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!
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)
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
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??
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
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
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!
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!
What you are working out here is actually the integral of |x!|^2 along the imaginary axis - as if it were a wavefunction...
That's a fascinating way to view it!
Can you pls explain
you stated your sentence as if there's some other way to do it
Very interesting integral and smart solution. Thanks for this video.
I'll try it myself tomorrow then i will watch the video
I wonder if this can be done by the Parseval's theorem for Mellin transform.
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
beautiful
The normalization version of Meixner-Pollaczek integral.
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 ?
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)
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
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
Couldn’t you make it so that the integrand is proportional to x/sinh(x) instead of x/sinh(pi*x)
Bro is scared of π being multiplied by x 💀💀💀
very nice
You could've used the evenness in the very beginning!
pi/2...ho usato la formula di reflection di hamma....thanks,ho visto che è corretta
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??
Can please make a video on transformation of variables in integrals, cause those are confusing to me?
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!!???
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
Noice!
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
When you don’t know what tf this is or even means🗿
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!