A Breathtaking Journey of Integration 

Wow, that’s an even nicer way of formulating it than on my video, great job! :)

The way he sounds so excited makes this video 10x better.

I appreciate the transparency regarding dead ends and the necessity of perseverence and stamina in taking on mathematical problems such as these.

Genuine application of gamma function.

You know you've explained it well when a maths noob can understand most of what you said, great video!!

i like the way how pi comes out of nowhere

Thank you for solving this, I was trying to solve a similar integral from past 7 years. I used ur method and I finally completed the damn integral after 7 years.

Omg, how does he do to see that kind of relations? Certainly impresive

What coincidence, blackpenredpen just proved this by showing the integral is gamma function of 2 times zeta function of 2. 😃

I especially love these integration puzzles. Keep up the amazing work, and thank you for your videos :)

An absolutely superb way to get to the solution with all the necessary inter step justification explained properly.

This is legitimately the greatest video I've ever seen

W H E R E I S T H I S P I C O M I N G F R O M ?

Great video for a greater integral. Thanks for your work.

Hello. :) The first thing I thought when I saw the integral was Bernoulli. There is another beautiful way to solve it in my opinion. If one knows that the integrand is the generating function of the Bernoulli numbers the solution follows immediately from some basic integral properties. :) Details: Integral 0 to inf x^(2n-1) / ( e^(ax) - 1) dx = beta(n) / 4 (2 pi /a) ^ (2n) Where beta(n) = (-1)^(n+1) * bernoulli( 2n ) Using n=1, a=1 => beta(1) = bernoulli(2) = 1/6 gives pi^2 over 6. :)

Beautiful!

Gamma FUNction! :D

I was grinning from ear to ear by the end of this video. Thanks!

Thank you for the video keep up the good work brotha man

Man! This is love! ❤❤❤❤❤❤

Amazing! I always have trouble with remembering all of these infinite series and their answers, but either way, I enjoyed this video to the fullest!

What a beautiful of explaining something. Thank you :)

Nice. And we are done!

Most impressive part of this is how quickly and nearly he is drawing with a click and drag mouse function .

So beautiful!!!!! Just love it

That ws amazing,Very clear and understandable,thanq

ِِAmazing solution ... Thanks

Impressive, man. You rock.

Wheeler: duel monster player and mathematician.

Very appropriate title! :) That was amazing

This is so awesome...it's going on my favorites list

Nice job x2

Nice job

I klicked the like button and it showed 2800 likes!! I love how many ways you are trying to wvaluate that integral 😘

You can also obtain the π^2/6 by taking a Fourier series of x, in the interval -π

Simple and clear presentation of the topics. wow !!

Well done. I love Euler 😊

thank you very much your are amazing

Awesome vid. One note: to avoid integration by parts, you can evaluate integral from 0 to ∞ of (xe^(-xn) dx) by differentiation under the integral as follows: I(n) = int 0 to ∞ of -e^(-xn) dx I'(n) = int 0 to ∞ of xe^(-xn) dx after differentiating with respect to n, which is what we want However, we can evaluate I(n) easily I(n) = int 0 to ∞ of -e^(-xn) dx = e^(-∞n)/n-e^(-0n)/n = -1/n I'(n) = 1/n^2 So we deduce that int 0 to ∞ of xe^(-xn) dx = 1/n^2, and basically all we had to do was integrate -e^(-xn)

Absolutely astonishing.

I love this. 😃

It is a very clean process, totally clear for the viewer. Concerning the final steps, when deciding over integration by parts or Gamma function, one could also consider the Laplace transform of f(x) = x, which is directly readable and there are no needed modifications to compute it, in contrast to the option of using Gamma function. Laplace transform is (I think) known by everyone having seen a first course on ODEs, so it should also be part of the "toolbox" available when solving integrals like this one.

This is absolutely incredible

Ingenious approach. Brilliantly done

Brilliant!

You must know that by this video , you make other nearly to the solution of the eternal prime number problem. Thank you .

holy mook a dang. Brilliant.

Bravo!!!

Beautiful

Great sir.......

Great job!

nice thanks sir..

Very smart.

Pay attention when you justify the change of the order of the summation and of the integration by referring to the uniform convergence of the series, it is way better to make use of the monotone convergence theorem that is studied in Measure Theory and Lebesgue Theory, as this example shows: pick the sequence of functions given by chi{[0,n]}/n, where chi{•} is the characteristic function of the set written between the brackets. It is really simple to see that this sequence converges uniformly to the constant zero function, nevertheless the integral extended over R+ or R is constantly 1, so you can't apply the uniform convergence theorem result. The reason is that there is another assumption that is made in that theorem, i.e. the fact that the measure of the set involved must actually have finite measure, otherwise situation like the previous example can occur. Since in our case you're integrating over R+, which has infinite measure, I would avoid to use the result about uniform convergent series of functions and I would rather notice that the series has only nonnegative functions, which means you have a monotonic increasing series of functions, so that monotone convergence theorem can be applied eventually to justify the inversion of the order of the operations of summation and integration.

Great stuff

I have used Lebesgue theory about majorant for the change suma And integral. x[(exp(-(N+1)x-1)/(exp(-x)-1)]≤x[ exp(-x)+1-1] ( for N=2), where N Is index S_N, (non-negative Borel-measurable functions guarantee the existence of the Lebesgues integrals).

Wasn’t too familiar with the Gamma function but the partial integral did just fine too to get (1/n)^2.

I never passed pre-calculus in high school haha but I love watching your videos!

excellent explanation

at 8:00 uniform convergence is a good argument for switching sum and integral, but actually it always holds if the terms are non-negative. (Beppo Levis theorem or Tonellis theorem) :D

Extraordinary

Haha, I really like your videos bro.

Best Solution 😊😊🇮🇳💓

i have a small trick for this integral x * exp(-xn) . you can write this as minus d/dn exp(-xn) . then put - d/dn out of integral . Then just calculate integral of exp(-xn) from 0 to infinity which is easy. its just 1/n . then let - d/dn act on 1/n and you have 1/n^2

Actually you can't refer to uniform convergence at all, because the sum does NOT converge uniformly. Here's why: for the sake of contradiction assume that the sum whose general term is exp(-nx) converges uniformly on R+. That would mean, by Cauchy criterion, that for a suitable N (chosen sufficiently large) we should have that the sup for x belonging to R+ of the modulus of the sum of the N-th,N+1-th,...N+p-th term would be smaller than a fixed quantity, say 1/2, FOR EVERY NONNEGATIVE INTEGER p>=0. This clearly implies that the general term of the series must actually converge to zero UNIFORMLY, but that's not the case, because FOR EVERY N, we have lim exp(-nx)=1 for x->0+. So the series does NOT converge uniformly, and the theorem couldn't be applied a priori anyway

The starting integral is the function f(2) where f is the ζ(s)Γ(s) which is equal to S[0->Infinity] (x^(s-1))/(e^x - 1)

AWESOME

Amazing

phenomal video

Nice integral, but always great complications !! we can do it easier : x/(exp(x)-1)=x*exp(-x)/(1-exp(-x)) then let u=exp(-x) the integral will be integ from 0 to 1 the fraction -ln(u)/u-1 known 1/1-u=1+u+u^2+u^3+...+u^n the integral will be minus somme from n=0 to infinity integral from 0 to 1 u^n*lnu du integration by parts and it's done somme from n=0 to infinity 1/(n+1)^2 equal to pi^2/6

Very intuitive

You should do some problems from the Calculus/Analysis section of the Berkeley Math Tournament

Nice explanation

Use Monotone Convergence Theorem rather than Uniform Convergence as it is much easier to show Monotone Convergence than Uniform Convergence which follows as the sum is a sum of positive quantities.

7:54 I'm not sure about this fact... Uniform convergence only allows to exchange sum/integral on a segment like [a,b], not an infinite set. In such case, you would need dominated convergence.

Thanks.

WaaaaaW❤❤❤❤❤

Thank you

The part when you use Taylor series of a/1-r is false because for the zero of the integral, exp(-x) is equal to one, which not permitted by the very condition of the radius of convergence

You made it easy.

The double integral from minus inf to positive inf of sinx/x siny/y sin(x+y)/(x+y) dxdy also turns out to be exactly this

Nice one :) which also btw leads to the conclusion that integ_from_0_to_inf(x*csch(x)) is (pi^2)/4, a rather neat answer :D

素晴らしい！ 非常に難しい問題ですが，とても勉強になりました．1/n^2からa/1-rを連想するとは．私自身，最初はe^xのTaylor展開に持ち込むのかと思っていました．

awesome

whoa, letssolvemathproblems fails along the way?!?! this is unprecedented! instant like and you put me in awe, lol. btw, is there something else i can call u by instead of letssolvemathproblems, perhaps your first name?

First impressions: I saw the connection to Zeta(2) immediately. :) Then I thought that the denominator could be rewritten as a geometric series. Going to watch the rest of the video now. Edit: I would have gone the integration by parts route, with the same result. The gamma function is much nicer tho!

5:21 summation from 0 to infinity 😁😁

I know I am too late( but I hope u see this LSMP) but actually it is a direct problem if one is aware of the relation between zeta and gamma function as: integral of (x^(s-1))/((e^ x)-1) from 0 to inf = zeta(s) * gamma(s) which in this case s=2 the answer is zeta(2)*gamma(2) =pi^2/6 *1!=pi^2/6

So, does this give us some kind of intuition into the mechanics of ζ(s)?

Omg stumpled across this on my own. By using the integral of x^-k between 0 and inf then messing around using leibniz rule and summing stuff. U sub to get the bounds to converge. Then transform the bounds again.

Excellent, I really like how you proved it

or let e^(-x)=y then the integral will be int(y=0 to 1) ln(y)/(1-y), use the series expansion for the denominator and you will get sum(n=1 to inf) 1/n^2 which is zeta(2) which pi^2/6

I literally clicked on the video hoping to see if the Basel Series pops in at some place

I see that I found what I'll learn next

Here is my approach Take e^x common in denominator and then substitute e^-x =t We will get integration of the form of Ln(t)/1-t now substitute 1-t=p to get integration of the form of -Ln(1-p)/p expand Ln(1-p) using mclaurin expansion and integrate the algebric function to get 1+1/2^2+1/3^2... =π^2/6

7:57 You don’t even need uniform convergence in this case. Everything is non-negative so it follows by monotone convergence theorem.

Nice approach but we could also show that this integral is ζ(2)×Γ(2)

pi^2/6 again? I can't escape that value. I keep running into it over and over. Something about the zeta function, but it shows up in the trigamma function and the 2nd(?) polylog when evaluated at 1 and it pops up all the time in integrals

Nice Video. But, there is some minor thing that bothed me. When we replaced the integrand with the infinite geometric sum, we assumed r to be smaller than one. Strictly speaking, this is only true when x is bigger than zero. The geometric series evaluates to 0 for x = 0 (a = 0,r=1), while the integrand is equal to 1 at x = 0. Can we neglect this fact, because changing the integrand at one point does not alter the value of the integral?

Man you didn't lie when you write a journey of integration :)⁩