# On the integral $\int\limits_{0}^{\infty}x^{s-1}\left(\frac{e^{(1-a)x}}{e^{x}-1}-\frac{1}{x}\right)\,dx$

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.
• Posted on the honor of welcoming back our friend @JackD'Aurizio Jul 16, 2021 at 22:34

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 $$(2)$$: $$\Gamma(s-m)\zeta(s-m,a)=\int_0^\infty x^{s-m-2}\left(\frac{xe^{-ax}}{1-e^{-x}}-\sum_{n=0}^m(-1)^n B_n(a)\frac{x^n}{n!}\right)dx.$$