WONDERFUL 𝜋ntegral! A Putnam Extravaganza [ Basel Problem Integral Representation ] 

The cheap physicist way of doing it: recognize it as the integral representation of the Bose-Einstein function g_s(z) with s = 2 and z = 1 (times an irrelevant Gamma(2), which is one). As is clear from the series definition of the functions, g_s(1) = Zeta(s). Hence the integral is Zeta(2)!

Exam day after tomorrow. Hm.... nah, 13 minutes of Papa Flammy can't hurt....

*Papa Flammy:* This integral is breathtaking! *Keanu Reeves:* YOU are breathtaking!!

I'm guessing the answer is π^2/6 (I haven't watched it yet), blame Dr.Peyam for that :D (he made a video of an integral very similar to this one but the x term is raised to the power of s, where s is a complex number, it's great

This is absolutely wonderful. I can already imagine sort of what is about to come and I can’t wait!!

Gauss or GauB xd

I feel so grateful that as an old man, i get to see this before I die. I can hear the symphony that you conduct. Thank you!

Do we have time for *_BONUS INTEGRAL_*?

Hey, I saw this same integral on Let's Solve Math problems. It was done almost the same exact way. Papa Flammy is doing so many integral videos I'm starting to see repeats from other integral solving RU-vidrs.

You remind me of our friend who said "...and wheeler!!!".

I’m quite new to mathematics of this level - this explanation was brilliant. I’m glad you deviate into simpler(ish) problems.

*Papaaaaaa* It was great, I want to watch it again and again! Great Thank you so much *My Dear Papa Flammy Mathy*

That was beautiful

2:21, actually, this can be converted to a geometric series starts at k=1(which is, the sum of z^k, from k=1 to infinity). Since the formula for that is z / (1-z). Then, the summation can be moved outside of the integral. This gives us the integral of x*e^(-k*x) from 0 to infinity. As you can probably notice, this is the Laplace transform of x. So, the integral can be replaced with 1 / k^2. Now, this gives us the sum of 1 / k^2 from k=1 to infinity. Which it just pi^2/6.

Gotta love the merch Papa Flammy is putting out. That’s not a meme, I genuinely love your merchandise

"We can actually Fubini this shit." XD

Did a change of variables after the geometric series trick, ended up with the Besel sum times an integral that went to 1 nicely, despite the fact that it involved zero times infinity.

I love your integeral science!

Hey, are you Jens Fehlau? I saw u on quora's recommendation for no reason. LMAO.

back in 17s if you where a master of this integarahl you would be an amazing physicist . this thing is everywhere in stat mech .

Love you papa, love me back

Wow man.....just loved it.....!!!!😘😘😘

very nice inthégral bro et loks layks goods

Boi what are you adding in that big sigma.

Can papa flammy bless me for my engineering entrance exam tmr???

It was just incredible, i hate it when i see a solution that i could figure out myself (but quitted trying too soon).

Beautiful

I have so much homework... so, one video of integrals arent be bad.

WOW Flammy is back! Welcome to anor video with the oiler macaroni consent.

Papa there is a simpler solution by making the substitution e^x=u. After simplifying the integral that we obtain we find that it equals intergal from 0 to 1 of ln u/u-1. This is equal to intergal from 0 to 1 of ln(1-u)/(-u). Expanding ln(1-u) by its Taylor series we easily get Zeta2 as the answer.

Thanks. Nicely Explained.

Go papa flamy. You inspire me.

Isn't the integral from 0 to infinity of (x*e^(-kx)dx) the laplace transform of x? Wouldn't that be an option? (It's 1/(k^2) too soooooo)

Nice video as usual, i was thinking of another factorisation : x/e^x * 1/(1-e^-x) then using taylor series, you get the same result

integral of x/(e^x+1) from 0 to infinity = pi^2/12 coool

PAPA could you do this bad boi ? Integral from 0 to infinity of (sinx)^2/(1+x^2)

Papa Flammy gonna prove the Riemann Hypothesis next video confirmed?

5:32 wait what the hell you can do that??

5:02 Very smart move.

You should also make vids about physics. Named Flammable Physics. Or even chemistry.

madlad

I enjoyed this integration as much as enjoy my favorite movie.

Me, watching this without even finishing limits: I can't understand shit but I like it

The captions at 10:15.

Just about to watch a Great Video

e to the negative teeth power /o/

You really should get into probability. I think you'd really enjoy it.

This can be a quick infinity boi, any integral of this form with x^(s-1) on top is Gamma(s)Zeta(s). xD

More integral equation pleaseeeeeeee!

Papa

I think that I love 😂❤️

That's actually the Bose Integral at n=2.

test: if papa flamy likes this, hes def using a bot

Papa bless

6:26 I'd've used papa feynman, but okay

1st anniversary of this video!

awesooooome

good

This video helped me solve the same integeral but with x^2, gonna start doing more integrals i think, getting kinda rusty during the holidays >.< Also need to learn some of the rules like interchanging summation & integration more uhm.. yuck.. "rigorously"

Integrals ❤️❤️

Where did you buy that watch? It looks great. Maybe you should contact the manufacturer and have them sponsor you 🤣

What a watch ma boi

GOD

It would have been a lot easier if you had set the integral from 5:54 to be equal to the derivative with respect to k of the integral of e^-kx, using Leibniz's rule, and then solve the integral (which is 1/k) and take the derivative and ends up with the Basel Summation.

Who else first saw the pi-loroid and thought that 3 blue 1 brown is here, then realised it isn't true, but still stayed for it?😀

yata desu ne!

What is that watch you have?

I have no idea how you were able to get e^-x as a numerator or as a or create a sum within the integral. you might want to explain that more

Why didn't you just taylor expand e^x, cancel the two 1's then cancel the x's on the top and bottom, and just take the a.derrivative of a power

Consider my breath taken

𝗪0𝗡de𝗥𝗙𝗨l

could you use complex analysis?

I am a simple person I saw RANDOLPH I clicked

Why dont you solve this using complex integration? instead of using x use z^2 (yes squared, else you wil get a zero) and chose a rectangular contour with hight 2*pi*i

Great video. I'm disappointed you don't take the steps to prove 1) the integral actually converge 2) you may express 1/1-exp(-x) as a serie and 3) you may exchange sum and integral signs. I guess it'd be boring and make a video too long, though. Anyway, I gave it a try myself and here's my approach (it's long), I give another more direct approach at the end. I = int (0, inf) x/exp(x)-1 dx f(x) = x/exp(x)-1 fonction f is continuous over ]0, inf[ => can be integrated over it I isn't improper at x=0 because lim 1/f(x) = lim exp(x)-1/x = lim exp(x)-exp(x0)/x-0 = exp(0) = 1 => f is continuous at 0 with f(0)=1 I is improper at x->inf x > 1 => f'(x) < 0 => f is strictly decreasing over [1, +inf[ x > 1 => f(x) > 0 let g(x) = 1/x2 lim (x->inf) f(x)/g(x) = lim x3/exp(x)-1 = 0 => f(x) = o(g(x)) f and g are of same sign (positif) for x > 1, g is riemann-integrable over [1, +inf[ with a convergent integral (Riemann) => a domination criterion is therefore met => I converges Calculation: I = int(0, inf) x.exp(-x)/1-exp(-x) dx we write 1/1-exp(-x) as a serie which is the geometric serie with ratio exp(-x) that converges for any x > 0 (the case x=0 is pesky) 1/1-exp(-x) = sum(k=0, inf) exp(-kx) I = int(0, inf) x.exp(-x).sum(k=0, inf) exp(-kx) dx we put back the term exp(-x) into the serie and we rescale the index I = int(0, inf) x.sum(k=0, inf) exp(-(k+1)x) dx = int(0, inf) x. sum(j=1, inf) exp(-jx) dx we put back the term x into the serie as well (the serie is still convergent as exp(-kx) is always negligible before x I = int(0, inf) sum(j=1, inf) x.exp(-jx) dx We then prove the serie to be uniformly convergent before applying the serie-integral inversion theorem The serie sum(j=1, inf) x exp(-jx) converges uniformly toward f because norme_sup (fn(x) - f(x)) = sup(abs(x.exp(-jx) - x/exp(x)-1) = sup(abs(x.exp(-jx)(exp(x)-1) - x // exp(x)-1) = sup(abs(x(exp(-jx)(exp(x)-1) - 1) // exp(x)-1) = 0 Let's exchange sum and integral signs I = sum(j=1, inf) int(0, inf) x.exp(-jx) dx let u=jx, x = u/j et dx = du/j, bounds don't change I = sum(j=1, inf) int(0, inf) u/j.exp(-u) du/j = sum(j=1, inf) 1/j2 int(0, inf) u.exp(-u) du we see int(0, inf) u.exp(-u) du = G(2) = 1! = 1 (Euler's Gamma function and factorial as you pointed) I = sum(j=1, inf) 1/j2 = z(2) = pi2/6 (Riemann's / Basel Problem as you also pointed) Another way let zeta(x, q) = sum(k=0, inf) (k + q)^-x for q natural integer and x real we prove that zeta(x, q)G(x) = int(0, inf) t^(x-1)exp(-tq)/1-exp(-t) dt immediately, it comes that I = zeta(2)G(2) with q=1 et x=2

I really want one of those infinity boi shirts, but it seems like the merch site can't ship to where I live (California, USA)? Can I still get one somehow?

すごい！ 登録しました。

Wow I'm here early...

i cant understand complex calculus

Can u evaluate thia integral but instead of the x in the numerator can u have x^2, I want to see what a general result would be....

Papa putting those "Putnam" in the title again :v

Papa Flammy you should do some question from the AIME exam!! Would be cool to see how you approach them.

🔥🔰-ʕ•ᴥ•ʔ-🗡💜! ALL COMMENTS = ENDORSEMENTS! I AM A COMMENT!

Halfway, it would be faster if you used the gamma function or Laplace transforms. To learn more visit the Mathematical Facts group on Facebook.

Can't you just use some complex analysis?

where do you learn all this??? do you have books or lectures that you could recommed for me to learn??

Bose Einstein integral for s=2.....!

mmm tasty basel boiiii

Can Papa Flammy bless me for my exam in 3 hours?

do you know how to solve that equation algebratically: (4x+2)^(1/x) = 2. thanks for that ;) great stuff anyway

Infinity Boi 8

Did you check the interval of convergence for the geometric series? ;)

>mfw sugoi desu

Hot. How about other values of zeta function times gamma function?

I got zero as the answer of this question

I'm in India, how can i buy it??

pi creatures!!!

Can you do a video on \zeta(4)

Why is there -1/12 on your tshirt?

Have you ever thought about doing videos about tetration and other 🅱️S like that?