An interesting generalised integration result: int 1/(1+x^s)^s from 0 to infinity

proof for the duplication formula:
• An elegant proof of Le...
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This was quite a fascinating integral and the results were fun to derive. Let me know in the comments section in case you find more cool results for this generalised result.

[@marcellomarianetti1770]

The golden ratio one having 1 as result caught me off guard

[@MatthisDayer]

hello complex man

[@rockthemegaman2760]

Played around a bit more, if you set s = a positive integer n > 1, You can get a fairly nice structure of a rational number * pi/sin(pi/n). so whenever sin(pi/n) has a nice value, so will the integral. You can get it down to { (n^2-1-n)(n^2-1-2n)(n^2-1-3n)...(n^2-1-(n-1)n) * pi } / [ n^n * gamma(n) * sin(pi/n) ]. I don't know a nice way to simplify the numerator without going back to the nasty gamma(icky)/gamma(yucky) structure you start with. Think my personal favorite is n = 6 giving the value of 124729pi/559872 which happens to be extremely close to 0.7

[@maths_505]

Now that's the kind of comment I wanted!

[@Abdalrhman_Kilesee]

Marvellous man I really appreciate you and what you’re doing

[@Maths_3.1415]

Cornel Ioan Vălean (Almost) Impossible Integrals, Sums, and Series With foreword by Paul J. Nahin Try this book for integrals

[@rockthemegaman2760]

Missed opportunity to simply the s = delta case to 1/sqrt(2) which is even prettier.

[@maths_505]

Oh that is nice too.

[@Kapomafioso]

The case for the golden ratio made me drop my jaw. Not even Mathematica simplifies that (the result I get is (1/phi)*Gamma(1/phi) / Gamma(phi), which it won't simplify further even if I do FullSimplify). But it is indeed equal to 1. Fascinating!

[@rockthemegaman2760]

Back at it several weeks later. But i realized randomly today why the phi and delta examples work so well. It's because they satisfy 1/s = s-1 for phi and 1/s = s-2 for delta and allow you to use the recursion on gamma to get rid of the nasty bits. So if we take the solution for 1/s = s-n for some positive integer n, we can get some nice results (can't do negative integers bc that would make s less than 1, which makes the integral diverge). Such solutions are s = [n+sqrt(n^2+4)]/2 (note we can't take the other solution bc it's negative and again, the integral diverges). This family of niceness gives rational answers for odd n and rational numbers times a square root for even n. Specifically, even n: integral = Sqrt(n^2+4) Gamma(n) 2^(n/2) / [(n^2+4)(n^2+4-2^2)(n^2+4-4^2)...(n^2+4-(n-2)^2)(n^2+4-n^2)] odd n: integral = Gamma(n) 2^(n+1/2)/ [(n^2+4-1^2)(n^2+4-3^2)(n^2+4-5^2)...(n^2+4-(n-2)^2)(n^2+4-n^2)] I'm 99.9% sure there's no more nice answers this time.

[@maths_505]

Nice one bro

[@manstuckinabox3679]

Sorry mister Maths 505, my brain is screaming keyhole contour, I must follow my complex instincts. edit: I tried it, didn't work for me.

[@Decrupt]

I had the same thought

[@Maths_3.1415]

13:46 now differentiate it

[@maths_505]

The proof is left as an exercise to the viewer "Laughs in evil math author"

[@thewolverine7516]

​@@maths_505Fermat's reference.... Lol

[@Maths_3.1415]

​@@maths_505lol 😂

[@ThAlEdison]

Just spitballing, might've made a mistake Let a=(n+sqrt(n^2+4))/2 be a solution to x^2-nx-1=0 where n is a natural number>1. Then I_a=(n-1)!(a^(n-1))/((a+1)...((n-1)a+1)) a^2=na+1 (ka+1)((n-k)a+1)=((n-k)ka^2+na+1)=((kn-k^2)a^2+a^2)=(kn-k^2+1)a^2=b_k*a^2 and (n/2)a+1=(n^2+4+nsqrt(n^2+4))/4=(sqrt(n^2+4)/2)(n+sqrt(n^2+4))/2=(sqrt(n^2+4)/2)a So for odd-n I_a=(n-1)!/(b_1*...*b_(n-1)/2) where b_k=kn-k^2+1 even-n I_a=(n-1)!/((b_1*...*b_(n-2)/2))*sqrt((n/2)^2+1)) I_(1+sqrt(2))=1/sqrt(2) I_((3+sqrt(13))/2)=2/3 I_(2+sqrt(5))=6/5sqrt(5) I_((5+sqrt(29))/2)=24/35 I_((3+sqrt(10))=2sqrt(10)/9 I_((7+sqrt(53))/2)=720/1001

[@minamagdy4126]

For s

[@francis6888]

8:38 you could have factored out the sqrt2 from the gamma in the numerator

[@anestismoutafidis4575]

Intg from 0 to infinite => Intg [(1+x^s)^(-s+1)/-s+1 •dx] +D={s(N)\ 0; 1}

[@ikarienator]

1+1/(1+sqrt2)=sqrt(2). In the end you get 1/sqrt(2)

[@natepolidoro4565]

I_delta = 1/sqrt(2) actually

[@jhulioux]

THE BEST VIDEO I'VE EVER WATCHED!

[@Maths_3.1415]

Nice video bro :)

[@saiajaygarine6382]

Me: wondering what is s here‽ Maths 505: cool, we did it.😅

[@maths_505]

This video was sponsored by Complex Man

[@holyshit922]

It looks like Beta function For certain values of s like s integer or s golden ratio can be calculated by parts sGamma(s) can be simplified further

[@parasgaur1856]

I ask of you to give me the list of the books you studied from. Pre-Pre-calc To Today. Please 😊

[@natepolidoro4565]

Cool video.

[@maths_505]

Thanks

[@shoshohoms5180]

How the limits of integration didn't change when s is a complex number

[@Kapomafioso]

BTW I'm pretty sure the original integral won't converge for s = i. The integral is fine around x = 0, but for x -> oo, you need Re(s^2) > 1. I guess the formula involving Beta/Gamma serves as the analytic continuation of the original function...

[@maths_505]

Ofcourse one can always take it to be the limit as Re(s) approaches zero from the right. That's pretty much what advanced calculus is all about; if something ain't working, pop in a limit😂

[@Kapomafioso]

@@maths_505 Yeah but Re(s) needs to be >1, so Re(s) approaching zero won't do much either I think...

[@maths_505]

Oh its perfectly fine don't worry....how else do you think we got the derivative of the Dirichlet eta function at s=1😂 That's a good video btw....check it out I'm sure you'll like it

[@maths_505]

It was part of the evaluation of int lnx/(1+e^x) from zero to infinity

[@Kapomafioso]

@@maths_505 Not sure how the Dirichlet eta relates to this integral, but since you mentioned it...from the structure of the series definition, it clearly converges for s > 0. You can use alternating series test to prove that. The derivative of the series also converges for s > 0 (you have to prove a very similar statement with log(k) / k^s instead of just 1/k^s, but the terms still go to zero while alternating in sign). From this, I don't see how s = 1 is a problematic point as far as the derivative goes, since it is deeply into the region of the convergence. The derivative is well defined and calculable from the series which converges at s = 1... ...unlike the integral in this video, which outright does not exist at s = i, so the result expressed via Gamma functions is an extension.

[@erfanmohagheghian707]

I sub delta simplifies to 1/sqrt(2)

[@giuseppemalaguti435]

I=(1/s)B(1/s,s-1/s)

[@MrWael1970]

This is Fascinating. Thank you.

[@nicogehren6566]

very nice

[@d.h.y]

Awesome!! Does this hold for whole s \in \mathbb{C} ?

[@extatix809]

0:09

[@d.h.y]

@@extatix809 Oh, I missed that part. Thanks!

[@thewolverine7516]

Hey bro, it's a request if possible please do upload complete theory lectures. I believe your conceptual understanding is extremely clear, I also wanna be like you. Or else if it's not possible, tell me some sources where I can learn and be like you.

[@maths_505]

I'm gonna start a second channel in a couple of months where I'll upload formal lectures in complex analysis and differential equations with more courses to come.

[@thewolverine7516]

@@maths_505Woohooo 🙌. I am desperately waiting....

[@herbertdiazmoraga7258]

@@maths_505 golasoo de xileee