WONDERFUL 𝜋ntegral! A Putnam Extravaganza [ Basel Problem Integral Representation ]
HTML-код
- Опубликовано: 14 окт 2024
- Help me create more free content! =)
/ mathable
Merch :v - teespring.com/...
shop.spreadshi...
2nd Channel: / @rockhardwooddaddy
Generalization:
Othe Putnam Int: • γ - A BRILLIANT Putnam...
Basel: • The Basel Problem & it...
Geo Series: • The Geometric Progress...
Gamma Fct: • DERIVING THE GAMMA FUN...
Today we are going to derive a crazy popazy integral representation for pi^2/6, namely the integral from 0 to infinity of x/(e^x-1). We are going to see, that it evaluates to zeta of 2 times the gamma function, it's going to be amazing! Just wait for the generalization, it's going to be gucci as hell! Enjoy :3
My Website: www.papaflammy...
Flammy's subreddit: / flammybois
Twitter: / flammablemaths
Facebook: / flammablemaths
Got some time to spare? Make sure to add captions to my videos! =) www.youtube.com...
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)!
Hahaha omg i literally recognized it from the thumbnail! Shit’s in my blood! 😂
ik this identity from blackpenredpen lol
Exam day after tomorrow.
Hm.... nah, 13 minutes of Papa Flammy can't hurt....
Same here lol
JEE mains in 2 days, why am i watching putnam exam solns? The result is absolutely beautiful tho, no doubt.
@@pranayvenkatesh8815 Same here boi. Which shift?
lolol #flammyflammily
@@pranayvenkatesh8815 me too i have my exams tomorrow 😂
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
*Papa Flammy:* This integral is breathtaking!
*Keanu Reeves:* YOU are breathtaking!!
This is absolutely wonderful. I can already imagine sort of what is about to come and I can’t wait!!
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_*?
Gauss or GauB xd
Let me help you, Gauss or Gauß
@@livedandletdie Yeah actually hace a spanish keyboard but i can do this, ñçÜl·l xd
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 RUclipsrs.
HMMMMM EXPOSÉ?
*Papaaaaaa*
It was great, I want to watch it again and again!
Great
Thank you so much *My Dear Papa Flammy Mathy*
I’m quite new to mathematics of this level - this explanation was brilliant. I’m glad you deviate into simpler(ish) problems.
You remind me of our friend who said "...and wheeler!!!".
Love you papa, love me back
Thanks papa
Gotta love the merch Papa Flammy is putting out.
That’s not a meme, I genuinely love your merchandise
Flammable Maths Papa, I know that me and my fellow mathematicians typically just meme in the comments but seriously, thank you for keeping my love of mathematics alive (despite my university’s maths department’s desperate attempts to make everyone hate maths). I always enjoy your videos, how you can educate and entertain at the same time. Thank you for what you do
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.
"We can actually Fubini this shit." XD
I love your integeral science!
:D
It was just incredible, i hate it when i see a solution that i could figure out myself (but quitted trying too soon).
That was beautiful
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.
back in 17s if you where a master of this integarahl you would be an amazing physicist .
this thing is everywhere in stat mech .
Hey, are you Jens Fehlau?
I saw u on quora's recommendation for no reason. LMAO.
WOW Flammy is back! Welcome to anor video with the oiler macaroni consent.
Beautiful
very nice inthégral bro et loks layks goods
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.
I too did it this way. Don't know why very few did this way.
Boi what are you adding in that big sigma.
I have so much homework... so, one video of integrals arent be bad.
Go papa flamy. You inspire me.
Thanks. Nicely Explained.
Me, watching this without even finishing limits:
I can't understand shit but I like it
integral of x/(e^x+1) from 0 to infinity = pi^2/12 coool
Wow man.....just loved it.....!!!!😘😘😘
PAPA could you do this bad boi ?
Integral from 0 to infinity of
(sinx)^2/(1+x^2)
it evaluates to 177013
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
Can papa flammy bless me for my engineering entrance exam tmr???
Mains? All the best!!
Lol. That's what the flammily does, watch an upload with jee mains tomorrow.
remember that Pi=e=2
Papa Flammy gonna prove the Riemann Hypothesis next video confirmed?
@@PapaFlammy69 Did you run out of margin space on the chalkboard?
I enjoyed this integration as much as enjoy my favorite movie.
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)
Where did you buy that watch? It looks great. Maybe you should contact the manufacturer and have them sponsor you 🤣
5:02 Very smart move.
This can be a quick infinity boi, any integral of this form with x^(s-1) on top is Gamma(s)Zeta(s). xD
yata desu ne!
You really should get into probability. I think you'd really enjoy it.
Probably
@@neilgerace355 i love you
Probability is the worst part of math.
@@griffisme4833
I love you
What is that watch you have?
1st anniversary of this video!
You should also make vids about physics. Named Flammable Physics. Or even chemistry.
Thanks sir.
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?
madlad
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"
Just about to watch a Great Video
e to the negative teeth power /o/
The captions at 10:15.
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?😀
5:32
wait what the hell you can do that??
Yup k and x are independent
I think that I love 😂❤️
where do you learn all this??? do you have books or lectures that you could recommed for me to learn??
More integral equation pleaseeeeeeee!
Papa bless
That's actually the Bose Integral at n=2.
sure
holy cow I got a heart and reply from papa!!! My day is made :)
Sure thing my Kizyzo boi :p
test: if papa flamy likes this, hes def using a bot
6:26 I'd've used papa feynman, but okay
Halfway, it would be faster if you used the gamma function or Laplace transforms. To learn more visit the Mathematical Facts group on Facebook.
@@PapaFlammy69 Nice.
Congrats on your channel. I am a mathematician from Brazil who loves solving integrals and series.
could you use complex analysis?
What a watch ma boi
Consider my breath taken
Papa
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
@@PapaFlammy69 is this really putnam btw?
@@PapaFlammy69 But great video as always
🔥🔰-ʕ•ᴥ•ʔ-🗡💜! ALL COMMENTS = ENDORSEMENTS! I AM A COMMENT!
Did you check the interval of convergence for the geometric series? ;)
awesooooome
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
so the function for the contour should be z^2/(e^z-1)
this technique will also work for every odd power of x i think :)
Papa putting those "Putnam" in the title again :v
good
do you know how to solve that equation algebratically: (4x+2)^(1/x) = 2. thanks for that ;) great stuff anyway
2^x=4x+2
k = -x - 1/2
2^(-k - 1/2) = -4k
2^(-k) = -4k*2^(1/2)
2^k = -1/(k*2^5/2)
k*2^k = -2^(-5/2)
k = W(-2^(-5/2)ln(2))/ln(2)
x = -W(-2^(-5/2)ln(2))/ln(2) - 1/2
No solutions in elementary functions as far as I know. The Lambert's W-function gives us two solutions here.
𝗪0𝗡de𝗥𝗙𝗨l
Integrals ❤️❤️
Papa Flammy you should do some question from the AIME exam!! Would be cool to see how you approach them.
ANIME exam?
@@OtiumAbscondita artofproblemsolving.com/wiki/index.php/AIME_Problems_and_Solutions
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
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....
Alreafy done! Check the integrals Playlist :)
Is it the zeta gamma extravaganza
GOD
すごい！ 登録しました。
Can't you just use some complex analysis?
i cant understand complex calculus
Hot. How about other values of zeta function times gamma function?
Why is there -1/12 on your tshirt?
Infinity Boi
8
I'm in India, how can i buy it??
My Merch? Over on my Teespring shop
>mfw sugoi desu
Bose Einstein integral for s=2.....!
x= 2
Can Papa Flammy bless me for my exam in 3 hours?
Can you do a video on \zeta(4)
mmm tasty basel boiiii
Wow I'm here early...
I got zero as the answer of this question
OMG WTF QAQ I don't no what you say??
pi creatures!!!
do u use a bot to like everything? excellent move!
Your channel is amazing! Where did you study math from?
Have you ever thought about doing videos about tetration and other 🅱️S like that?
@@PapaFlammy69 yes^yes^yes^... = -W(-ln(yes)) /ln(yes)