2
$\begingroup$

On the integral representation of Hurwitz Zeta Function inside the critical strip,

Show that: $$ \Gamma(s)\zeta(s,a)=\int_{0}^{\infty}x^{s-1}\left(\frac{e^{(1-a)x}}{e^{x}-1}-\frac{1}{x}\right)\,dx \tag{1} $$ for $\, 0\lt \Re\{s\}\lt1 ,\,\, a\in\mathbb{R}^{+} \,$.


I am also seeking for a proof (or disproof) of the more general representation (claim):
$$ \Gamma(s-m)\zeta(s-m,a)=\int_{0}^{\infty}x^{s-m-2}\left[\frac{xe^{(1-a)x}}{e^x-1}-\left(\sum_{n=0}^{m}B_{n}\frac{x^n}{n!}\right)\right]\,dx \tag{2} $$ for $\,0\lt\,\Re\{s\}\,\lt1 ,\,\, a\in\mathbb{R}^{+} ,\,\, m\in\mathbb{N} ,\,\, B_{n}={\operatorname{Bernoulli}}_{\#} ,\,\, B_{1}=-1/2 \,$. Thanks.
$\endgroup$
1
  • $\begingroup$ Posted on the honor of welcoming back our friend @JackD'Aurizio $\endgroup$ Jul 16, 2021 at 22:34

1 Answer 1

2
$\begingroup$

The formula $(2)$ needs some correction: for small $x$ (actually for $|x|<2\pi$) we have $$\frac{xe^{(1-a)x}}{e^x-1}=\frac{xe^{-ax}}{1-e^{-x}}=\sum_{n=0}^\infty(-1)^nB_n(a)\frac{x^n}{n!},$$ where $B_n(a)$ are the Bernoulli polynomials. Both $(1)$ and $(2)$ [fixed] can be deduced from $$\zeta(s,a)=\frac{\Gamma(1-s)}{2\pi i}\int_\lambda\frac{z^{s-1}e^{az}}{1-e^z}\,dz\qquad(s\notin\mathbb{Z}_{>0},\Re a>0)$$ where the contour $\lambda$ encircles the negative real axis (but not the poles $z=2n\pi i$, $n\in\mathbb{Z}_{\neq0}$ of the integrand). In turn, one proves this for $\Re s>1$ by "squeezing" $\lambda$ closely to the negative real axis: $$\int_\lambda\frac{z^{s-1}e^{az}}{1-e^z}\,dz=\int_\infty^0(xe^{-\pi i})^{s-1}\frac{e^{-ax}\,d(-x)}{1-e^{-x}}+\int_0^\infty(xe^{\pi i})^{s-1}\frac{e^{-ax}\,d(-x)}{1-e^{-x}}\\=2i\sin s\pi\int_0^\infty\frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx=2i\sin s\pi\sum_{n=0}^\infty\int_0^\infty x^{s-1}e^{-(n+a)x}\,dx\\=2i\sin s\pi\,\Gamma(s)\zeta(s,a)=\frac{2\pi i}{\Gamma(1-s)}\zeta(s,a),$$ and by analytic continuation elsewhere, since the integral is an entire function of $s$.

Now, for $0<\Re s<1$ and $m\in\mathbb{Z}_{\geqslant 0}$, we have $$\int_\lambda\frac{z^{s-m-1}e^{az}}{1-e^z}\,dz=\int_\lambda z^{s-m-2}\left(\frac{ze^{az}}{1-e^z}+\sum_{n=0}^m B_n(a)\frac{z^n}{n!}\right)dz$$ since $\int_\lambda z^\alpha\,dz=0$ for $\Re\alpha<-1$; "squeezing" $\lambda$ the same way, we get a fixed version of $(2)$: $$\Gamma(s-m)\zeta(s-m,a)=\int_0^\infty x^{s-m-2}\left(\frac{xe^{-ax}}{1-e^{-x}}-\sum_{n=0}^m(-1)^n B_n(a)\frac{x^n}{n!}\right)dx.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.