Let ˜Γ be an analytic continuation of Γ on C∖(−N0). Show that the function
˜Γ(z)˜Γ(1−z)−πsin(πz)
can be analytically continued to an entire function.
I do assume that the analytic continuation is the classical ˜Γ(z)=Γ(z+n)z(z+1)⋅…⋅(z+n−1)
for z∈C∖(−N0),Rez>−n with residues Res−n(˜Γ)=(−1)nn! with n∈N0 which I had to deduce in the excersice leading to this problem. I stumbled upon explanations on how to compute ˜Γ(z)˜Γ(1−z) using the Beta function I am not familiar with. Now I am curious as to what I have to do exactly with the given function and which methods there are available to do so.
Answer
The "standard" analytic continuation leads to to Euler's limit product formula, that leads to the Weierstrass product:
Γ(z)=e−γzz∏n≥1(1+zn)−1ez/n
from which:
zΓ(z)Γ(−z)=1z∏n≥1(1−z2n2)−1
and you may finish the work by recognizing the (reciprocal) Weierstrass product for a sine function in the RHS.
As an alternative, you may just prove that
Γ(z)Γ(1−z)−πsin(πz) has no singular point,
since both terms have simple poles at the same points with the same
residues: Res(Γ(z)Γ(1−z),z=−n)=(−1)n,
Res(πsin(πz),z=−n)=limz→nπ(z−n)sin(πz)DH=cos(πn)=(−1)n.
Yet another way is to prove the red equality:
d2dz2log(Γ(z)Γ(1−z))=ψ″
through Fourier series or other means.
No comments:
Post a Comment