore the product is going to zero. So the limit will be zero. So this means that this evaluation yields zero.
21:19:510Paolo Guiotto: Okay, so we have… I sigma square.
21:23:710Paolo Guiotto: Then we have… this is 0, minus… we have a minus here. A minus here becomes plus.
21:31:140Paolo Guiotto: So I do not write anything. I carry out DI, but I can also carry out DC, because this is not the integration variable. The integration variable is X, and this is C, is just a constant with respect to X.
21:46:520Paolo Guiotto: So I can carry outside that, it becomes xi integral on R of what? Of E minus X squared 2 over 2 sigma squared, then E2 minus iCX.
22:01:180Paolo Guiotto: And this is exactly what initially was defined as F hat of C.
22:07:00Paolo Guiotto: This is F at of C.
22:12:620Paolo Guiotto: Okay, so at the end, we have E times… I times i is minus 1, minus…
22:19:550Paolo Guiotto: sigma squared C, F at C. This is the calculation, what the calculation of the derivative is after this integration by part. So, let's write the conclusion.
22:33:810Paolo Guiotto: So, this long calculation is the derivative with respect to C of DF hat. So, the derivative with respect to C of F hat, C,
22:45:730Paolo Guiotto: is equal to minus sigma square C f hat of C. So that's what we found for every C real.
22:57:30Paolo Guiotto: Now, step 3, which is the final step, we solve…
23:06:760Paolo Guiotto: For this F at of C, so the differential equation. Because as you can see, this… yeah, I know it's gonna be the partial, but here C is a real variable, so that's the derivative, no?
23:20:690Paolo Guiotto: So, it is saying the delivery of this function f is equal to minus sway of Z times the function.
23:28:250Paolo Guiotto: No? So, it is like… F hat of C.
23:35:380Paolo Guiotto: Verifies…
23:40:20Paolo Guiotto: with more familiar notation, something like Y prime of C equal minus sigma square CY of C.
23:50:130Paolo Guiotto: Now, I find out very quickly that in the informal way, it's not 100% correct, but I could say that Y prime of C divided Y of X
24:03:380Paolo Guiotto: is equal to minus sigma squared Xi.
24:07:450Paolo Guiotto: then this is the derivative of log of YX, Equal that.
24:17:660Paolo Guiotto: And, well, of course, this is not completely 100% formally correct, because in principle, this YC is that function f hat, which comes from a complex interval, so it could be a complex number.
24:33:420Paolo Guiotto: So, it should be understood what is the log of YX. But we have a… you can do formally with the standard method, so this is a linear first-order equation, if you look at the equation. However, what we get is that the log of Yxi
24:53:160Paolo Guiotto: is if the derivative is sigma minus sigma squared c, it means that the log of Y is a primitive of that in the variable x, so it means minus sigma squared c squared over 2 plus a constant, possibly.
25:10:00Paolo Guiotto: a constant K.
25:16:30Paolo Guiotto: So this means that, our Y of T, which is,
25:20:620Paolo Guiotto: the F hat of C should be something like E minus sigma squared.
25:28:360Paolo Guiotto: C squared over 2 plus constant.
25:32:750Paolo Guiotto: Or, if you want.
25:34:730Paolo Guiotto: I can split the exponential. E to K is another constant, so I still call K.
25:41:480Paolo Guiotto: If you prefer K tilde, E minus sigma squared c squared over 2.
25:50:90Paolo Guiotto: So, at this point, we have almost found the F hat, because we found f hat modular this factor k tilde, which is, in any case, a constant, okay? So, we can say that F hat of C
26:08:350Paolo Guiotto: Which was, by definition, the integral on R, E minus X squared over 2 sigma squared, E minus iCX,
26:20:20Paolo Guiotto: This is equal to k tilde E minus sigma squared over 2C squared.
26:28:710Paolo Guiotto: Now, it remains to determine k tilde.
26:42:330Paolo Guiotto: And the… This can be done, for example.
26:47:680Paolo Guiotto: If we are able to find some value of this function.
26:51:690Paolo Guiotto: And this can be done if we take the value, yeah.
26:55:980Paolo Guiotto: Because if you evaluate that t equals 0,
27:04:140Paolo Guiotto: We have, at the left, the integral of the quotient, E minus X squared over 2 sigma squared dx.
27:12:80Paolo Guiotto: And as we said, this is the Gaussian integral, whose value is 2 pi sigma squared.
27:19:940Paolo Guiotto: At right, when you put X equals 0, you get exactly the constant k tilde.
27:26:230Paolo Guiotto: So, the conclusion is…
27:33:520Paolo Guiotto: We found that the F at of x, which is the Fourier transform of that Gaussian, is root of 2 pi sigma squared exponential minus sigma squared over 2C square.
27:51:640Paolo Guiotto: forever exceed in R.
28:00:980Paolo Guiotto: So, in contemporary mathematics, as you probably know, the fact that we are able to compute exactly quantities is much less important, okay?
28:13:590Paolo Guiotto: This, of course, is a calculation of an old-fashioned mathematics, if you want.
28:20:810Paolo Guiotto: But, sometimes having a… especially the caution is quite important in probability, also in analysis, because it is related, as we will see, probably you have already
28:32:350Paolo Guiotto: learned this in differential equations. It is related to the solution of the heat equation. That's the fundamental solution of the heat equation.
28:41:820Paolo Guiotto: So, it's an important function, okay? Knowing explicitly, it's much better. You find the quotient in many problems in finance.
28:52:210Paolo Guiotto: Because in finance, you, maybe you won't have the Gaussian variables.
28:58:280Paolo Guiotto: Because if you want to model a price, price normally are positive quantities, and Gaussian variables are distribution on the interior line.
29:08:490Paolo Guiotto: But, for example, typically, a model for prices is the price is modeled by the exponential of aggression.
29:18:510Paolo Guiotto: And that is called Deloeb Normal Distribution. So, Gaussians can be found everywhere, and having exact… for example, if you go in the financial world.
29:29:880Paolo Guiotto: they want formulas, analytical formulas, because this makes the difference. You can really, you can really sell something to a businessman who has to deal with this kind of objects. So, a generic function is… or numerical is not often
29:49:860Paolo Guiotto: A good, the good choice.
29:52:940Paolo Guiotto: Okay, so, now here at the end of the chapter, there is a number of exercises, which have more or less all the same shape. This is a typical exercise that you will find in many of the exams.
30:08:50Paolo Guiotto: By the way, I think that by the end of the week, I will start publishing a file with a collection of previous past exams.
30:20:520Paolo Guiotto: And,
30:22:480Paolo Guiotto: I'm still thinking, because I don't know how physically to do this, but I'm still thinking to now start
30:29:880Paolo Guiotto: with the homework. So it depends on how many of you are going to think to participate to this, because if there is a good participation, we can do. If there is the participation of 3, 4 of you.
30:42:530Paolo Guiotto: it doesn't make sense. However, we will see. Probably I will give you an assignment, and then we will see, how to, by the end of the week, for the end of the next week. However.
30:53:980Paolo Guiotto: But let's see some of the exercises. So, let's take the exercise 731.
31:02:590Paolo Guiotto: we will, we will see more or less, there is, the same kind of problems, similar techniques, but not, of course, straightforward applications, of the same procedure. So here is, we have a let, capital F of X,
31:21:560Paolo Guiotto: be defined as the integral from 0 to plus infinity of…
31:27:280Paolo Guiotto: We have E minus Y sine of X times Y divided YDY. Of course, here, as you can see, DY is the integration variable, the X is the parameter, okay? We must be a bit flexible.
31:44:470Paolo Guiotto: So, the exercise asks, this, so, show, that, capital F is… well defined.
31:58:10Paolo Guiotto: for every X in R, Diana.
32:04:300Paolo Guiotto: compute
32:09:10Paolo Guiotto: the derivative of F, And… detail, I mean…
32:21:350Paolo Guiotto: A bit more guided, but more or less, the questions are like this.
32:27:30Paolo Guiotto: So…
32:34:540Paolo Guiotto: By the way,
32:37:560Paolo Guiotto: So this… in the Feynman idea, the idea would be to compute the integral of E minus y. This problem would start like that. Compute the integral of E minus Y sine y over y, which is impossible to do by elementary methods.
32:53:290Paolo Guiotto: So what he says is, let's put an X into the argument and differentiate, and at the end you will be able to compute the value of the initial integral. So it's a very ingenious way.
33:04:590Paolo Guiotto: Okay, however, we have to show that this is well-defined, so we recognize that F of X is integral between… on 0 plus infinity of a function of XY. I'm sorry I used this order, but, I mean, let me be a bit flexible. It is clear that Y is indication variable, X is the parameter.
33:24:740Paolo Guiotto: So, to be… to check if this function, capital F, is well-defined, we have to check, if this function here is integral as a function of Y,
33:37:20Paolo Guiotto: for possibly every value of X. This is the conclusion, okay? So, let's check, let's start checking.
33:45:100Paolo Guiotto: Let's discuss.
33:50:340Paolo Guiotto: integrability of F, the parameter is X as function of Y,
33:58:50Paolo Guiotto: In L1, the integration domain is 0 plus infinity. Let's see what can be said.
34:04:180Paolo Guiotto: So…
34:06:350Paolo Guiotto: Well, you see that there is a little, difference, there is a value at which, things,
34:14:40Paolo Guiotto: are easy, it is X equals 0, because sine vanishes, everything ventures. We can say that F0 sharp is constantly equal to 0, so clearly it is L1 on 0 plus infinity.
34:31:920Paolo Guiotto: If X is different from 0, then what happens?
34:37:90Paolo Guiotto: Well, if we look at this function as a function of Y, we see that we have a problematic at all, because of that denominator.
34:45:980Paolo Guiotto: divided by Y.
34:48:750Paolo Guiotto: And, then, the function of Y is a continuous function from 0 to plus infinity, so to discuss negative, we discussed the behavior of this thing when y goes to plus infinity.
35:01:650Paolo Guiotto: Okay, so basically, we have these two problems. So we noticed that.
35:08:590Paolo Guiotto: we notice.
35:12:200Paolo Guiotto: That.
35:14:300Paolo Guiotto: F of X, say, sharp, is a continuous function on 0 excluded plus infinity, So… We, have… to…
35:32:790Paolo Guiotto: discuss. So, in other words.
35:35:770Paolo Guiotto: at 0 and at plus infinity, we have a generalized integral here, if you want, no? And if you consider the condition that says that this is a 1,
35:48:590Paolo Guiotto: you remind that it is… it must be integral, absolutely integral in generalized sense, okay? So, we have to discuss absolute.
36:02:400Paolo Guiotto: integrability… Act.
36:07:170Paolo Guiotto: Y equals 0, and… Y equal plus infinity.
36:13:870Paolo Guiotto: So at Y equals 0, so let's take models of FXY. When y goes to 0, this is absolute value E minus y sine XY divided by Y.
36:29:730Paolo Guiotto: when Y goes to zero?
36:33:100Paolo Guiotto: So, if you send Y to 0, well, we can use the asymptotic comparison. So, it is asymptotic at zero, so let's be precise, when Y goes to zero. If you want B to be even more precise when 0 plus, that doesn't change. So, what about the exponential? This is asymptotic, too.
36:54:40Paolo Guiotto: 1, because it goes to 1,
36:57:190Paolo Guiotto: Then we have a sine XY.
37:00:320Paolo Guiotto: Since Y is going to 0, XY is going to 0, and sine is asymptotic, too.
37:09:430Paolo Guiotto: sine of T, when T goes to zero, is asymptotic to T.
37:16:720Paolo Guiotto: Exactly. So we are here. If you want the absolute… we don't need the absolute value for 0, but in any case, let's write XY,
37:25:760Paolo Guiotto: And Y is asymptotic to himself, so let's keep Y. So, as you can see, this simplifies, so it means that our function is asymptotic to X, which is a constant for Y.
37:38:170Paolo Guiotto: So, this function is definitely integral at zero. It can be extended by continuity, because you can compute the limit easily, and it comes to X, no?
37:49:500Paolo Guiotto: So…
37:50:900Paolo Guiotto: This means that the function f is absolutely integral fxy in the variable Y at 0. I write this, you see, with this integral without the final endpoint.
38:04:920Paolo Guiotto: modulus of F, X, sharp, or better F.
38:12:520Paolo Guiotto: absolutely integral.
38:15:660Paolo Guiotto: Absolutely intimidable.
38:18:540Paolo Guiotto: at Y equals 0. It's a way to write.
38:22:800Paolo Guiotto: We may notice that this would be equivalent. We could… Also, Notice…
38:34:540Paolo Guiotto: that, actually, we can compute the limit, it's the same, basically, that when we send the variable Y to 0, and we compute FXY,
38:46:990Paolo Guiotto: part of the function e to minus y remains e to minus y, so it goes to 1. Then we have sine XY divided Y, but when y goes to 0, sine t over t goes to 1. So if you multiply, divide by X, you easily see that it'll go to X.
39:05:220Paolo Guiotto: And therefore, if… F of X sharp.
39:11:950Paolo Guiotto: can be… Extended… Oz.
39:20:820Paolo Guiotto: a continuous function at Y equals 0. So again, integral
39:31:580Paolo Guiotto: Y equals 0, so we can consider as if it is really a function without problems at zero, okay?
39:40:50Paolo Guiotto: Now, at class infinity, the situation is a bit more complex, because when we go to class infinity with the Bible Y,
39:50:450Paolo Guiotto: We consider absolute value of F. Now, this goes to zero. This fraction goes to 0 because, we are side, this is between minus 1 and 1.
40:02:180Paolo Guiotto: of a Y, which is pushing bluefin, so this step backward, you know. However, we do not have an asymptotic behavior here. Both sine as not an asymptotic behavior when the argument goes too clean. So I cannot say it's asymptotic to is not correct.
40:18:140Paolo Guiotto: But, I can say that… modulus of F, X, Y, Is, in any case, bounded
40:28:450Paolo Guiotto: by… I throw away the sine term, which is of no help at plus infinity.
40:36:170Paolo Guiotto: No? Because it's just an oscillating factor. And so you have E minus Y over Y,
40:43:590Paolo Guiotto: Which is, by the way, a function I cannot use at 0 now, because at 0, this is asymptotic to 1 over Y, which is a non-integrable function. So I cannot say, this is, for every y, then it is controlled by an L1 function on 0 plus infinity piece integral. You see this point?
41:02:220Paolo Guiotto: So this, bound holds for every X, for every Y,
41:07:410Paolo Guiotto: Okay, is correct. It is a good bound at plus infinity of this function, is integral at plus infinity, but it's not a good bound to say that function is integral of zero plus infinity.
41:20:850Paolo Guiotto: Because the spine zone could not be integrated in zero.
41:25:50Paolo Guiotto: Okay, so that's why the other part of the discussion
41:29:430Paolo Guiotto: is… is necessary, okay? So, but this function is integral, at plus infinity.
41:38:260Paolo Guiotto: I think we can accept this, no?
41:40:850Paolo Guiotto: We don't need to verify, no?
41:43:740Paolo Guiotto: So… And this holds for every value of X, which is here the parameter.
41:52:250Paolo Guiotto: So the conclusion is that, putting together everything, we say… we have seen that the function fxy is integral at 0, I forgot to mention this, for every value of X, okay?
42:10:590Paolo Guiotto: It is integral at plus infinity for every value of X. Then, the function f of x, sharp, belongs to L10 plus infinity.
42:22:240Paolo Guiotto: And this holds for every value of X real.
42:27:160Paolo Guiotto: And this means, sir, then, that since the capital F is the integral of this little f, that capital F is well-defined for every
42:37:590Paolo Guiotto: Experiential, okay?
42:40:480Paolo Guiotto: So… capital F of X equal integral 0 plus infinity F.
42:48:100Paolo Guiotto: XY.
42:50:190Paolo Guiotto: DY, exists this for every X real.
42:58:100Paolo Guiotto: The question might be posed in the form of, for example, determine the domain of definition of capital F. What is the domain of definition of a function? A set of values for which the function makes sense.
43:13:160Paolo Guiotto: And what is the condition to make… to have this one we were defined here, that this integral makes sense, so the function f must be integral in Y for some value of X. The values of X for which
43:26:240Paolo Guiotto: That function is integral r dx in the domain of this quantity, R dx for which this quantity is well divided, no? So, if I ask you to tell me the domain of this, what does it mean this?
43:39:240Paolo Guiotto: means to determine the X for which this thing is well-defined, so XY is L1 divided by 1.
43:46:680Paolo Guiotto: Okay. Now… By the way, We want to take a break?
43:53:650Paolo Guiotto: Okay.
43:56:720Paolo Guiotto: So, we discussed the…
44:01:100Paolo Guiotto: the first… the first, part of this question, no? Show that F is well-defined for M1. Now we have to compute F prime.
44:10:490Paolo Guiotto: So, here, it is where we will apply the differentiation.
44:14:780Paolo Guiotto: theory. So, to… Computer… F prime, we… Apply.
44:27:550Paolo Guiotto: Differentiation, and integral.
44:35:50Paolo Guiotto: sign.
44:37:110Paolo Guiotto: that says that the derivative of this thing with respect to X is the integral of 0 plus infinity of the derivative with respect to X of this function of XY.
44:49:110Paolo Guiotto: Now, provided this little f verifies Two conditions provided.
45:00:30Paolo Guiotto: Number one, of course, there exists the derivative with respect to X, of the function F,
45:09:330Paolo Guiotto: So, do not memorize things. Try to…
45:13:50Paolo Guiotto: You see, normally this, this follows a logical path, no, so this must be…
45:21:130Paolo Guiotto: for every X for which I want to compute the derivative, it is not written, or maybe it is written, because sometimes, you know, the question could be open. Determine the set of X for which there exists F prime of X.
45:33:580Paolo Guiotto: So you don't know. Now, here I'm not saying anything, probably. So, let's see. For every X in R, maybe it's not true, maybe the differentiability domain is smaller, I don't know yet.
45:48:220Paolo Guiotto: or for every X on D containing in R, where this will be the domain where the derivative makes sense.
45:56:590Paolo Guiotto: In any case, this would be almost every Y in 0 plus infinity.
46:04:210Paolo Guiotto: And number two, that we have a dominance of this, so there exists a function G. Function of the integration variable, which is, in this problem Y, so function G of Y, which is an L1 on the integration domain 0 plus infinity.
46:23:60Paolo Guiotto: Such that we have that the derivative with respect to X of that FXY
46:29:360Paolo Guiotto: is controlled by that G of Y. And again, this will be almost every Y.
46:35:990Paolo Guiotto: In the range, 0 plus infinity.
46:39:340Paolo Guiotto: And for every X in the domain D, where we have the derivative. Okay, so let's start with 1.
46:49:750Paolo Guiotto: Now, the derivative with respect to X normally is an easy problem, because we have to do derivative with respect to X of F, go back to F,
47:00:760Paolo Guiotto: you see that X is just inside this sine. So Y is a constant.
47:07:560Paolo Guiotto: for X. So this means that the derivative will be… E minus… E minus Y over Y.
47:15:720Paolo Guiotto: Then there is the derivative with respect to X of sine XY.
47:21:00Paolo Guiotto: This is E minus Y over Y.
47:24:600Paolo Guiotto: derivative of sine is cosine, same argument. Then I have to do the derivative with respect to X of the argument, which is XY, so differentiating with respect to X, I get Y.
47:38:480Paolo Guiotto: This simplifies this one.
47:41:550Paolo Guiotto: Well, formally, I cannot take Y equals 0, because at Y equals 0, there is no definition here, so I should say for Y positive, and this means almost every Y in 0 plus infinity, as desired.
47:58:880Paolo Guiotto: And actually, as you can see, there is no restriction for X, no? So, for every X in R.
48:07:10Paolo Guiotto: So it seems that we are going to apply the differentiability on R.
48:12:170Paolo Guiotto: Now, here I will show, and I don't know if this is the case of this example, but sometimes you have to be flexible. Let's see what this means.
48:22:400Paolo Guiotto: I don't know if it happens with this example. However, number two, we have to find this dominance, okay?
48:29:920Paolo Guiotto: We have to find this function G that dominates the derivative with respect to X. Now, if you take the absolute value of DXFXY,
48:40:710Paolo Guiotto: you get this is E minus Y cost.
48:45:80Paolo Guiotto: XY, so here we are particularly lucky, because, of course, E minus Y is positive, then we have the absolute value, XY. You see that X is just in that cosine.
49:00:180Paolo Guiotto: So there is an actual way to throw away that X, which is…
49:06:770Paolo Guiotto: 1, no? So we bound Y1, and that's less or equal E minus Y, which is a good function, G of Y, that belongs to L10 plus infinity. Now, this holds
49:20:880Paolo Guiotto: Of course, in terms of why there is no conditions, apart from the fact that at Y equals 0, formally speaking, we could not talk about the derivative, so we can say almost every Y in 0 plus infinity.
49:36:640Paolo Guiotto: and the for every X, real.
49:40:760Paolo Guiotto: So it means that we can really apply the theorem We can.
49:47:960Paolo Guiotto: Apply.
49:51:480Paolo Guiotto: D.
49:53:410Paolo Guiotto: Differentiability.
49:55:740Paolo Guiotto: TRM.
49:58:80Paolo Guiotto: on… domain D, which is for parameters, the interior line.
50:05:40Paolo Guiotto: We get…
50:08:570Paolo Guiotto: that the derivative with respect to X of function f of x is equal to the integral on… no, sorry, the domain is 0 plus infinity.
50:20:870Paolo Guiotto: of the derivative of the little f, which is, at the end, it is equal to E minus y cos XY, right?
50:29:340Paolo Guiotto: So, E minus Y cos… X, Y, D. Y.
50:37:10Paolo Guiotto: Okay, now we computed the derivative. At least.
50:41:540Paolo Guiotto: Not explicitly, it depends on what we mean by explicit. Because this is explicit, no? It's something that… Now, we can explicit a bit better this, because we can compute this integral, perhaps.
50:57:800Paolo Guiotto: This is… belongs to a standard family exponential type process. There is a formula, I do not remember. However, we do by integrating by parts, so we write this as the derivative with respect to Y of minus E to minus y.
51:13:820Paolo Guiotto: So, by parts…
51:17:420Paolo Guiotto: you have to do a couple of integrations. We get the evaluation of minus E minus y cos XY,
51:26:500Paolo Guiotto: Between y equals 0, y equals plus infinity.
51:33:40Paolo Guiotto: minus integral 0 plus infinity, then we move the derivative on the other factor, so we have minus C minus y times derivative of cosine, which is minus
51:48:80Paolo Guiotto: minus sign XY.
51:51:530Paolo Guiotto: And the derivative with respect to Y of the argument is X.
51:59:660Paolo Guiotto: Integrating in Y?
52:02:180Paolo Guiotto: About the evaluation. At plus infinity, the exponential kills the cosine.
52:08:390Paolo Guiotto: Because E to minus Y goes to 0, cosine is bounded, so 0. At 0, exponential is 1 cosine plus 0, which is 1, so that product is 1 with minus in front, minus 1, but remember that this is minus.
52:24:620Paolo Guiotto: So all the value is,
52:26:820Paolo Guiotto: plus 1. So this evaluation is plus 1. Then we have 3 minus, so it's at the end, minus. We can carry outside the X. Integrals 0 to plus infinity, E minus Y sine XY.
52:47:360Paolo Guiotto: Okay, now, this is the calculation of that integral. Normally, this is a standard thing, we have to do it
52:56:710Paolo Guiotto: couple of integrations by parts to get developed. So we repeat the integration by part, maybe we give the minus to this one, such that this is the derivative with respect to Y of E minus Y. So we get 1 plus X times… we again have to do the evaluation.
53:14:870Paolo Guiotto: Minus E minus Y sine XY.
53:20:980Paolo Guiotto: Again, between Y equals 0, y equals plus infinity.
53:26:220Paolo Guiotto: Minus integral 0 plus infinity.
53:31:230Paolo Guiotto: Then we have to move the derivative on sine, so this will be E minus y, derivative of sine is cos XY, derivative is respect to Y, so the argument yields another X here.
53:47:590Paolo Guiotto: About the evaluation, at plus infinity, again, the exponential kills the sign, you see?
53:55:800Paolo Guiotto: So, E minus Y.
53:58:50Paolo Guiotto: sine XY, when you send Y to plus infinity.
54:05:110Paolo Guiotto: this guy goes to zero. This is bounded.
54:08:600Paolo Guiotto: It has not a limit, but it is bounded, so it goes to zero.
54:12:620Paolo Guiotto: At 0, you get 0, because exponential is 1, sine is 0, so at the end, here you have 0.
54:20:380Paolo Guiotto: So, we have 1 plus X,
54:24:270Paolo Guiotto: Then we have, parenthesis. The X can be carried outside, so minus X integral 0 plus infinity of E minus y cos XY.
54:40:40Paolo Guiotto: DUI, So, as you can see, we came back to the initial integral, this one.
54:49:190Paolo Guiotto: Okay? So, this is a sort of equation that can be used to determine the value of the integral, you see?
54:59:240Paolo Guiotto: So, let's say that integral was the derivative of capital F. So, what we obtained is f prime of x is equal to 1
55:10:770Paolo Guiotto: minus X squared, this integral, which is F' of X.
55:17:940Paolo Guiotto: Do you see this?
55:21:440Paolo Guiotto: Okay, remind that, F prime of X,
55:24:920Paolo Guiotto: As being computed using the differentiation.
55:28:450Paolo Guiotto: And the integral sine, and it turned out that it is E minus y integral of this.
55:36:190Paolo Guiotto: So now you can extract F prime.
55:40:360Paolo Guiotto: You get f prime of X, carry on the left-hand side, 1 plus X squared.
55:46:790Paolo Guiotto: is equal to 1, and this will give F prime of X equal 1 over 1 plus X squared.
55:56:320Paolo Guiotto: So the derivative of this function, capital F, is 1 over 1 plus X squared, which is the derivative of the tangent.
56:04:830Paolo Guiotto: So, now we conclude that F of X must be the arctangent of X plus a constant.
56:15:980Paolo Guiotto: We don't know if it is exactly the arctangent or not, but there must be a constant. So, the last step is to determine the value of the constant.
56:25:340Paolo Guiotto: And here is where we need to, it is sufficient to know one value of F. Sometimes it's not a value, but rather a limit, let's see, here. So this is saying that, going back to the beginning, the integral from 0 to plus infinity of E minus y
56:44:490Paolo Guiotto: of a Y sine XY.
56:49:130Paolo Guiotto: BY is equal to arctangent of X plus constant for every X real.
56:59:480Paolo Guiotto: Because the calculation. So what would you do?
57:13:10Paolo Guiotto: Absolutely.
57:14:140Paolo Guiotto: Yeah, in this case, if we plug x equals 0, at left, we get 0.
57:21:660Paolo Guiotto: So, integral of 0.
57:24:230Paolo Guiotto: which is 0. At right, we have an arctangent of 0. So, again, 0 plus k. So, at the end, k is equal to 0. So, the conclusion is that,
57:38:220Paolo Guiotto: the integral from 0 to plus infinity of E minus y of y, or as it was written initially, e to minus y sine
57:50:200Paolo Guiotto: XY.
57:52:160Paolo Guiotto: over Y, DY, is arctangent of X for every X real.
58:01:510Paolo Guiotto: So, as Feynman would have started, we would have started by computing that integral with X equals 1.
58:09:410Paolo Guiotto: So, for example, here you get integral from 0 plus infinity minus y sine y over y.
58:17:710Paolo Guiotto: DY is the octangent of 1, which is pi over 4.
58:24:200Paolo Guiotto: Which is a result that you hardly achieve… would achieve by doing direct calculations and
58:30:850Paolo Guiotto: And, you see how tricky it is? You introduce this little parameter, you plug this little
58:37:750Paolo Guiotto: grain of sand into the integral, and this whole story leads you to compute the value of that.
58:47:600Paolo Guiotto: I wanted to show… Another example, so let me…
58:56:750Paolo Guiotto: Probably this will come with the next one.
59:00:00Paolo Guiotto: So, do you have any questions on this?
59:03:400Paolo Guiotto: Well, the point is that the techniques are more or less clear, no? If you want to compute the derivative of something which is an integral, you have to apply the differentiation theorem, so you have to verify a hypothesis. Then, the remaining part, this calculation.
59:19:130Paolo Guiotto: As relatively to do with that part, okay?
59:22:580Paolo Guiotto: This is by the calculus.
59:24:740Paolo Guiotto: Bob.
59:25:920Paolo Guiotto: I want to show… let's see if this happens with the next example, 732, that sometimes you must be a little bit flexible in the application of the
59:38:30Paolo Guiotto: differentiation theorem. So, here we have a problem which is similar. f of x is defined as the integral 0 plus infinity
59:46:650Paolo Guiotto: E2 minus XT, minus E to minus T divided by TDT.
59:58:260Paolo Guiotto: So, one… F of X, well… defined… for every X paused.
00:09:290Paolo Guiotto: Number 2?
00:12:10Paolo Guiotto: show that there exists F prime of x, When I reacts positive.
00:18:800Paolo Guiotto: And computer number 3.
00:21:720Paolo Guiotto: compute.
00:25:30Paolo Guiotto: F of X, that you won't do by doing directly the calculation Because that's not…
00:32:430Paolo Guiotto: easy. I don't know if it is possible, but…
00:37:690Paolo Guiotto: So, number one, again, as you can see, we can,
00:43:40Paolo Guiotto: set, the problem as a problem where we have to integrate a function f that depends on two variables, X and T. X here is the parameter, and T is the integration variable, okay?
00:56:760Paolo Guiotto: So… Let's check.
01:03:900Paolo Guiotto: when the function f x fixed as a function of T belongs to L10 plus infinity.
01:15:30Paolo Guiotto: So, first of all, about, so we freeze X with the…
01:20:980Paolo Guiotto: a fixed value, and we look at that function as a function of T. What do we see?
01:26:880Paolo Guiotto: We see that there is a problem at t equals 0 because of that denominator.
01:32:820Paolo Guiotto: And for T positive, there is no more any problem, and the function is good, it's continuous. So we can say that F of X sharp
01:43:510Paolo Guiotto: I don't care what happens for T negative, because it's not part of this problem. The function is, of course, continuous also for T negative, but who cares? Here we may say, this is continuous on 0 plus infinity.
01:58:280Paolo Guiotto: So, definitely, it is a measurable function on 0 plus infinity.
02:03:870Paolo Guiotto: Maybe we can see if it is continuous at zero, so we eliminate the problem.
02:10:470Paolo Guiotto: Okay, concerning the integrability, we notice… That… the limit…
02:21:590Paolo Guiotto: when t goes to zero of FXT is the limit
02:28:440Paolo Guiotto: when T goes to 0, E minus, XT Yes. E minus T… divided by T.
02:39:830Paolo Guiotto: Now, when t goes to 0, both exponentials go to e to 0, 1, so minus 1, 0. Numerator goes to 0, denominator goes to 0, 0 over 0, it is an indeterminate form.
02:53:360Paolo Guiotto: Here, we could use… He beloved the hospital rule.
03:00:690Paolo Guiotto: We may say this is the limit when t equals to 0, derivative of t is 1,
03:06:320Paolo Guiotto: We are doing limit in t, so we will compute derivatives in T, okay?
03:12:120Paolo Guiotto: And then we have derivative of the numerator, exponential, derivative with respect to T. So this is minus X E minus Xt.
03:23:730Paolo Guiotto: minus delinity of E minus T, it is minus E minus T, so it becomes plus E minus T.
03:31:470Paolo Guiotto: Now, when t goes to 0, the exponentials both go to 1,
03:38:100Paolo Guiotto: So, numerator now goes to 1 minus X.
03:44:30Paolo Guiotto: denominator is a constant, goes to 1, so there is no problem now, and the limit is 1 minus X. So we can say that, F…
03:54:30Paolo Guiotto: of X sharp, let's say, can… B. Extended.
04:04:850Paolo Guiotto: as a continuous function.
04:12:460Paolo Guiotto: of P.
04:14:870Paolo Guiotto: at t equals 0. So this means in particular that F is integral at… T equals 0.
04:26:120Paolo Guiotto: So the unique real problem is what happens at t equal plus infinity.
04:32:700Paolo Guiotto: Okay?
04:34:270Paolo Guiotto: So… F of X.
04:37:670Paolo Guiotto: shot.
04:38:700Paolo Guiotto: will belong to L10 plus infinity, Ethan, Oliv… F, X, Sharper.
04:49:750Paolo Guiotto: is integral at… plus infinity.
04:55:770Paolo Guiotto: Now, what can be said here? Well, as you can see, well, if you want…
05:06:730Paolo Guiotto: Well, the exercise, we may say, the exercise asks to discuss for X positive.
05:15:410Paolo Guiotto: But suppose that we don't know this, no? Suppose that the question is slightly more open. It says, for which values of X is this thing well defined?
05:27:140Paolo Guiotto: So, we don't know for which X. Here, it… you see the difference. Here, it is saying, check that F is well-defined for X positive. So, you take X positive and you go straight, you have to check that the function is integral. But what if the question were different? So, like.
05:42:770Paolo Guiotto: For which values of X is this F well-defined? No? Or determine the domain of definition of that capital F? This would mean that you have to discuss
05:53:610Paolo Guiotto: all values of X, positive, negative, and see whether or not this function is integral. So let's try to do this kind of discussion, okay? As if we don't know that it is required only for X positive.
06:09:140Paolo Guiotto: You understand what I'm saying? Okay. Okay, so let's discuss… the point is that as you look at this function here.
06:18:260Paolo Guiotto: the behavior change with X, for X negative, positive, or zero.
06:25:110Paolo Guiotto: Because for X positive, that's a negative exponential, so it's something which is going to zero infinity very fast, but no? Sufficiently fast to be integral. When x, for example, is negative, so put X equals minus 1, what happens? That becomes 3 to T, you see?
06:44:780Paolo Guiotto: If you put X equals minus 1 there, you're taking X negative.
06:50:640Paolo Guiotto: This becomes X minus 1 would become e to t plus e to minus T divided by t. So in this case, when t goes to infinity, this… the numerator goes, like, e to t, but t is the t at the denominator, so it's going to expose at infinity and want to be integral.
07:09:00Paolo Guiotto: You see, this seems to say that for X positive, we have a story, for X negative, another, for X0, another again. So, if…
07:19:330Paolo Guiotto: X is negative.
07:21:70Paolo Guiotto: The function F, XT, is…
07:24:830Paolo Guiotto: E2 minus XT minus E2 minus t divided T. Now, this minus X is positive, okay? That's the important fact.
07:34:830Paolo Guiotto: So, this exponential is going to zero. This one is going to plus infinity when we send the t to plus infinity, no?
07:44:390Paolo Guiotto: So, when we send t to plus infinity.
07:47:650Paolo Guiotto: If you look at the numerator, the numerator is made by two quantities, but one is big and the other is small. So the big one will dominate, and in fact, I can say the numerator
07:57:850Paolo Guiotto: will be asymptotic to e to minus Xt divided by t, because the denominator remains T.
08:05:850Paolo Guiotto: And I can conclude that this quantity even goes to plus infinity when T goes to plus infinity, so definitely won't be integral, okay?
08:15:00Paolo Guiotto: So… There cannot be any integral at plus infinity, but not even in a generalized sense.
08:23:830Paolo Guiotto: Be careful, because I know that you are thinking that the function is integral, even though if the function goes to zero at infinity. This is false. We will return on that, but think about, because…
08:36:40Paolo Guiotto: this is… this is not… a function can be even unbounded and be still interval. Of course, must be unbounded on a small set. On a small set means if it is a measure of zero set, it is trivia, because we can throw away a measure of zero set, but it can be…
08:52:330Paolo Guiotto: Even unbounded on a positive measure set. However, we will return on this a bit later.
08:59:390Paolo Guiotto: Okay, so this, is equal to plus infinity, so the function f…
09:06:290Paolo Guiotto: X sharp is definitely not in L1.
09:11:620Paolo Guiotto: Okay? What if X is equal to 0? In that case, we can see exactly what is F.
09:20:20Paolo Guiotto: When we plug x equals 0, the function is written there.
09:24:890Paolo Guiotto: You see that the first exponential becomes E to 0, so 1,
09:29:170Paolo Guiotto: minus E2 minus T divided by t. And this, when t goes to plus infinity, is asymptotic to…
09:42:779Paolo Guiotto: 1 over T. 1 over T, and what do you know about 1 over T?
09:48:340Paolo Guiotto: It is not integral at plus infinity, no? The integral of 1 over t is the log of t, no?
09:55:70Paolo Guiotto: So, it goes to infinity, okay?
09:58:390Paolo Guiotto: is not in L1. Well, remind this, these are a few general facts. About the integral mediat plus infinity, something like 1 over, let's use the letter T,
10:08:470Paolo Guiotto: T2 alpha is integral at plus infinity if and only if the condition is the same of numerical series, like 1 over n2 alpha harmonic series.
10:18:990Paolo Guiotto: Alpha must be greater than 1.
10:21:500Paolo Guiotto: Okay, then you have another class, simple class is the class of the exponential. E to minus alpha t is integral at plus infinity, if and only if, in this case, we need the data alpha be positive.
10:37:600Paolo Guiotto: So the exponential decay to zero.
10:41:160Paolo Guiotto: There are more general classes, but then you have at minus infinity, and for example, in 0, another thing that should be reminded is that in 0,
10:53:370Paolo Guiotto: This one behaves the opposite. This thing is integral, so there exists the integral here, sorry.
11:02:360Paolo Guiotto: If and only if, in this case, the alpha must be less than 1.
11:06:420Paolo Guiotto: Okay? Because if it goes too fast at infinity 0, it is not integral. That's the idea. Alpha less than 1.
11:14:250Paolo Guiotto: Okay, so this is the discussion for X negative. For x0, that says there is no integral for X less or equal than 0. If x is positive, finally.
11:26:140Paolo Guiotto: Now, our function, FXP… is asymptotic to water? Well, this… the answer… well, let's wait.
11:35:890Paolo Guiotto: E minus Xt minus E to minus T divided by T.
11:41:220Paolo Guiotto: You see that in the numerator.
11:43:990Paolo Guiotto: The answer is, it depends, because it takes explicit bangs. Put X equals 1, what happens? The numerator punches, so it is 0.
11:55:640Paolo Guiotto: But in that case, the function would be zero, so it is integral, right?
12:00:860Paolo Guiotto: Put x equal 2, this would be e to minus 2t versus e to minus T. Which one is bigger? E to minus 2T, or e to minus t?
12:12:620Paolo Guiotto: Beginning.
12:23:880Paolo Guiotto: E to minus P, because minus to P, the external negative, the smaller. So you see that when x is greater than 1, it is this one leading there, and when x is less than 1, it is this one leading there.
12:40:370Paolo Guiotto: When X is one, both are equal, and they pass in the difference.
12:44:810Paolo Guiotto: Sure, if I want to have an exact discussion, I will show a shortest way. So I could say that for X less than 1, but still positive.
12:57:730Paolo Guiotto: The leading term, the biggest, is e to minus Xt, so I could say that… well, sorry, let's put asymptotic… asymptotic when t goes to plus infinity.
13:09:200Paolo Guiotto: So the numerator is E minus Xt divided T,
13:13:430Paolo Guiotto: When x is equal to 1, we have a 0 over T. Well, actually, it is equal to 0.
13:22:360Paolo Guiotto: And when X is larger than 1,
13:25:420Paolo Guiotto: It is now the other exponential that leads.
13:29:690Paolo Guiotto: So, I have minus T to minus T divided t, but whatever is the case, these functions are all integrals, because
13:39:180Paolo Guiotto: the exponential keys… well, actually, that key is going to help, but alone is not sufficient, okay? 1 over P is not integral, so you need the exponential.
13:49:730Paolo Guiotto: These are integrals. If you want, we can use this kind of information. So this is in L1, this is in L1, this is in L1 because it's constant equal to 0.
14:03:320Paolo Guiotto: In altema team?
14:06:130Paolo Guiotto: an alternative, argument, in this case easier, would say you take the absolute value of Xt, which is the absolute value of that fraction, E minus Xt, minus E to minus T divided T.
14:23:680Paolo Guiotto: Now, here we have to…
14:26:450Paolo Guiotto: The information is that we have to say if the function is interval, so we don't need to be extremely precise. So we can say this is less or equal.
14:35:700Paolo Guiotto: You split the fraction.
14:37:970Paolo Guiotto: modulus of some differences less than sum of models, everything here is positive, so less or equal than E minus Xt divided T plus E minus T divided t.
14:48:840Paolo Guiotto: It doesn't matter, because for X positive, this is in L1.
14:53:840Paolo Guiotto: And this is in L1, because there is no X, it is e to minus T.
15:00:00Paolo Guiotto: at plus infinity. Be careful, because at zero, of course, none of this is good, okay? These are…
15:08:430Paolo Guiotto: all of those functions at 0 for the integral. But we already discovered the discussion at zero, because we said the function could be extended by continuity, so we don't need to check integrity in 0. Yeah, but you know, zero problem is, is it integral and continuity?
15:25:100Paolo Guiotto: So we need good downside utility, doesn't matter if they are not counted at zero.
15:30:220Paolo Guiotto: I already discussed zero.
15:34:270Paolo Guiotto: Okay.
15:35:440Paolo Guiotto: So, at the end of this, we can conclude that the function f is well-defined for every X positive.
15:43:260Paolo Guiotto: So, conclusion… F of X.
15:49:310Paolo Guiotto: Well, defined… for every X positive.
15:55:770Paolo Guiotto: Question 2.
15:58:340Paolo Guiotto: Now, question 2 is the health of the problem, derivative.
16:03:310Paolo Guiotto: So, we use the differentiability theory, okay?
16:06:840Paolo Guiotto: So… We… use… Differentiability.
16:14:620Paolo Guiotto: I'm there.
16:16:890Paolo Guiotto: Integral, sine…
16:21:140Paolo Guiotto: So that f prime of X turns out… will… turns out to be the integral of the derivative with respect to X of the little f x.
16:30:980Paolo Guiotto: BT.
16:32:190Paolo Guiotto: We can do that.
16:34:520Paolo Guiotto: Provided…
16:38:680Paolo Guiotto: we can apply the, assumptions of the… we can check, verify the assumptions of the theorem. Number one.
16:48:350Paolo Guiotto: There existed the derivative with respect to X of the function F, XT,
16:56:80Paolo Guiotto: Now, this derivative, must exist for the X for which you want to compute the derivative.
17:03:370Paolo Guiotto: So, in principle, we could say for every X positive, because…
17:08:640Paolo Guiotto: Hopefully, the derivative will exist for X positive.
17:12:950Paolo Guiotto: But as you will see, the story here.
17:15:540Paolo Guiotto: Can be a little bit difficult.
17:18:690Paolo Guiotto: And, almost every T.
17:23:310Paolo Guiotto: positive, that's for the integrability. Number two.
17:27:420Paolo Guiotto: there exists a function G, function of the integration variable, which is T, which is an L1 function on 0 plus infinity.
17:39:20Paolo Guiotto: For which we have that the derivative with respect to X of F,
17:44:120Paolo Guiotto: Xt is less or equal than this G of T,
17:48:360Paolo Guiotto: Again, for every X for which we want the derivative, so let's say for the moment X positive.
17:55:240Paolo Guiotto: Do you understand why I'm saying X positive?
18:00:720Paolo Guiotto: Because F is defined for X positive. So, let's see if there is a derivative for all the X for which there is F, okay?
18:08:690Paolo Guiotto: But in general, the domain of definition of the derivative, this is not a question of these type of functions in general, the domain of the derivative is smaller than the domain of the definition of the function, okay?
18:23:380Paolo Guiotto: That cannot be bigger than… because it's like, say, there is the tangent where there is not the function, so it does not make any sense, you see. Okay, and almost every team.
18:34:310Paolo Guiotto: Now, let's see what happens. Number one, derivative with respect to X of F, this is easy, because we have to differentiate with respect to X, that e to minus XT minus E to minus t divided by t. Now, t is a constant for X.
18:51:920Paolo Guiotto: So we have 1 over T, then we have derivative with respect to X of E minus XT minus E2 minus T.
19:02:830Paolo Guiotto: So, an error that can happen here is that you do a mess with the variables, okay?
19:10:980Paolo Guiotto: You must be clear who is the parameter, who is the integration variable, and do not confuse the two.
19:18:200Paolo Guiotto: Okay, so here, we are differentiating with respect to the parameter, which is X in this game, and therefore, whatever depends on T is a constant. So this means that, for example, when I do the derivative of this, I get 0.
19:31:750Paolo Guiotto: Okay? I do not get minus E to minus T. So I will have 1 over T, then I have derivative of that exponential, E minus Xt, times derivative of the exponent with respect to X.
19:45:140Paolo Guiotto: So I get…
19:49:880Paolo Guiotto: minus D. Okay, so at the end, I have minus E to minus XD.
19:56:600Paolo Guiotto: Now… For which acts for which day?
20:00:330Paolo Guiotto: For X, you see, there is no condition, no? So, I would say for every X, actually also for X negative, okay? But in this problem, X negative, X0, who cares, because we already discarded those.
20:14:800Paolo Guiotto: X, so for every X positive, and for every T, not exactly because of this denominator, not T equals 0, even if at the end you don't see divided by T, you have to remind that this calculation was made on this formula, so with T. So I would say for every T positive, but who cares? Because this means almost every
20:37:510Paolo Guiotto: T in 0 plus infinity. Just one single T is out.
20:43:40Paolo Guiotto: Okay?
20:44:520Paolo Guiotto: So, good. Number two, the bound.
20:49:610Paolo Guiotto: Now, here, the boundary is almost immediate. There is not so much to do, because you see the derivative of FXT
21:01:380Paolo Guiotto: Is absolute value of minus C to minus XT.
21:06:440Paolo Guiotto: So, definitely this is E2 minus XTE, and now you see.
21:12:850Paolo Guiotto: What is the bound for this? The reminder, the bound must be a function of P only, you don't see any mixer.
21:21:410Paolo Guiotto: And must be valid for all values of X that you are considering.
21:26:800Paolo Guiotto: Now, what can be the boundary effects is a bit more important.
21:33:350Paolo Guiotto: What movie is about?
21:37:100Paolo Guiotto: If your low X could be any positive number, 3 months.
21:43:440Paolo Guiotto: You say E to minus 2.
21:46:870Paolo Guiotto: That's right, who…
21:48:970Paolo Guiotto: I know, because this is, this, this example is simple, with calculation not particularly complicated, but it's, it is something.
21:58:250Paolo Guiotto: George, is that correct?
22:03:160Paolo Guiotto: So, can I say that this is corrected?
22:07:00Paolo Guiotto: for every X positive, because remember, we are applying on the domain… there is a condition, no? You see.
22:15:40Paolo Guiotto: Go back to the statement, no? This says, you want to have…
22:20:190Paolo Guiotto: The derivative, I forgot to wrote here, is differentiable for every value of parameter in the domain.
22:28:960Paolo Guiotto: Which domain? The domain on which you apply this, so you see this domain for brand everywhere, no?
22:39:30Paolo Guiotto: Now, here what you have is that, can you say that this is true for every X positive and almost every T?
22:48:560Paolo Guiotto: Yes or no?
22:50:630Paolo Guiotto: Why not? Because, we, acts between Nero and one isn't… Hey, that's the problem.
22:59:530Paolo Guiotto: So, again, what is the best bound that you can write here for FEX positive? 1. Is that a good bound?
23:09:510Paolo Guiotto: Yes.
23:11:10Paolo Guiotto: Yes.
23:14:690Paolo Guiotto: and visa.
23:16:20Paolo Guiotto: Optional.
23:22:500Paolo Guiotto: So, the problem here is that… so this is not a good bound. This bound works… this works for… for which checks?
23:39:740Paolo Guiotto: That's the… Okay, greater or equal than 1. Okay?
23:49:190Paolo Guiotto: So this would say, okay, we have a bound if X is greater than or equal than 1, and this would be a good bound. This would be an L1.
23:59:710Paolo Guiotto: But if we pretend to take all the X,
24:03:530Paolo Guiotto: There cannot be any bound, because when you send X to 0, that exponential goes to 1.
24:11:20Paolo Guiotto: So, what could we say at this point?
24:14:130Paolo Guiotto: Now, at this point, we could say, okay, we can apply the theorem
24:19:420Paolo Guiotto: for this X. For this X, we have…
24:23:430Paolo Guiotto: everything, because the previous step works for every X positive, so it works for every X greater or equal than 1.
24:31:550Paolo Guiotto: So we have the partial derivative, and we have the bound on the derivative.
24:35:920Paolo Guiotto: So we would conclude that there exists a derivative of F for X greater than 1, not for X positive.
24:42:910Paolo Guiotto: So it's like a data for values of X,
24:46:260Paolo Guiotto: So we start from 0 to plus infinity. With this argument, we would be able to say that here.
24:54:460Paolo Guiotto: On this interval, there exists the derivative. And what about this part here?
25:04:670Paolo Guiotto: Can we do anything better than this?
25:11:260Paolo Guiotto: So how can we solve this? Okay, this is not a wrong idea, it's not a bad idea. It is, in fact, the right idea, but we have to work a bit more in general.
25:21:270Paolo Guiotto: So, because you have that. This is a good bound when X is greater than 1. And if I would have… if I want a bound that works for X greater than A,
25:33:150Paolo Guiotto: What is the bound?
25:38:540Paolo Guiotto: What?
25:40:230Paolo Guiotto: Yeah, but you are… you are… oh, look at this function, E…
25:44:720Paolo Guiotto: to minus XT, no, is a decreasing function. So… This is E2 minus Xt.
25:54:530Paolo Guiotto: If X is greater than A, This is bounded by…
26:07:890Paolo Guiotto: Exactly. E to minus AT.
26:12:820Paolo Guiotto: Notice that this is not depending on X.
26:16:320Paolo Guiotto: When X is in that interval, this works, okay? So I call it, let's say, GA of T,
26:25:90Paolo Guiotto: Which is an L1 function on 0 plus infinity.
26:29:720Paolo Guiotto: So I can say that.
26:31:840Paolo Guiotto: So…
26:33:580Paolo Guiotto: The argument… and you have to learn something from this, because you have to adapt this kind of idea. So, I am able to prove that for every A positive.
26:46:170Paolo Guiotto: What I'm doing, I apply the theorem on the interval a to plus infinity.
26:52:830Paolo Guiotto: on interval from A to plus infinity, which is exactly this.
27:02:110Paolo Guiotto: So, I have still to be away from zero, because A must be positive, but the A is a bitter in this argument. I have… we have…
27:13:670Paolo Guiotto: Number one, there exists the partial derivative with respect to X at… sorry, the other was XT.
27:25:100Paolo Guiotto: And this is minus E to minus XT. This also, for every x positive, so in particular, it will be for every X greater or equal than A, and almost every T in 0 plus infinity.
27:39:310Paolo Guiotto: And number two, we also have that modulus of DXFXT.
27:46:520Paolo Guiotto: is controlled by this function, E minus AT, that we baptized GA of T, which is L1.
27:55:370Paolo Guiotto: 0 plus infinity.
27:59:310Paolo Guiotto: And this works for every X, greater or equal than A, and almost every T is 0 plus infinity.
28:07:880Paolo Guiotto: So this means that our D
28:10:880Paolo Guiotto: Our D is a DA, is the interval 0, sorry, A to plus infinity.
28:19:360Paolo Guiotto: So, on this domain, The differentiation theorem applies.
28:32:630Paolo Guiotto: And we get that there existed the derivative of capital F,
28:38:520Paolo Guiotto: And it is equal to integral 0 plus infinity of the derivative of the little f.
28:43:510Paolo Guiotto: Which is, by the way, minus C to minus XT DT.
28:49:310Paolo Guiotto: This happens for every X in that domain, that means for every X greater or equal than A.
28:56:520Paolo Guiotto: Okay?
28:58:940Paolo Guiotto: Now, I can say that, actually, this holds for every X positive.
29:04:310Paolo Guiotto: Why?
29:06:290Paolo Guiotto: Because whatever is, this is zero plus infinity.
29:12:240Paolo Guiotto: Imagine you pick an X, somewhere, from 0 to plus infinity.
29:17:330Paolo Guiotto: You take an A, small enough, in such a way that your X will be in the offline from A to plus infinity.
29:24:870Paolo Guiotto: In that offline, Theorem applies
29:29:950Paolo Guiotto: And it says there exists the derivative, and the derivative is that integral. That formula, of course, does not depend on A.
29:37:260Paolo Guiotto: It is almost the same. So this says that since you can do that for every X possible negative, it means that the derivative exists for every X possible objection.
29:47:540Paolo Guiotto: It's easy to take care.
29:49:310Paolo Guiotto: So sometimes it may happen that you are not able to prove on a full domain that this is derivative, because you have problems with bounds. But maybe you can arrive to the full domain
30:03:370Paolo Guiotto: by a sort of limit, no? In a smaller domain, you can have good bounds, and that you can show that the domain can be enlarged and clean to fill all the initial domain, like here. So the conclusion is that
30:18:860Paolo Guiotto: What is it?
30:22:00Paolo Guiotto: So… there exists F prime of X,
30:26:680Paolo Guiotto: equal to integral 0 plus infinity of minus C2 minus XTT, this for every x positive.
30:36:760Paolo Guiotto: So at the end, we've got the same conclusion, even if we cannot apply directly the theorem on that interval. Now, we can, in a second, we finish the calculation, because now it is easy. So this is the derivative with respect to T of
30:53:620Paolo Guiotto: E minus Xt, perhaps it missed something. The derivative of this, it is E minus Xt, then I have times the derivative of the exponent with respect to t.
31:04:610Paolo Guiotto: which is minus X, so I need to have an X here. I multiply, divide by X, which is positive, so KSL.
31:13:810Paolo Guiotto: So I have 1 over X,
31:17:100Paolo Guiotto: integral 0 to plus infinity of derivative with respect to t of e minus Xt dt.
31:25:580Paolo Guiotto: So that's fundamental theorem of integral calculus, 1 of X, evaluation of E minus XT,
31:33:290Paolo Guiotto: from t equals 0 to t equals plus infinity.
31:39:320Paolo Guiotto: Remind that in this game, X is positive, so when I evaluate that plus infinity is exponential, goes e to the minus infinity 0,
31:50:400Paolo Guiotto: minus the value at t equals 0 into 0, 1, so the value is minus 1.
32:02:560Paolo Guiotto: Who we have.
32:04:170Paolo Guiotto: minus 1 over X. That's F prime of X.
32:09:840Paolo Guiotto: And from this, we can finally…
32:12:980Paolo Guiotto: determine F of X. F of X is minus log of x plus a constant.
32:24:590Paolo Guiotto: And again, we have to determine the value of the constant. Here we have integral 0 plus infinity of E minus XT minus e to minus t divided by t.
32:37:870Paolo Guiotto: So what would you do to determine K?
32:49:540Paolo Guiotto: Ceco ?
32:50:710Paolo Guiotto: X equals 1.
32:52:130Paolo Guiotto: If we take X equals 1 here, we integrate 0, so we have integral 0 plus infinity of 0, which is equal to 0, so we get that k minus log of 1 must be equal to 0. Log of 1 is 0, so k is 0.
33:10:640Paolo Guiotto: And therefore, all this calculation was to discover that F of X is minus log of X.
33:20:350Paolo Guiotto: That seems to be… trust me, it seems to be quite impressive, because you can do this calculation directly, computing the integral with the first-year calculus tools, okay?
33:34:460Paolo Guiotto: Okay, so we stop here, I leave you the exercises to do.
33:40:120Paolo Guiotto: Do exercises number… so we did the first two, so do 7, 3, 3, 4, 5,
33:51:530Paolo Guiotto: Well, you can do everything, until the end. All the exercises are similar.
33:58:880Paolo Guiotto: Tomorrow is, probably… I won't, do class, but I will confirm you by tonight, because maybe they will change, once again, the appointment, okay? So, please check the email by tonight, I will send you the confirmation if there is class or not tomorrow.
34:18:350Paolo Guiotto: Okay, let's stop.