An integral involving the Riemann zeta function

Question

Let $c\in \mathbb{R^{+}}$, $0<\delta<1$, and $s\in \mathbb{C}:\;\Re(s)+c>1$. We have the integral : $$\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\frac{\zeta(s+z)}{z(z+1)}\left((1+\delta)^{1+z}-\delta^{1+z}\right)dz$$ I tried shifting the line of integration to the left of zero, picking up the residues at $z=1-s$ and $z=0$, but i have no idea on how to do the integral : $$\frac{1}{2\pi i }\int_{c^{'}-i\infty}^{c^{'}+i\infty}\frac{\zeta(s+z)}{z(z+1)}\left((1+\delta)^{1+z}-\delta^{1+z}\right)dz\;\;\;\;c<0$$ Any help is appreciated.

Answer 1

We have : $$\frac{1}{z+1}\left((1+\delta)^{1+z}-\delta^{1+z}\right)=\int_{\delta}^{1+\delta}x^{z}dx$$ and by Perron's formula : $$\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\zeta(s+z)\frac{x^{z}}{z}dz=\acute{\sum_{n \leq x}}\frac{1}{n^{s}}=\zeta(s)-\zeta(s,x+1)$$ where $\zeta(\cdot,\cdot)$ is the Hurwitz zeta function. Thus : $$\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\frac{\zeta(s+z)}{z(z+1)}\left((1+\delta)^{1+z}-\delta^{1+z}\right)dz=\int_{\delta}^{1+\delta}\left(\zeta(s)-\zeta(s,x+1\right)dx$$ $$=\zeta(s)-\frac{(\delta+1)^{1-s}}{s-1}$$