AI Assistant
Transcript
00:23:150Paolo Guiotto: Okay, so I apologize for the delay.
00:27:70Paolo Guiotto: It's due to traffic, heavy traffic to… down.
00:33:210Paolo Guiotto: And, since now it's about 9, almost 9, 10,
00:39:720Paolo Guiotto: We won't do the break, we will… we will just go straight, yeah.
00:45:740Paolo Guiotto: So, let's start with the sum of the exercises left.
00:52:890Paolo Guiotto: For example, I take, the…
00:59:110Paolo Guiotto: Yeah, number 6 I do, but it's not particularly relevant.
01:06:670Paolo Guiotto: It is the integral on… Yes, I do, one second.
01:11:560Paolo Guiotto: Exponential minimum between X squared and Y squared, the X DY, right?
01:21:260Paolo Guiotto: So… Here, we have the solute… the function FXY, Is exponential of minimum.
01:32:390Paolo Guiotto: between X… It's maximum.
01:37:660Paolo Guiotto: Yeah.
01:40:240Paolo Guiotto: It's not… it's not Saturday first.
01:44:120Paolo Guiotto: Maximum.
01:47:580Paolo Guiotto: Y-square.
01:50:710Paolo Guiotto: Now, this, well… There is a fact that we can accept that remark.
02:02:840Paolo Guiotto: When you do the maximum between two functions, let's say H of XY, this is for any, actually, any function of vector variable, H of X,
02:17:480Paolo Guiotto: Geomax, huh?
02:21:150Paolo Guiotto: And these, H and G.
02:24:400Paolo Guiotto: continuous.
02:26:110Paolo Guiotto: Then, this is, if F is this, F is continuous.
02:34:730Paolo Guiotto: So this is just to say that our function is continuous.
02:38:900Paolo Guiotto: On the domain, which is 01 square, which is compact.
02:45:290Paolo Guiotto: So, compactor. So, we are sure that there exists the integral of F.
02:51:750Paolo Guiotto: on the domain.
02:53:190Paolo Guiotto: I'm sure that I will, but the problem is with the calculation of this integral.
02:57:290Paolo Guiotto: Now, the integral on 0, 1… Square.
03:01:870Paolo Guiotto: of F.
03:04:440Paolo Guiotto: Of course, we, want to apply the reduction formula.
03:08:910Paolo Guiotto: will be a iterated integral of the function E, maximum, between, X square and Y square.
03:21:710Paolo Guiotto: let's say that it's indifferent because the situation is completely symmetric in X and Y, so we may start integrating in X and
03:30:240Paolo Guiotto: Finishing in Y. About the domain, it's not difficult to determine sections, because the domain is 0, 1 squared. This is the set of points XY, with X between 0 and 1, and Y between 0 and 1.
03:46:960Paolo Guiotto: So, if you decide to integrate first in X and then in Y, you have to say that for Y between 0 and 1, X must be between 0 and 1, and for other Y, there is no X. So this means that this splits just into these double integrals.
04:04:730Paolo Guiotto: Now, how do we compute that? Well, we observe that the maximum between X square.
04:14:470Paolo Guiotto: X squared and Y squared. It is one of the two. It will be X squared when X squared is greater or equal than Y squared. It will be Y squared when Y squared is greater or equal than X squared.
04:28:990Paolo Guiotto: Now, this happens if and only if reminds that our X and Y are in 01.
04:39:270Paolo Guiotto: Each, so they are in this.
04:41:650Paolo Guiotto: Rectangle. So the condition X squared, greater or equal than Y square, is equivalent to…
04:49:960Paolo Guiotto: X greater or equal than Y.
04:53:290Paolo Guiotto: And the second one is when Y is greater than or equal than X.
04:59:790Paolo Guiotto: So, now, if I look at the innermost integral, this one.
05:05:260Paolo Guiotto: the integral in X, so let's write this.
05:08:400Paolo Guiotto: Integral between 01 of e to maximum x squared Y square.
05:16:680Paolo Guiotto: DX.
05:17:930Paolo Guiotto: So, this is an integration in X. So if I look at that maximum as a function of X, I have that the maximum is X squared when X is greater than Y. It is Y squared when X is less than Y, so this means that for X, I have to split that integral into
05:36:330Paolo Guiotto: One is the integral from 0 to Y.
05:38:850Paolo Guiotto: plus the other will be the integral from Y to 1. Remind that the Y is varying between 0, 1, because of this.
05:46:790Paolo Guiotto: So when X is between 0Y, it means we are, in this case, X is less than Y, and you read there, E2, the maximum will be Y squared.
05:59:980Paolo Guiotto: In the second case, X is between Y and 1, so it means that it is greater than Y in particular, and the maximum is E2X squared.
06:10:420Paolo Guiotto: So that's what we have to compute.
06:14:370Paolo Guiotto: Okay?
06:18:260Paolo Guiotto: Good. Now, the first one is easy, because you see that the function you have inside is a constant for X, so you can carry outside, it remains a factor of 1 inside, so we got EY squared integral 0 toy of 1 dx.
06:37:260Paolo Guiotto: Well, for the second integral, it seems to be practically impossible, because, you know, this function is an example of a function for which we cannot compute
06:48:680Paolo Guiotto: a primitive, so we don't know what is the function, whose derivative is this one, even if it seems very simple, but this is a particular situation. However, let's keep for a second the syntax. Let's… we can do anything, so let's leave this integral…
07:05:820Paolo Guiotto: As with this notation.
07:10:450Paolo Guiotto: So the first one is now YE to Y squared, plus this second one is the integral to… from Y to 1 of E
07:20:750Paolo Guiotto: X squared. And let's see what we can do, because we have still to do another integration.
07:26:200Paolo Guiotto: This is the innermost integral, then we have to integrate from 0 to 1 this thing in variable Y. So…
07:35:350Paolo Guiotto: The integral on domain D of F is the integral from 0 to 1 of this thing that goes here.
07:43:520Paolo Guiotto: And so we got Y, E2Y squared plus integral from Y to 1, E2X squared dx dx, all this in DY.
07:57:130Paolo Guiotto: Okay, now the first part is easy, because E to E square is not an easy derivative, but this is an easy derivative.
08:06:20Paolo Guiotto: Because you see, this is the derivative with respect to Y, more or less, of e toy squared. In fact, if we do the derivative, we get e toy squared times 2y. So to have the derivative, I need a factor 2, so I can put a 2 here and 1 half here.
08:23:280Paolo Guiotto: So I will have 1 calf.
08:26:350Paolo Guiotto: the integral of this, which is the derivative of e to y squared, so this is the evaluation e to y squared between y equals 0 and y equals 1, and that's over. Okay, now we do the calculation.
08:41:940Paolo Guiotto: What remains is a second integral, integral 01, of integral from Y to 1 of e tox squared dx, then dy.
08:51:60Paolo Guiotto: And the problem is that we don't know directly how to compute this.
08:55:650Paolo Guiotto: But now, here's the trick. First, we finish that part. There's one half. When I put y equals 1, I get e to 1, e. When I put Y equals 0, I get e to 01.
09:08:190Paolo Guiotto: So, what is the trick to do this, huh?
09:10:800Paolo Guiotto: Well, I integrate by parts. I imagine that here there is a factor 1, which is the derivative of Y,
09:18:650Paolo Guiotto: times the integral for y to 1 of E2X squared dx in dy.
09:25:340Paolo Guiotto: Now, if we do, let's focus on this integral, yeah, okay?
09:30:420Paolo Guiotto: If we do the integration by parts.
09:36:150Paolo Guiotto: So I have the evaluation of the product this time this, so y times integral from Y to 1, e tox squared dx, to be evaluated from y equals 0 to y equals 1. And let… now we see that this is easy.
09:54:530Paolo Guiotto: Minus integral from 0 to 1, the derivative moves on this factor.
09:59:670Paolo Guiotto: And look at what happens. So, let's write just derivative with respect to Y of the integral from Y to 1 of e tox squared dx.
10:09:650Paolo Guiotto: then all this will be wild.
10:12:640Paolo Guiotto: Let's first solve this. When y is 1, you get 1 integral from 1 to 1, whatever, is 0. You see?
10:21:220Paolo Guiotto: 1 times integral from 1 to 1 EX squared dx, this quantity here is 0.
10:28:910Paolo Guiotto: So the value will be 0, minus.
10:31:400Paolo Guiotto: 0 times integral from 0 to 1 of e tox squared. We don't know what is this integral, because this cannot be computed, but it is multiplied by 0. So again, we get 0. So at the end, this evaluation is 0.
10:47:840Paolo Guiotto: And now, what is the other part? It is minus integral from 0 to 1 of y, and now what is the derivative with respect to Y of this thing? Integral from Y to 1E to X squared dx.
11:04:270Paolo Guiotto: Do you recognize
11:10:340Paolo Guiotto: What kind of object is this?
11:18:20Paolo Guiotto: Okay, maybe you don't have to recognize, so let's flip the order of the two endpoints by changing the size, so this is the same of DY of minus integral from 1 to y e tox square dx. Do you recognize better now?
11:40:880Paolo Guiotto: No, you have to go back to first year.
11:46:640Paolo Guiotto: This object is something like, I'm doing the derivative with respect to Y of integral from constant, say, to y, of a function, let's call, with another name, G of T dt.
12:00:40Paolo Guiotto: What is the name of this thing?
12:08:70Paolo Guiotto: You don't have to remind. This is called the integral… function… Oh.
12:17:570Paolo Guiotto: G.
12:18:730Paolo Guiotto: And what is the remarkable property of this integral function?
12:23:100Paolo Guiotto: It is in a very important theorem, because it is called the fundamental theorem of Integral calculus.
12:30:90Paolo Guiotto: Okay I will repeat 100 of times, at least, because
12:35:940Paolo Guiotto: There is a reason why this theorem is so important. The main reason that you have seen is because it allows to compute integrals practically, because it's from this theorem that you have the formula. The integral, if you have a primitive, you do final value minus initial value. That formula comes from the fact that
12:54:500Paolo Guiotto: When you do the derivative of this, You got what?
13:00:890Paolo Guiotto: If the function G has some minimal conditions, so, like, continuous.
13:05:640Paolo Guiotto: Then the derivative is just a function. That's a primitive of the… of the function you have in the integral.
13:13:870Paolo Guiotto: Okay, this is the fundamental… DRM… Integral.
13:21:00Paolo Guiotto: calculus.
13:23:390Paolo Guiotto: Fundamental. Fundamental, fundamental. It's fundamental, okay? It's so important. So, this is G of Y.
13:34:300Paolo Guiotto: And therefore, when we apply to this, well, the minus is outside the minus, we have the derivative, which is E2Y squared.
13:43:870Paolo Guiotto: So the derivative of that guy is minus C2Y square, and therefore, going back to our problem, we have to compute integers from 0 to 1 of Y times that derivative. The derivative is now
13:56:610Paolo Guiotto: minus E toy squared.
13:59:540Paolo Guiotto: That's what we need to put here.
14:02:650Paolo Guiotto: integrated in DY. And this integral can be computed. It's the same integral we computed a minute ago. It was with the…
14:12:850Paolo Guiotto: Where is it? It was here. You remind, integral 0 to 1 of YEY squared, this is 1 half E2 minus E minus 1, so this will be minus, minus, plus 1 half E minus 1.
14:31:410Paolo Guiotto: So, we have now the conclusion. Conclusion.
14:38:700Paolo Guiotto: the integral on D of the function F, so let's go back where we started, is this one.
14:45:640Paolo Guiotto: Which is equal to that.
14:48:10Paolo Guiotto: we have 1 half E minus 1 time plus this, but this is, going down here, another 1 half E minus 1, so 2 times 1 half E minus 1, it is E minus 1.
15:01:870Paolo Guiotto: And that's the final answer.
15:05:730Paolo Guiotto: Okay, so this was tricky, not because of the calculation of the setup of the multiple integral, but because of the
15:14:140Paolo Guiotto: of the… Nature of the function.
15:23:80Paolo Guiotto: I do not remind if we have done… have we done any one of these 7, 8, 9, 8, 9, no?
15:29:960Paolo Guiotto: So let's do the number 7, which is, easy.
15:35:190Paolo Guiotto: So, we have integral, and the domain is 0 plus infinity, Cartesian product, 1 plus infinity.
15:44:940Paolo Guiotto: of E minus XY square DXDY.
15:52:70Paolo Guiotto: So… solution.
15:54:980Paolo Guiotto: So the function F, is…
15:59:150Paolo Guiotto: for which we want to complete the integral is E minus XY square, which is clearly continuous everywhere, and in particular, on the integration domain, D.
16:10:430Paolo Guiotto: The main D, which is this set, 0 plus infinity, so as you can see, it is unbounded for the X, and it is also unbounded for the Y.
16:19:430Paolo Guiotto: So, this is, easily closed down.
16:22:930Paolo Guiotto: Because you can see this is the set where X is greater or equal than 0, and Y is greater or equal than 1. It's a set of points that verify these two conditions, so it is closed.
16:35:170Paolo Guiotto: But… I'm bounded.
16:40:630Paolo Guiotto: So, we cannot say immediately this, function is integral, okay? So, to check… integrability.
16:57:610Paolo Guiotto: We… apply… tonality… DRM that says…
17:10:670Paolo Guiotto: If one of the, two iterated, two in this case, two iterated integrals for the model also, If…
17:18:640Paolo Guiotto: one of…
17:20:890Paolo Guiotto: So we have, one-off means you do first integration next, and then white, it doesn't matter you write the appropriate domains, it's not the case here, or…
17:33:470Paolo Guiotto: The other one is, for the integration, first in Y and then in Xer, is find it.
17:44:250Paolo Guiotto: then there exists the integral on domain D of F. Of course, if you wanted notations, this… since this is the integration in first in X, we have the Y sections, and these are the Ys for which the Y section is non-empty. This is on the X sections, and the second integration is on D.
18:04:290Paolo Guiotto: on the X for which the DX section is gonna… okay?
18:08:170Paolo Guiotto: Now, the function here is positive, because, you see, it's exponential of something, Since,
18:17:850Paolo Guiotto: F is positive, modulus of F coincide with F. So, once we compute one of these two iterated integrals, we have done the iterated integral for F.
18:29:920Paolo Guiotto: And so, automatically, once we, we know that, the…
18:35:900Paolo Guiotto: function is integral, to compute the integral, we will apply the reduction formula to F. Let's say that this is, for example, the iterative integral of F, integrated first in X and then in Y, same domains, so this is DY and DY non-empty.
18:56:70Paolo Guiotto: or the same with the other iteration, DYDX. So we don't have to repeat the calculation, because the calculation, the calculation is already done.
19:10:420Paolo Guiotto: Okay?
19:12:730Paolo Guiotto: So now the question is, which one should be the choice.
19:18:40Paolo Guiotto: We have two problems. One is the description of the… these sections. We should choose,
19:27:340Paolo Guiotto: That's not a criteria, but we should probably choose
19:31:950Paolo Guiotto: For, the one which is easier.
19:36:120Paolo Guiotto: But in this case, both descriptions are the same, because you see how the domain is made. You have X is greater or equal than 0 independently of Y, and Y greater or equal than 1, independently of X.
19:50:640Paolo Guiotto: So this says that if you choose the first one, the DY section will be X greater or equal than zero.
19:57:940Paolo Guiotto: when Y is greater or equal than 1. And for this, right choice, we have Y greater than or equal than 1, the innermost integral, and the X integral.
20:09:550Paolo Guiotto: But it's, you know, yeah, so it's, not a problem.
20:13:150Paolo Guiotto: The second point is that we have to integrate modulus of F, which is F, in X, in Y. Now, if we look at this function, there is… it seems there is a little difference, an important difference, when we have to do the first integration.
20:28:210Paolo Guiotto: Because, if you imagine integrating first in X, you have something like E2 constant times X, which is easy.
20:36:550Paolo Guiotto: while, if you integrate first in Y, you have constant Y square, which is impossible. So, in this case, the choice, is just a one way. We should integrate first in X and then in Y. So let's great, yeah.
20:52:600Paolo Guiotto: suppose.
20:55:260Paolo Guiotto: Okay.
20:57:580Paolo Guiotto: However, it doesn't change, okay? It's even worse. Okay, so let's do first the integration in X. So let's take, say, this one, okay? We choose this.
21:11:310Paolo Guiotto: So, the iterated integral of modulus of F, dx, dye.
21:18:50Paolo Guiotto: As we said, with this choice, fortunately, the domain is easy. X will be between 0 and plus infinity when Y will be between 1 and plus infinity. This is the iterated integral.
21:32:600Paolo Guiotto: So it is the integral from 1 to plus infinity of integral from 0 to plus infinity.
21:38:340Paolo Guiotto: of E minus XY power 4, dx, then DY.
21:44:70Paolo Guiotto: Of course, you cannot see what is going on on the second integration unless you are able to compute in your mind the first integral, and then, let's see, maybe there is a problem in the second step.
21:57:40Paolo Guiotto: But, so let's start, to do this. Now, this, as function of X, is the derivative of what? This comes
22:06:450Paolo Guiotto: Well, you know that exponential comes from exponential, so probably I should compute this derivative, the derivative of this. In fact, this is the same exponential times the derivative of the exponent with respect to X, which is minus y to power 4.
22:25:70Paolo Guiotto: So, since I have this, I have to divide by this factor, which is a constant for the derivative, so I can carry in and out of the derivative, nothing changes.
22:36:280Paolo Guiotto: And, so I can say that that function is the derivative of this one. So this will be the evaluation of that function, so let's write E minus XY power 4 divided by Y power 4,
22:50:720Paolo Guiotto: between, X equals 0 and X equals plus infinity, then I will have to integrate this in Y.
23:00:180Paolo Guiotto: Now, this is a factor, so we carry outside the integral from 1 to plus infinity minus 1 over Y24, and then we have this evaluation, E minus XY power 4 at X equals 0 plus infinity, and then the difference.
23:17:250Paolo Guiotto: At plus infinity, we have to take the limit when x goes to plus infinity.
23:22:390Paolo Guiotto: So, when X goes to plus infinity, this quantity.
23:26:290Paolo Guiotto: So, when X goes to plus infinity.
23:30:850Paolo Guiotto: what it does, we have this is minus, this Y power 4Y is greater than 1, so it's definitely positive. So the exponent goes to minus infinity. So, e to minus infinity, which is 0.
23:46:970Paolo Guiotto: So the value at plus infinity is 0, the value at x equals 0, E minus XY power 4 at x equals 0 is e to 0 equals 1.
23:58:30Paolo Guiotto: So at the end, that evaluation gives 0 minus 1, so it is minus 1.
24:04:830Paolo Guiotto: So, and since we have another minus here, at the end, we, boil down to this integral, 1 over Y to power 4DY, which is easy.
24:17:120Paolo Guiotto: Now, this is a power. It is Y to exponent minus 4.
24:23:370Paolo Guiotto: Since the exponent is not minus 1, the primitive is a power, which is y to the exponent minus 4 plus 1, so minus 3 divided that number, minus 3, to be evaluated between y equals 1, y equals plus infinity.
24:38:900Paolo Guiotto: At Y equal plus infinity, we get 0.
24:42:510Paolo Guiotto: at Y equal 1, we get minus 1 third. So, doing the difference, I get a plus 1 third as the final value.
24:52:380Paolo Guiotto: This says at once.
24:54:890Paolo Guiotto: This is the iterated integral for the modules. So, it is finite, we have that, this is the Tonelli.
25:03:80Paolo Guiotto: theorem, that the function f is integral on the domain DE, And the…
25:12:370Paolo Guiotto: the value of the integral on D of F by the reduction formula is the value of the either-it integral that I just computed, because the function F being positive coincides with its modulus.
25:26:860Paolo Guiotto: And therefore, this is equal to 1 third, so I don't have to repeat real. This is because the function has constant sine.
25:36:70Paolo Guiotto: Otherwise, wouldn't be true.
25:40:810Paolo Guiotto: Okay, I want to show… Before we continue, something.
25:50:240Paolo Guiotto: I want to show you something which is, but he moved.
25:58:160Paolo Guiotto: That should be somewhere in this example.
26:02:210Paolo Guiotto: Yeah, okay. It is the example, in notes, example…
26:08:320Paolo Guiotto: 539, which is actually a warning.
26:17:460Paolo Guiotto: So, something you should be aware.
26:21:590Paolo Guiotto: That is… okay, we said, and we have seen here the machinery, that…
26:29:860Paolo Guiotto: I've not repeated, I've said, but let's return on the statement here.
26:41:910Paolo Guiotto: Okay, this theorem says if one of the iterated integral of the models is finite, the function is integral, and then you can compute the integral by using the reduction formula.
26:54:440Paolo Guiotto: Be careful, because it may happen that the two iterated integral of F, not models of F, can be fined
27:04:450Paolo Guiotto: but different.
27:06:770Paolo Guiotto: Okay? And in that case, the function f cannot be integral.
27:12:190Paolo Guiotto: So I cannot replace F modulus of F with F, because with that is false.
27:18:00Paolo Guiotto: You say, but you just did this, you used F, but that's because modulus of F is F, not because you can replace modulus of F with F. So, it may happen.
27:36:840Paolo Guiotto: That.
27:38:920Paolo Guiotto: the iterated integrals for F…
27:43:290Paolo Guiotto: on the appropriate domains, so I do not repeat, I do only once, so here you have the Y sections, and here the Y, which is different from empty, and the other one that's integrating first in Y and then in X,
28:01:200Paolo Guiotto: booth find it.
28:04:740Paolo Guiotto: So, apparently, it seems that this says that the function has an integral, because if it has an integral, it is one of these two, no? It's the reduction formula that tells this. But…
28:17:860Paolo Guiotto: There is no integral of F.
28:22:320Paolo Guiotto: So, how is possible? So, let's see first the example, then we think about this. Take this function.
28:33:960Paolo Guiotto: The function is, not particularly complicated, it takes a little bit of calculations.
28:43:230Paolo Guiotto: However, elementary calculations, X minus Y divided X plus Y cubed for XY on the square, 01 square, which is our domain.
28:58:380Paolo Guiotto: Okay?
28:59:620Paolo Guiotto: Now, let's do the calculation of the first iterated integral. As you will see, the second one, we don't have to repeat the calculation. So, here…
29:13:170Paolo Guiotto: We have, huh?
29:17:500Paolo Guiotto: that if I do the iteration integral, let's say, first in X, FXY.
29:24:490Paolo Guiotto: DXDY. Now, what are the range for XY? Now, the domain D is X points XY, where X is between 01 and Y is between 0, 1, right?
29:37:790Paolo Guiotto: So again, this is a simple domain. When Y is between 0 and 1, X will be between 0 and 1. So this is the iterated integral I have to do.
29:48:600Paolo Guiotto: So let's do. It takes a little bit of time, but, whoever.
29:53:140Paolo Guiotto: As a function of X, this is a rational function, so a ratio between polynomials, so we have a sort of algorithm to do this calculation.
30:01:840Paolo Guiotto: So we have integral 01, 01 of X minus y divided X plus Why? Cube.
30:10:950Paolo Guiotto: DX first, and DY then. So let's focus first on this guy.
30:17:390Paolo Guiotto: So this is integral 01x minus y divided X plus Y cubed
30:25:220Paolo Guiotto: And you have to forget about Y. Y is a number here. X is the variable.
30:30:410Paolo Guiotto: So, I can compute easily this by… with some trick. For example, I recreate the denominator, adding and subtracting a Y here, so in such a way that I have… I split, then, this… the fraction, so I will have X plus Y divided X plus Y.
30:50:150Paolo Guiotto: Cuba.
30:51:890Paolo Guiotto: Then I have, you see that this minus Y with the other minus Y, it's minus 2Y, so minus 2Y divided the X plus Y cubed.
31:05:280Paolo Guiotto: Now, this is inexo.
31:10:120Paolo Guiotto: Okay, so the first one simplifies a bit, so we get integral 1 over X plus Y squared minus 2Y divided X plus Y cubed.
31:25:820Paolo Guiotto: These two integrals are both easy, because look at the first one. Integrals 01 of 1 over X plus Y squared.
31:36:560Paolo Guiotto: What is this?
31:38:740Paolo Guiotto: Now, look at the function. The function is a power. It's X plus Y to exponent minus 2.
31:46:100Paolo Guiotto: So, should… this should come… doing the derivative with respect to X of…
31:54:40Paolo Guiotto: Well, let's… let's do this to minus 1, then we adjust. The derivative of this is…
32:00:850Paolo Guiotto: Minus 1, you take down the exponent, then you decrease by 1 unit, so this becomes to minus 2, then you do the derivative of the argument with respect to X, and that's 1.
32:11:880Paolo Guiotto: So basically, it's the same, apart for the minus 1. So now, with the minus here, we have… let this disappear. So, it is the derivative of that quantity. And therefore, this integral will be the evaluation of that quantity that let's rewrite as 1 over.
32:29:190Paolo Guiotto: X plus Y between X equals 0 and X equals 1. So this yields the minus outside. When X is 1, I get 1 over 1 plus Y, or Y plus 1, minus when X is 0, 1 over Y.
32:45:340Paolo Guiotto: So, at the end, the first result is 1 over Y minus 1 over Y plus 1.
32:53:410Paolo Guiotto: This is for this part. Then we have the other guy, the 2Y. But the 2Y is a constant for that integration. Now, you see that here the integration variable is X, while here we have this Y. Y is a variable, yes, but not for the integral. For the integral, it is a constant.
33:11:290Paolo Guiotto: So when I take this second integral, 0 to 1 of minus 2y divided X plus Y to exponent 3,
33:21:520Paolo Guiotto: in the variable x, I can carry the minus 2Y outside the integral 0 to 1, and then I have this, more or less the same thing, as the previous integral, no? Because now we have X plus Y to minus 3.
33:35:50Paolo Guiotto: But what is this? This comes by doing the derivative with respect to X of… you have to increase the exponent by 1, so it becomes to minus 2.
33:45:980Paolo Guiotto: If you do the derivative, this is…
33:50:820Paolo Guiotto: minus 2, X plus Y to exponent minus 2 minus 1 minus 3 times 1, which is the derivative. So basically, you see that it is convenient to carry the minus 2 back to the interior, inside the integral, so let's keep the minus 2 here, and we have a derivative, right?
34:09:70Paolo Guiotto: So now we have Y times evaluation of this X plus Y2 minus 2, or 1 over X plus Y squared.
34:19:139Paolo Guiotto: Again, between X equals 0 and X equals 1, which is Y times. When X is 1, I get 1 over Y plus 1 square. When x is 0, 1 over Y squared.
34:35:800Paolo Guiotto: Which is multiplied all this by Y, so Y divided Y plus 1 square minus 1 over Y.
34:45:20Paolo Guiotto: That's the second integral. Now, let's put all this together, so let's return here.
34:53:520Paolo Guiotto: So this is the value of this integral, okay? So let's write the value of that integral. Integral from 0 to 1 of X minus y divided X plus Y cubed in variable X is equal to…
35:11:180Paolo Guiotto: So we… we wrote that integral as this one.
35:15:30Paolo Guiotto: The first part is the integral of 1 over X plus Y squared is 1 over Y minus 1 over Y plus 1, so let's copy. 1 over Y minus 1 over Y plus 1.
35:28:80Paolo Guiotto: plus… then there is a second part, which is the integral of minus 2Y divided by blah blah blah. That integral is here, and the final value is this one. So I have to add this value. Y divided by 1Y plus 1 squared minus 1 over Y,
35:48:20Paolo Guiotto: which simplifies this 1 over Y, and so at the end, we get this. I wouldn't do the common denominator, because I have still to do another integration, and it is more convenient
35:59:970Paolo Guiotto: No.
36:02:260Paolo Guiotto: It is better if we decompose this again by doing the trick of adding and subtracting 1, or whatever I can do later, because in this way it becomes 1 of minus 1 over Y plus 1 plus
36:18:730Paolo Guiotto: When you do, you split this numerator with Y plus 1, divided by 1 plus 5.
36:24:200Paolo Guiotto: square, so it becomes a 1 over Y plus 1, which cancels this one.
36:30:310Paolo Guiotto: minus… 1 over Y plus 1 square. So that's, at the end, the value of 13.
36:40:700Paolo Guiotto: Okay, we have not yet finished, because we computed the…
36:44:810Paolo Guiotto: V value of this box here.
36:47:900Paolo Guiotto: As you can see, it does not contain any X, okay? That cannot be X. So let's go back here, let's put a star. Maybe I will just copy it down here.
37:00:730Paolo Guiotto: So… The iterated integral of F.
37:05:380Paolo Guiotto: first in X, then in Y. We say that this is between 01 and 01, is equal to the integral 01 in Y of this guy, minus 1 over Y plus 1 square DY,
37:20:440Paolo Guiotto: You see?
37:22:300Paolo Guiotto: Okay?
37:23:440Paolo Guiotto: And this one is easy, because, this is exact…
37:27:250Paolo Guiotto: create the derivative respect to Y
37:30:260Paolo Guiotto: of Y plus 1 to exponent minus 1, right? The derivative is minus 1, y plus 1 to exponent minus 2 times 1. Exactly that quantity.
37:42:710Paolo Guiotto: So, the conclusion is that we get the evaluation of y plus 1 to exponent minus 1 between y equals 0 and y equal 1.
37:55:980Paolo Guiotto: And this, finally, is the value when y is 1.
38:01:340Paolo Guiotto: We gathered the value, 2 to minus 1, 1 half.
38:06:940Paolo Guiotto: minus when the value Y is 0, get 1. So the conclusion is minus 1 half.
38:13:190Paolo Guiotto: Okay, so let's write this. The iterated integral.
38:17:240Paolo Guiotto: between 0 and 01 of the function FXY,
38:22:580Paolo Guiotto: first in X, then in Y, turns out to be minus 1F. So it is finite, and this is the bet.
38:30:740Paolo Guiotto: What happens if we switch the order of the two integration?
38:36:810Paolo Guiotto: Now, you may imagine that we have to repeat this long calculation, but we don't need it. Let's see why.
38:42:180Paolo Guiotto: Because if you do these, other integrations, first in… now in Y, and then in X, of course the range for X and Y is the same, because the condition is the same. Now, let's write down this.
38:56:420Paolo Guiotto: is, explicitly. This was X minus Y divided X plus Y cubed, right?
39:03:900Paolo Guiotto: This is in DY in the X.
39:06:590Paolo Guiotto: But let's look at what is this guy. This is the integral from 0 to 1, explicitly.
39:12:740Paolo Guiotto: integral 0 to 1 of the same thing, X minus y divided X.
39:18:540Paolo Guiotto: plus Y cubed. Now, this is in X first in Y step.
39:24:670Paolo Guiotto: Do you see anything?
39:37:830Paolo Guiotto: Yeah, they are the same apart for the sign, and you can see, because letters are letters, you agree with me, no? So if you now change the letter X with the letter Y, change the letter, not doing anything else, we write this by
39:53:600Paolo Guiotto: by writing Y where you see X and X where you see Y, you would see Y minus X divided Y plus X cubed
40:06:370Paolo Guiotto: DXDY. I'm not doing the switched integration, I'm just changing the letters. Lattice are letters, it doesn't matter how I call the variables. And if you look at this one, it's exactly that one, apart for the sign, you see?
40:22:710Paolo Guiotto: there is the denominator X plus Y is the same of Y plus X, so denominator is the same, but numerator is different by the size. So, I can say that since this one has value minus 1 half, the other one will have value plus 1 half.
40:40:190Paolo Guiotto: So, you see that at the end, this saturated integral has value plus 1 half.
40:49:520Paolo Guiotto: So what we obtained is that if we do the iterated integration first in X, and then in Y, this is value minus 1 half.
40:59:700Paolo Guiotto: If we do the iterated integration of F, first in Y, then in X, this yields value plus 1 half, and they are not the same.
41:09:100Paolo Guiotto: So what does it mean? This means that,
41:17:310Paolo Guiotto: that, there cannot be the integral of F on D.
41:23:790Paolo Guiotto: Why?
41:24:970Paolo Guiotto: Because, we say that there is a general factor, general theorem.
41:29:920Paolo Guiotto: Which is, the reduction formula.
41:36:420Paolo Guiotto: I don't know where is this statement, so let me…
41:39:720Paolo Guiotto: Yes. That says, if the function f is integral, then the integral of F is one of the two iterated integrals, so this is the version with the first integration in Y, and then the second integration in X, but we say the
41:58:550Paolo Guiotto: Maybe in the other class.
42:02:180Paolo Guiotto: Maybe in this one.
42:04:160Paolo Guiotto: Let's see…
42:17:500Paolo Guiotto: Well, okay, here it is written, no? If there is the integral, then the integral of F is equal to one of these two, and they are the same. They must be the same.
42:28:380Paolo Guiotto: So, if they are different, it means that F cannot be integral. Otherwise, the integral of F would be the same value of these two. In particular, these two should be the same number. If they are different.
42:42:840Paolo Guiotto: This formula cannot be valid, and therefore the assumption cannot be verified. That's the point.
42:50:540Paolo Guiotto: Okay? So this means that there is no integral of F, otherwise… reduction formula. Would… Apply… And…
43:08:360Paolo Guiotto: the integral of F would be the same of the two iterated integrals.
43:15:180Paolo Guiotto: The first one with the integration first in X and then in Y, the second one with the opposite order. And in particular, the two should be the same, should be equal.
43:25:70Paolo Guiotto: But that's not true for our case.
43:28:00Paolo Guiotto: So, this example shows that it is not sufficient to check that the iterated integral of F
43:37:350Paolo Guiotto: one of the two is finite, because they could, too, deduce that f is integral, because they could be both finite and different. In that case, you would have that the function cannot be integral.
43:50:420Paolo Guiotto: So you have to pass through the absolute value. If you are lucky that the function has a constant sign, so that means that you have basically to do just the check of the iterated integral for the absolute value for the function… where is it? I know it's not here.
44:12:950Paolo Guiotto: Yes, it's written here. You have to do one of the two, integration, check it is finally, and then, since you have done the calculation, actually, with F, not with modules of F,
44:23:120Paolo Guiotto: better. Modules of F coincide with that. You don't have to repeat, okay, to the application of the reduction formula, because you have already done one of these two iterated integrals, okay?
44:36:810Paolo Guiotto: If the function has not constant sine, then it means that you, in principle, should do two checks. Number one, you check the iterated integral for the absolute value.
44:48:260Paolo Guiotto: If once this is finished, you pass through the calculation, and in that case, the calculation probably will be different. But you are sure that you can apply the reduction formula.
44:58:170Paolo Guiotto: However, I don't think we will deal with such kind of situations, but it's important that you have clear that the condition to be checked is this one with the absolute value, and not just directly with F.
45:11:590Paolo Guiotto: Of course, I repeat once more, if F has constant sine, positive or negative, modulus, if F is positive, it's F itself. If F is negative, it's minus F, you understand that
45:24:490Paolo Guiotto: Putting a plus or a minus in front of a function does not change
45:28:210Paolo Guiotto: the value of the integral, okay? This is not meaning that it's the same if the function has variable sign. This is constant sine function at each point positive or negative.
45:40:320Paolo Guiotto: Okay, so we have to move a bit forward.
45:43:990Paolo Guiotto: Now, we have seen the reduction formula and all the story, let's say, for double integrals. So, for integrals are functions of two variables.
45:54:320Paolo Guiotto: What happens when we have to deal with a function of 3 or more variables? I will show you the case of function of three variables, so you can understand how it works in general, okay?
46:07:180Paolo Guiotto: So… The question is, how… reduction formula.
46:14:750Paolo Guiotto: walks… for… functions… F… of 3 variables, X, Y, Z.
46:27:900Paolo Guiotto: Well, the idea is that we should keep the, let's say, the philosophy, the same philosophy. So, we have this fact. The first fact is the Fubini theorem.
46:44:810Paolo Guiotto: Which is the… also the… called the reduction formula.
46:54:170Paolo Guiotto: This says, if you already know that the function is integral, the reduction formula applies, okay? So, if…
47:03:490Paolo Guiotto: there is the integral on the domain D. Now, the function F is function of three variables, XYZ, DXDYDZ,
47:14:210Paolo Guiotto: Then… this integral, on the… can be computed.
47:23:660Paolo Guiotto: by doing one of the following iterated integrations. Be prepared, because there are six different formulas, okay? But you understand the first two, and you understand the remaining four.
47:37:540Paolo Guiotto: Because the… what is the philosophy of the reduction formula with the two variables?
47:43:800Paolo Guiotto: You have here, no? It says…
47:46:160Paolo Guiotto: The idea is that you start integrating with a variable, and then you integrate with the remaining variable. Now, in the case of functional variable, there are only two variables, so the choice is black. Either you start with X and you end with Y, or you start with Y and you end with X.
48:04:220Paolo Guiotto: Now, there are 3 variables.
48:06:830Paolo Guiotto: So you could say, suppose that I start index.
48:11:540Paolo Guiotto: The second integration will be on the remaining variables, so it would be a double integration, because there are still two variables, not one.
48:19:210Paolo Guiotto: So, I could say that this is an iterated integration, where I do first the integration.
48:27:140Paolo Guiotto: let's say, in Excel.
48:30:370Paolo Guiotto: And then, what remains is an integration in what are the remaining variables, which are Y and Z. So the second step is a double integration, DYDZ.
48:41:810Paolo Guiotto: What about the domains? The domains are the same, because you see this one, you take this domain here. This is a domain of X, because we are integrating in X. So, it's a set of X, such that this point, X, Y,
49:00:870Paolo Guiotto: Z belongs to the domain D, otherwise I go outside of the integration domain. Also, the idea is the same.
49:09:470Paolo Guiotto: Now, this is a set of abshesa of points where the other two coordinates, the Y and the Z, are fixed. So, these are like parameters. We can put this down here, YZ, and we call this the YZ section, again.
49:26:840Paolo Guiotto: Y. Z section… off.
49:33:780Paolo Guiotto: D.
49:35:510Paolo Guiotto: So this is the domain we have to put here, DYZ, which is a domain of X. It is always a domain of the remaining variable. So, in this case, you see that there is no X, that's a domain of X.
49:50:200Paolo Guiotto: And, on which YZ will you integrate? On the YZ for which that set is non-empty. So, DYZ…
49:58:590Paolo Guiotto: Different from empty.
50:02:170Paolo Guiotto: So at the end, you can see it's the same… the same idea that you have here. So you integrate first here on the set DY, which is a set of X, and then for Y, you take the Ys for which this is non-empty.
50:17:110Paolo Guiotto: Okay, it's the same, no?
50:20:930Paolo Guiotto: But now, the interesting thing is that you could switch this order and say that, why not? We start doing the integration in
50:30:550Paolo Guiotto: Y and Z?
50:33:430Paolo Guiotto: Okay, so the pair… so the two variables are X and the pair YZ. So I start integrating in X, and I finish integrating YZ, or conversely, I start integrating in YZ, so this will be a double integral.
50:53:880Paolo Guiotto: the function F, X, Y, Z, and I finish integrating what remains, which is the variable X.
51:02:970Paolo Guiotto: You see, since the variables are 3,
51:05:580Paolo Guiotto: If I have divided in two steps,
51:09:190Paolo Guiotto: Either I choose the first step is one variable, and the second step is two variables, or the first step is two variables, and the last step is one variable.
51:21:550Paolo Guiotto: Because you cannot divide one variable in half. You give half of y to X and half of y of Z. This is nonsense, no?
51:30:00Paolo Guiotto: And here, what is the domain? This is the domain of YZ, so it will be the domain of points YZ, such that this point, XYZ, belongs to D. So, we will call this the X section.
51:48:820Paolo Guiotto: Hmm?
51:50:230Paolo Guiotto: And this is a set of Cartesian points in a plane, you know, because it's a set of YZ. So this is the X section, and about integration in X, it will be again on the X for which
52:06:870Paolo Guiotto: This thing is non-empty.
52:10:550Paolo Guiotto: Okay?
52:13:260Paolo Guiotto: Well, this is the choice where I start integrating X, and I finish in YZ, or vice versa. I start YZ and finish in X.
52:22:410Paolo Guiotto: But why not? I could start in Y, and finishing in XZ, and I have another two formulas. So I won't write in details, but you have a curated integration.
52:35:200Paolo Guiotto: XYZ, if you decide to start integrating in Y, because for certain reasons, that's better for you, you will finish in what remains, which are the other two variables, D, X to Z.
52:48:570Paolo Guiotto: Or, again, you could decide to start integrating in X and Z, this will be a double integral, and you finish with the remaining variable.
53:02:220Paolo Guiotto: And finally, you could also decide to start with Z, so the first integration is same function, XYZ,
53:11:740Paolo Guiotto: in the variable Z, and you finish in what remains. The other two are X and Y.
53:18:490Paolo Guiotto: Or conversely, you could decide to start in X and Y, and finish.
53:28:230Paolo Guiotto: In the set.
53:30:360Paolo Guiotto: So that's why, at the end, you see, you get 6 different formulas. You don't have to remind the 6 different formulas. You have to understand the macros.
53:39:940Paolo Guiotto: You will decide which is the first integration, that in this case can be either a one-variable integration or a double integration, like this, and then automatically you have the second integration decided.
53:56:520Paolo Guiotto: Of course, if you decide to start from this, this is a double integral, and it's not forbidden to apply the reduction formula to this one, to split again this into one variable integrals. So at the end, you will do probably three one-variable integrations.
54:13:920Paolo Guiotto: But the situation can be more articulated, because as we will see, when we introduce the change of variable, there could be,
54:23:290Paolo Guiotto: could be the case where we compute, we… instead of computing this double integral, we change variables, and we simplify this. So, we have to get ready for this kind of…
54:35:150Paolo Guiotto: situations. So, let's illustrate an example.
54:40:620Paolo Guiotto: of the Eastern.
54:43:780Paolo Guiotto: So, here you have, It's a size 572, which is,
54:49:300Paolo Guiotto: So it's just a longer version of the previous problems.
54:53:940Paolo Guiotto: So, exercise… Well, wait, wait, wait, wait.
55:01:320Paolo Guiotto: We say that this is the reduction formula that says if the function is integral, then the integral is this one.
55:09:200Paolo Guiotto: How do you check that the function is integral? Well, if function is continuous and domain is compact, this is granted. But what if this is not the case? There is the check of the iterated integrals, exactly as for the, for the case of double integrals.
55:28:660Paolo Guiotto: So, D, let's say that the reduction formula requires…
55:40:190Paolo Guiotto: So, no.
55:43:370Paolo Guiotto: that… F is integral.
55:49:480Paolo Guiotto: on… D.
55:54:600Paolo Guiotto: this… is true.
56:00:980Paolo Guiotto: For example, Eve.
56:06:510Paolo Guiotto: F.
56:07:570Paolo Guiotto: is continuous. On D, And D is compact.
56:15:790Paolo Guiotto: So we don't need to verify anything, because knowing that the function is continuous, and the domain is compact is sufficient to know that the integral exists.
56:25:540Paolo Guiotto: If the domain is not compact, which is the main problem, If, D… is not.
56:35:120Paolo Guiotto: compactor.
56:39:730Paolo Guiotto: to check.
56:42:580Paolo Guiotto: integrability.
56:49:610Paolo Guiotto: We… But… the… Turner, Ethereum.
57:00:870Paolo Guiotto: That, does the same, things that we…
57:05:810Paolo Guiotto: we discussed above. So, if you have a function continuous on the N, One.
57:12:920Paolo Guiotto: All.
57:14:830Paolo Guiotto: the, I would say, 6… iterated.
57:22:410Paolo Guiotto: Integrals.
57:27:470Paolo Guiotto: for the absolute value of F, is finding it.
57:35:150Paolo Guiotto: So, for example, just to write one, for example…
57:41:490Paolo Guiotto: the first one we wrote, you take the absolute value of F, XYZ, you integrate first in X, and then you finish in YZ.
57:52:380Paolo Guiotto: So this will be on the YZ section, and this will be on the YZ, for which that section is non-empty.
58:02:730Paolo Guiotto: So, for example, this one is finite.
58:06:930Paolo Guiotto: Then… the function f is integral on domain B.
58:14:530Paolo Guiotto: Okay?
58:16:770Paolo Guiotto: Again, if the function has constant sine, this check is the iterated integral of F, and it is automatically also to define it, the value of the integral.
58:29:380Paolo Guiotto: So, in particular.
58:34:600Paolo Guiotto: if… F is positive.
58:37:890Paolo Guiotto: or F negative. It's the same, because you just changed the sign, yeah?
58:44:660Paolo Guiotto: this… Check.
58:51:370Paolo Guiotto: yields.
58:55:130Paolo Guiotto: devalue.
58:57:580Paolo Guiotto: Also.
58:59:120Paolo Guiotto: It's the value.
59:02:100Paolo Guiotto: of the integral on the oven.
59:05:660Paolo Guiotto: Okay?
59:06:800Paolo Guiotto: So let's do just one example to fix ideas.
59:11:370Paolo Guiotto: It is, from 5-7.
59:15:10Paolo Guiotto: To… the number 1… We have to compute this integral on the domain, which is 1 plus infinity cubed.
59:25:150Paolo Guiotto: So this means each of the coordinates is greater or equal than 1,
59:29:790Paolo Guiotto: Now, see, this is the set of points, X, Y, Z, such that X is greater or equal than 1, and Y is greater or equal than 1, and
59:39:870Paolo Guiotto: Z is greater or equal than 1, so you see the three coordinates are independent, basically, no, because X is not conditioned by Y, and Y is not Z, and Y is not conditioned by X and Z, and so on.
59:53:580Paolo Guiotto: The function is, Y cubed.
59:58:690Paolo Guiotto: Z to power 8, and there is an exponential minus X.
00:06:40Paolo Guiotto: Y square… And, Z Cube.
00:19:460Paolo Guiotto: So as you can see here, we are exactly in the case where the domain is not compact, because it's clearly unbounded. So this is the integration domain.
00:30:890Paolo Guiotto: So we may say that clearly the function f is a continuous function, domain B, D is closed.
00:41:660Paolo Guiotto: And, I'm bounded.
00:45:940Paolo Guiotto: So… In particular, not… Mom pas.
00:51:270Paolo Guiotto: So I cannot say that the function is integral because it is continuous.
00:58:40Paolo Guiotto: However, even if the function itself
01:02:330Paolo Guiotto: It wouldn't be necessarily positive, no, because Z to power 8 is positive, the exponential is positive, but Y cubed is not positive.
01:12:810Paolo Guiotto: However, on this domain, being Y greater than or equal than 1, that quantity is positive. So I can say that, I notice…
01:23:350Paolo Guiotto: that F is positive, so constant sine, on the domain B.
01:29:420Paolo Guiotto: So, to check integrability, I will check the iterated integral for the,
01:36:430Paolo Guiotto: absolute value that coincides with the integral of F, in fact. So in practice, if this is finite, I will also automatically add the value of the integral at the end.
01:48:360Paolo Guiotto: So, now the point is that, let's… Check it.
01:57:260Paolo Guiotto: that, tonnelli… DRM… applies.
02:08:800Paolo Guiotto: So, we have to decide now how to set what is the right choice.
02:14:300Paolo Guiotto: And, so, which of these six versions of the iterated integral for the absolute value of F have we to choose? Now, the choice should be made in function of
02:30:680Paolo Guiotto: The shape of the domain, how much is complicated to describe the domain, so… the sections.
02:37:300Paolo Guiotto: And how much complicated it is to describe the function. Now, for the sections, it's very easy, because the conditions are independent, so I could decide to start in X, or in Y, or in Z, or in a pair X, Y, it doesn't matter, it's very easy. What makes the difference, maybe, is the function.
02:57:440Paolo Guiotto: So now, try to look at this function as a function of, for example, one single variable. Which integration would be
03:04:940Paolo Guiotto: Would look easier, apparently.
03:07:900Paolo Guiotto: in the X, because that's, like, DY cubed Z power 8 is a constant, so who cares? Exponential of constant times X, that's easy, so…
03:18:310Paolo Guiotto: we should start with integrating in X and leave Y and Z for the future. So, we do this, so we decide to start with the absolute value of F in the X, and then we will do the other integration in Y and Z.
03:32:970Paolo Guiotto: This, by the way, coincides with F, because F is positive. So, now let's start to set properly this integration.
03:41:310Paolo Guiotto: Now, what is the range for XYZ? Well, it is written here. X must be greater than 1,
03:47:650Paolo Guiotto: But when Y and Z are both greater than 1? So here I will put 1 to plus infinity, and here I do not put 1 to plus infinity, because that one is a double integral, so the second one is I keep the domain y greater than or equal 1, z greater or equal to 1 for the moment.
04:04:880Paolo Guiotto: Then, of course, I will apply the reduction formula again to that one, but let's see.
04:10:830Paolo Guiotto: Now, this is the integral for Y greater or equal than 1, z greater or equal than 1, integral from 1 to plus infinity. Now, let's put the function, which is y cubed, Z power 8, exponential minus X,
04:25:380Paolo Guiotto: Y squared Z cube, so we said the DX here, DYDZ.
04:31:280Paolo Guiotto: later.
04:32:940Paolo Guiotto: Now, we carry outside this part that is independent of X.
04:38:350Paolo Guiotto: So it's a constant for that integration. So we get integral for Y greater or equal than 1, z greater or equal than 1 of Y cubed Z to power 8, integral from 1 to plus infinity, E minus XY square Z cubed VX.
04:57:180Paolo Guiotto: than DYDZ.
04:59:270Paolo Guiotto: Okay, let's do now the innermost integral. That's easy because this is more or less the derivative with respect to X of the same exponential.
05:07:770Paolo Guiotto: But we have, of course, to adjust because there is some constant. If we do this derivative, we get E minus XY squared Z cubed times
05:18:300Paolo Guiotto: Minus?
05:21:670Paolo Guiotto: Why Square?
05:23:770Paolo Guiotto: Z cubed, okay? So I do not have this Y square, Z cube, and the minus, but I can borrow, for example, from this, I leave Y squared here, I also leave a Z cube here.
05:39:150Paolo Guiotto: I don't have a minus, I can put a minus in and out, and now this becomes minus integral for Y greater or equal than 1, z greater or equal than 1,
05:49:300Paolo Guiotto: what remains here is YZ to power 5. Then we have the evaluation of E minus XY squared Z cubed.
05:59:410Paolo Guiotto: between X equals 1 and X equal plus infinity, right?
06:05:40Paolo Guiotto: DYDZ.
06:06:880Paolo Guiotto: Once we have done the evaluation, we don't see any more X.
06:10:790Paolo Guiotto: Okay? So, when X is plus infinity, since,
06:15:580Paolo Guiotto: our Z and Y are greater than 1, so Y squared would be positive, Z cubed, not necessarily, but Z is greater than 1, so definitely this is a positive quantity. With the minus, it means that when X goes to plus infinity, that exponent goes to
06:32:790Paolo Guiotto: exponential to minus infinity, no? So this gives value 0. Minus, when X is 1, we get e to minus Y square Z cubed.
06:43:660Paolo Guiotto: So at the end, with the minus outside, we have an integral y greater or equal than 1, z greater or equal than 1, YZ to power 5, exponential minus Y square Z… sorry, there is a cube here. Z cubed.
06:59:400Paolo Guiotto: DYDZ. So now X has disappeared, and this is a double integral.
07:05:940Paolo Guiotto: And now you repeat, you know? You say, okay,
07:11:140Paolo Guiotto: to establish if this is, in principle, I should do this algorithm. I now have to check if this is positive, right? It's fine. Now, the function here is positive.
07:21:730Paolo Guiotto: So, it coincides with the modulus of itself on that domain, and therefore, I can apply the tonality theorem to check if this is fine. It's saying that I will check an iterated integral of YZ power 5e minus y squared
07:40:890Paolo Guiotto: Z cubed. Now we have to decide, should we integrate first in Y or in Z?
07:48:700Paolo Guiotto: In Y, because Y is more or less the derivative of Y squared, so it is reasonable to start with Y and to finish with Z. For the application, Y will be from 1 to plus infinity, Z from 1 to plus infinity.
08:04:690Paolo Guiotto: Okay, so now this goes outside or better. We check that derivative with respect to YE minus y squared Z cubed is equal E minus Y squared Z cubed. We have to close quickly, so let's see what comes. It is minus 2YZ cubed.
08:22:740Paolo Guiotto: So I need the Z, the cube, here, I will have Z squared here, a minus and a minus.
08:29:680Paolo Guiotto: So this becomes integral minus 1 to plus infinity, z squared, the evaluation of D minus y square z cubed between y equals 1, y equals plus infinity.
08:44:960Paolo Guiotto: And then integrated inside.
08:47:200Paolo Guiotto: This quantity comes, again, for Y going to plus infinity, you get 0 easily, and then it is minus Z cubed. So at the end, we have to integrate from 1 to plus infinity Z squared E minus Z cubed.
09:03:870Paolo Guiotto: Which is, more or less a derivative.
09:07:189Paolo Guiotto: Because if you do the derivative respect to Z of E minus z cubed, you get E minus z cubed times the derivative of minus Z cubed, which is minus 3Z squared. So if you put the minus 3 and we divide by minus 3,
09:22:290Paolo Guiotto: We got minus 1 third.
09:27:160Paolo Guiotto: the evaluation of E minus z cubed, again, between 1 and plus infinity. At plus infinity, you get 0. At 0… at 1, you get e to minus 1. So at the end, we have,
09:41:520Paolo Guiotto: 1 over 3E.
09:44:590Paolo Guiotto: This says that this is finite, therefore going back to this point, this one, this iterated integral at the top of the board is finite. Therefore, we applied the tonality theorem to this double integral, this one is finite, but this one
10:00:870Paolo Guiotto: was the application of the Torelli Theorem to this one. So this one is finite.
10:06:920Paolo Guiotto: This says that f is integral, and the value of the integral, because f is positive, is that value there.
10:14:410Paolo Guiotto: Okay?
10:16:880Paolo Guiotto: Okay, so, do the remaining, exercises of, BC572.
10:30:860Paolo Guiotto: There are two integrals.
10:33:40Paolo Guiotto: for your joy.
10:35:700Paolo Guiotto: Okay.
10:37:00Paolo Guiotto: See you on Friday.