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I am currently studying the book "Higher Trascendental functions" by Erdélyi et al. On page 27 he defines the function:

$$\Psi(z,s,v)=\sum_{n=0}^{\infty}(v+n)^{-s}z^n$$ and then claims that it can be represented as a contour integral:

$$2\pi i \Psi(z,s,v)=-\Gamma(1-s)\int^{0,+}_{\infty}(-t)^{s-1}e^{-vt}(1-ze^{-t})^{-1}dt$$ for $\operatorname{Re}(v)>0$, $|\arg(-t)|\leq \pi$ and assuming that the contour does not enclose any of the points $t=\log z\pm 2n\pi i$ $(n=0,1,2,\ldots)$.

In the book I can't seem to find any proof and searching online I can't even find anyone else referring to this sum as $\Psi$.

Does someone know of any text that contains this proof, or of some way to prove it?

Gary
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Zackury
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1 Answers1

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The details, using Hankel's contour integral representation, valid for any $s\in\mathbb{C}$ (the idea is here): $$ \frac{2\pi i}{\Gamma(s)}=\int_\lambda z^{-s}e^z\,dz, $$ where $\lambda$ encircles the negative real axis (the branch cut of $z^{-s}$ taken in the principal value sense).

Take $w\in\mathbb{C}$ with $\Re w>0$, and substitute $z=wt$ in the integral: $$ \frac{2\pi i}{\Gamma(s)}w^{s-1}=\int_\lambda t^{-s}e^{wt}\,dt $$ (by Cauchy's theorem, thanks to $\Re w>0$, the contour $\lambda$ may be left as is).

Now take $z,v\in\mathbb{C}$ with $\Re v>0$ and $|z|<1$. Choose $\lambda$ so that $|ze^t|<r<1$ for $t\in\lambda$ (that is, let $\lambda$ be close enough to the negative real axis). Then, after replacing $s$ with $1-s$ in the formula above, we obtain $$ \Psi(z,s,v)=\frac{\Gamma(1-s)}{2\pi i}\sum_{n=0}^\infty z^n\int_\lambda t^{s-1}e^{(v+n)t}\,dt=\frac{\Gamma(1-s)}{2\pi i}\int_\lambda t^{s-1}e^{vt}(1-ze^t)^{-1}\,dt $$ ($|ze^t|<r<1$ ensures the validity of moving the summation under the integral sign). After $t\mapsto-t$, this is the claimed formula (analytic continuation beyond $|z|<1$ leads to the conditions for the contour).

metamorphy
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