I really like this integral

[@farfa2937]

When I have a child I'll tell them the natural numbers go: 1, one over pi squared times the integral from 0 to infinity of the natural log squared of x over squared root x times one minus x all squared, 3, 4, and so on.

[@jadonjones4590]

What about 0?

[@farfa2937]

@@jadonjones4590 It's not natural

[@bjornfeuerbacher5514]

@@farfa2937 Depends. In some parts of maths, it's considered to be a natural number, in others, it isn't.

[@farfa2937]

@@bjornfeuerbacher5514 Yeah there're exceptions to every rule, but I didn't feel it was necessary since it was just a joke...

[@Walczyk]

dont

[@goodplacetostop2973]

0:40 Lemma 5:03 Main result 18:15 Good Place To Stop

[@jkid1134]

14:03 Fake homework

[@CTJ2619]

Thanks!

[@richardstone5096]

Really nice integral, it's always cool when you can use sums to help solve integrals.

[@عمرانآلعمران-و7خ]

In the last step we could not use the geometric series, instead we can visualize the integral as a second derivative of beta function which involves some digamma function values.

[@Vladimir_Pavlov]

1. Replacement in the original integral t=lnx. We get (the given multiplier 1/π^2 will be taken into account at the end): ∫ (from -∞ to ∞) t^2*e^t*dt/(1-e^t)^2. (1) We divide this integral into two (from -∞ to 0) and (from 0 to ∞). In the first one, we do the replacement t→-t. After simple transformations (1) is reduced to the form ∫ (from 0 to ∞) t^2*[e^(-t/2) +e^(-3t/2)]*dt/[1-e^(-t)]^2 = (*) 1/[1-e^(-t)]^2 = - e^t * d/dt [1/(1- e^(-t))] =- e^t*d/dt[1+∑(n from 1 to ∞) e^(-nt)]= =∑(n from 1 to ∞) n*e^[-(n-1)t]. (*) = ∑(n from 1 to ∞) n*∫ (0 to ∞) t^2*[e^(-(n-1/2)*t)+e^(-(n+1/2)*t)]. (2) 2. Using Feynman's trick, we obtain that if J(a)=∫ (from 0 to ∞)e^(-at)dt =1/a, a>0, then ∫ (from 0 to ∞)t^2*e^(-at)dt = d2J(a)/d2a = 2/a^3. (3) From (2) taking into account (3) (for a= n-1/2 and a=n+1/2) we get 2*∑(n from 1 to ∞) n*[1/(n-1/2)^3+1/(n+1/2)^3]= = 8*∑(n from 1 to ∞) {[(2n-1)+1]/(2n-1)^3+[(2n+1)-1]/(2n+1)^3} = 8*∑(n from 1 to ∞) {[1/(2n-1)^2+1/(2n+1)^2 +1} =16*∑(n from 1 to ∞) 1/(2n-1)^2 =16*(π^2/8)=2*π^2. Answer: 2*π^2/π^2=2.

[@orionspur]

16:20 "That's because all numbers are either even or odd." Homework lemma. :)

[@manstuckinabox3679]

Beautiful Timing just as I'm taking real analysis, plus I've never heard of an even irrational, emphasis on never heard.

[@manstuckinabox3679]

@@orionspur exactly, the question isn't very well-defined, the lemma, imo, might go like this: for all { x ; x belongs to Z), (X is even xor X is odd). can be done with basic first-order, or even just propositional.

[@manstuckinabox3679]

@@orionspur I think zero is fairly even though, it can’t be written in the form, 2k+1... but then again that depends on the set of axioms that define odd-ness. Edit: It really depends on the field we’re working in.

[@guilhermepimenta_prodabel]

The most beautiful integral I've ever seen!

[@manucitomx]

I just love it when after a long trek through math you end up with 2. Thank you, professor.

[@6612770]

Hahaha. Reminds me of times in high school when I would write out a full page of work on solving an integral, only to end up with the value of Zero. I remember always remarking to myself : "Gosh, I did all of that work for Nothing..." 🙂

[@Noam_.Menashe]

I did the same thing except I substituted sqrt(x) before the integral because I didn't want to deal with fractions in the denominator (It doesn't change much at all).

[@emersonschmidt7882]

As others said, it's always nice when geometrical series and stuff like that shows up in integrals.

[@romajimamulo]

3:45 it might be my minor color blindness, but the yellow and orange are hard to tell apart EDIT: Later on it was easy to tell, must have been a lighting issue in that shot

[@scp3178]

Agree. Michael's yellow looks like a lightgreen (yellowish) to me. But that could be a color difference between the video cam and the final edited video.

[@kmlhll2656]

A very nice demonstration.

[@minwithoutintroduction]

رائع كالعادة. واصل يا أستاذ.احتراماتي.

[@The1RandomFool]

In my attempt before watching the video, I used contour integration twice. The first time was a half circle in the upper half plane (and avoiding the origin) to put this integral in terms of a different integral, but quite similar. The second was a proper evaluation of that with a keyhole contour. I think the square root in the denominator made this easier to evaluate.

[@timotejbernat462]

13:18 forgot to write the integral sign for the sum term, which made it really hard to follow that step and the next, the rest is very well put together

[@aidarosullivan5269]

Should have put 1/(2 п^2) before the integral ig.

[@bl8r1ner]

Makes me think of Rube Goldberg machine. Very nice solution.

[@KundanKumar-oh8qb]

Today this question appeared in my test

[@dorong3232]

Heres a nice question- What irrational numbers have a continuous simple fraction expansion {a0,a1,a2,a3,...} So that their decimal representation b0+b1/10 + b2/100+.... Is such that b0=a0, b1=a1, b2=a2,.... What about any other representation? (Binary, hex,...)

[@hypergraphic]

Nice video!

[@jyotisingh4109]

Great video sir

[@Patapom3]

Love it!

[@wolfmanjacksaid]

Gradshteyn and Ryzhik anyone? I wonder if the Integral Suggester knows this book. Bought it for my grad level modeling class and barely cracked it since. Some of the integrals had no closed-form solution as indefinite integrals, only as definite. Really interesting stuff!

[@bezobrazie7607]

Wolframalpha said that answer is 1.57.....

[@manstuckinabox3679]

Hiya Doc! It always fascinates me, but how do you know which lemma, or tool each integral requires? I mean, not you specifically, but how does one know?

[@spearstip9812]

After playing around and reaching the point where for example you have int{ x^a ln(x) dx} you calculate it then and once you find the answer you rewrite the problem in a more presentable fashion. It's rare to know a priori what you need to solve a problem like this :b

[@leekon_]

5:01 no1 can tell me that that isnt a good transition

[@ubncgexam]

It's a very "easy" way to calculate 1+1. 😂😂😂

[@ericthegreat7805]

So technically the goal integral is 2pi^2? Can this be related to Gaussian functions?

[@gennarobullo89]

I dont get why if x=1/y, then later y turns back to an x

[@rocky171986]

Dummy variable

[@jkid1134]

This maybe provides a bit of clarity: go to like 8:17 and imagine substituting x=t into the left integral and y=t into the right integral. Now you can add them together, and even just work the whole thing in t if you want to. The variable of integration can be changed to anything, since it never appears in the final result. It is like the k in summations you often see, simply used to index some set, a "dummy variable"

[@dodokgp]

Spoiler alert : the answer is the only prime number that can be written as the sum of two cubes ;)

[@hypergraphic]

I guess that generalizes all the way up right? 1^n + 1^n = 2?

[@UltraMaXAtAXX]

Thumbnail lacks pi-squared.

[@goodplacetostop2973]

Then the answer is 2π 😂

[@bjornfeuerbacher5514]

@@goodplacetostop2973 Or rather 2π². ;)

[@Keithfert490]

@@bjornfeuerbacher5514 no, there's a pi in the thumbnail; it's just not squared

[@zahari20]

Clumsy! To prove the lemma just differentiate inegral of x^a twice.

[@bertrandviollet8293]

You're great

[@pow3rofevil]

Para cuando un problema de Ramanujan ?

[@mathunt1130]

To prove the lemma, note that x^a(log(x)))^2=d^2/da^2(x^a).

[@michaelguenther7105]

The lemma is easy by taking d^2/da^2 of the integral from 0 to 1 of x^a.

[@Mathskylive]

phương pháp tích phân từng phần chọn u, dv. Cảm ơn.

[@nasim09021975]

I want a T-shirt that says "... and that's a good place to stop" :D

[@BridgeBum]

Check his shop, he sells them helpfully translated into many languages.

[@skylardeslypere9909]

For the first integral (the lemma), I first thought of Leibniz' Rule for integration, as d/da (x^a) = x^a ln(x), but that wouldn't work, as it would just increase the power of ln(x). However, then I integrated the expression in the lemma w.r.t. a two times, and I got the integral of x^a from 0 to 1, equaling 1/(a+1). Differentiating w.r.t. a two times again gives the correct result. My question is, how valid is this? I think we could argue that integration is just a fancy limit, meaning we could actually do this. But I'd love to hear a second opinion.

[@khaledjebari1874]

I ve sended u many requests of integrals Please can u do the one of the result zeta(3) Thanks in advance

[@jesusandrade1378]

The Maple Calculator gives the definite integral as 2π^2, so twice the value you got.

[@leif1075]

For heavens sake if YOU SONT KNOW the lemma or thst fsct can't you still solve the original with trying integration by parts..and or maube some u substitution like u is x minus 1 or u is ln x ..surely thst will work

[@zzz942]

Or just use complex analysis...

[@pauldraghici8623]

This is most likely none of my business, but is everything ok? You look very tired in the video.