Tuesday, 5 May 2015

complex analysis - Analytic continuation of Euler's reflection formula with the Gamma function




Let ˜Γ be an analytic continuation of Γ on C(N0). Show that the function
˜Γ(z)˜Γ(1z)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+n1)

for zC(N0),Rez>n with residues Resn(˜Γ)=(1)nn! with nN0 which I had to deduce in the excersice leading to this problem. I stumbled upon explanations on how to compute ˜Γ(z)˜Γ(1z) 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)=eγzzn1(1+zn)1ez/n
from which:
zΓ(z)Γ(z)=1zn1(1z2n2)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(1z)πsin(πz) has no singular point,
since both terms have simple poles at the same points with the same
residues: Res(Γ(z)Γ(1z),z=n)=(1)n,
Res(πsin(πz),z=n)=limznπ(zn)sin(πz)DH=cos(πn)=(1)n.




Yet another way is to prove the red equality:
d2dz2log(Γ(z)Γ(1z))=ψ

through Fourier series or other means.


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