A satisfying gamma function integral

[@herbertdiazmoraga7258]

nice result as always! I recommend to you to find the norm of the Hilbert Matrix, the result it's very surprising and it can be obtained with the reflection formula (I think, I tried to do it but y did not achieve it) Keep the awesome integrals coming!

[@mikeoffthebox]

What you are working out here is actually the integral of |x!|^2 along the imaginary axis - as if it were a wavefunction...

[@DepozidoX]

That's a fascinating way to view it!

[@ShanBojack]

Can you pls explain

[@NoNameAtAll2]

you stated your sentence as if there's some other way to do it

[@putianyi8889]

I wonder if this can be done by the Parseval's theorem for Mellin transform.

[@MrWael1970]

Very interesting integral and smart solution. Thanks for this video.

[@Unidentifying]

beautiful

[@jehejhdhdj1135]

I'll try it myself tomorrow then i will watch the video

[@cameronspalding9792]

Couldn’t you make it so that the integrand is proportional to x/sinh(x) instead of x/sinh(pi*x)

[@maths_505]

Bro is scared of π being multiplied by x 💀💀💀

[@Hidensoul1]

An unexpected nice and simple result for this one. When you write 1/(1-x) as a sum for k of x to the k, is it a Taylor serie ?

[@andy_lamax]

It is more like a geometric series of ratio x that converges. S = 1 + x + x^2 + x^3 + . . . is a geometric series with first term (G1=1) and common ration (r = x) The sum to infinity of a converging geometric series (A geometric series converges if its common ratio is less that one) is given by S = g1/(1-r) plugin everything back in S = 1 / (1 - x)

[@maximebree4360]

Taylor series is a first approach when you learn series expansion, after Taylor series you see series expansion, which is for some functions the Taylor series at the +infinity order, but for other function when u can't get the n-th derivative it exist other ways

[@Hidensoul1]

Thanks a lot for your answers. Geometric serie is the first i should had in mind if course ! Not a big deal then. I've got it now

[@illumexhisoka6181]

I tried to solve it before watching the video I used the exacte same approache Be instead of substituting I used integration by parts at the last

[@UnknownGhost97]

Question: Did ramanujan invented or used this gamma function theory which now indians are praising him now a lot as the greatest mathematician of all time??

[@nguyenquangkiet2103]

The normalization version of Meixner-Pollaczek integral.

[@albertohart5334]

You lost me at S = (pi^2)/8 like shouldnt the odd values of 1/n^2 be closer to half of the og sum instead of like the vast majority!!???

[@ambiguousheadline8263]

The terms are getting smaller fast so the early terms contribute most to the value of the sum. 1/(1)^2 = 1 which is already more than half the sum's value in the first term. So clearly the odd terms contribute significantly more to the final result

[@nicogehren6566]

very nice

[@erfanmohagheghian707]

You could've used the evenness in the very beginning!

[@giuseppemalaguti435]

pi/2...ho usato la formula di reflection di hamma....thanks,ho visto che è corretta

[@krishnenduchakraborty7000]

Can please make a video on transformation of variables in integrals, cause those are confusing to me?

[@davidblauyoutube]

Noice!

[@maxvangulik1988]

gamma(1+ix)=ix•gamma(ix) gamma(ix)gamma(1-ix)=pi•csc(pi•ix) i•csc(ix)=csch(x) I=pi•int[-♾️,♾️](x•csch(pi•x))dx b=pi•x db=pi•dx I=int[-♾️,♾️](b•csch(b))db/pi b is an odd function of b csch(b) is an odd function of b odd•odd=even I=2/pi•int[0,♾️](b•csch(b))db I=4/pi•int[0,♾️](b/(e^b-e^-b))db I=4/pi•int[0,♾️](be^-b/(1-e^-2b))db I=4/pi•int[0,♾️](be^-b•sum[n=0,♾️](e^-2nb))db I=4/pi•sum[n=0,♾️](int[0,♾️](be^-(2n+1)b)db) r=(2n+1)b dr=(2n+1)db I=4/pi•sum[n=0,♾️]((2n+1)^-2•int[0,♾️](re^-r)dr I=4/pi•sum[n=0,♾️]((2n+1)^-2) sum[n=0,♾️]((n+1)^-2)=pi^2/6 sum[n=1,♾️]((2n)^-2)=pi^2/24 I=4/pi•(pi^2/6-pi^2/24) I=pi/2

[@GirlyOrbs]

When you don’t know what tf this is or even means🗿

[@angeldude101]

Setting aside the use of the gamma function over the pi function and the use of pi instead of tau, I burst out laughing when the integral somehow managed to reduce to 1!. It was also then that I realized we had long since left the ℂomplex plane. That was actually really cool!