Multivariable Calculus | Changing the order of integration.

[@shohamsen8986]

Small typo. you wrote (x^2-y^2)^2 in the denominator when discussing the other iterated integral.

[@MichaelPennMath]

Thanks! I pinned your comment so everyone else can see the correction (at 7:11)

[@shohamsen8986]

@@MichaelPennMath Thanks. :)

[@michelebrun613]

Minute 11:00. But if a/(x^2+a^2) is positive, the presence of the minus makes the contribution negative (-Arctan(x/a)), so you cannot say that it is greater or equal

[@bobzarnke1706]

To understand exactly why changing the order of integration gives two different results, I set the region of integration to [a,b]x[a,b], instead of [0,1]x[0,1]. Then the first part of the proof above leads to the intermediate result Ia - Ib, where Ia = ∫[a,b] a/(x²+a²) dx and Ib = ∫[a,b] b/(x²+b²) dx, for one order of integration, and Ib - Ia, for the other order. When a > 0, and even for lim(a→0), Ia - Ib = Ib - Ia = 0, so that the integral is well-defined. (Intuitively, the integral should be 0, since f(y,x) = -f(x,y) and the ranges of integration of x and y are equal.) However, when a is exactly 0, Ia is also 0, leaving the result as Ib or -Ib (π/4 or -π/4 when b = 1), depending on the order of integration.

[@regulus2033]

I've noticed an interesting thing. If we replace x->y, y->x in integral (let's call it A) then the numerator changes its sign and the denominator and limits doesn't change at all. So A=-A =>A=0. Of course we could say that if the integral converged.

[@ДенисЛогвинов-з6е]

Thanks. Good video. Good luck your channel

[@rafael9285]

Up!

[@erfanmohagheghian707]

Can you please by any chance do some tough examples on changing the order of triple integral?

[@michaelempeigne3519]

but how to know when the absolute value of the integral converges ?

[@roman_roman_roman]

11:07 - i guess it's a mistake

[@matthewryan4844]

It should say the integral is less than or equal to Integral of 1/2x dx = -lna/2 -> minus infinity i.e. it diverges

[@SirBartolomew]

@@matthewryan4844 it's the limit as a goes to zero, not infinity. So - ln(a)/2 does converge to +infinity

[@sensei9767]

How is -a/(x^2+a^2) always positive?

[@MichaelPennMath]

The a/(x^2+y^2) is always positive (without the minus sign). Then we are subtracting something that is always positive, so if we remove it we have the stated inequality.

[@sensei9767]

@@MichaelPennMath But why would removing the -a/(x^2+a^2) make the integral smaller

[@MichaelPennMath]

ahh, I noticed my mistake! This makes the inequality go the wrong way. This doesn't change the outcome though as this portion of the integral converges.

[@megauser8512]

@@MichaelPennMath That makes sense (cuz infinity - a finite number still = infinity).

[@MrRyanroberson1]

To fix the mistake: x >= a for all x in that integral, therefore replacing x with a in the denominator will make a/(x^2+a^2) larger, thus making the integral smaller, and the term is specifically a/2a^2 = 1/(2a). evaluating that integral from a to 1 we get (1-a)/(2a) = 1/(2a) - 1/2 which blows up to infinity alongside the -ln(a)/2

[@pacolibre5411]

So if you cannot change the order of integration, does that mean that the integral diverges?

[@coc235]

It does

[@EebstertheGreat]

Sort of. If we are using Riemann integrals, then yes, I think that is true. The reason is that the value of the integral will depend on the partition used, so it isn't defined. For Lebesgue integrals with unusual measures on spaces that are not σ-finite, or for nonmeasurable functions, or for other cases, this won't necessarily hold. That is, the multiple integral might converge, but it simply doesn't equal either of the iterated integrals (or equals just one of them). For the "normal" case, Fubini's theorem will only fail if both the positive and negative parts of the function have infinite integrals, so it's a case of ∞ - ∞. It's the same thing that happens with conditionally convergent series: the sum depends on the order of the terms. So if there isn't a natural order, then there is no meaningful way to define the sum.

[@lindsaywaterman2010]

The first two integrals are made much easier if we let x = rcos(theta) and y - rsin(theta) and da y r dr dtheta.

[@cicik57]

so the result pi/4 is not valide?

[@adrianamor8472]

It follows directly from Fubini's theorem that if both iterated integrals are different, then the double integral must diverge. No need to calculate further. I think the explanation is a bit misleading and this makes the viewer think that the if condition of the theorem is reversed when it's not.