I need your help on the following.
(1)First we are to find the residue of $\frac{x^s}{s}$ at $s=0$. Since $s=0$ is the pole of order 1, so we get
Res$(\frac{x^s}{s},s=0)=\frac{1}{2\pi i}\int_{C_1^+(0)} \frac{f(s)}{s}ds$ where $f(s)=x^s$. In other words,
Res$(\frac{x^s}{s},s=0)=f(0)=1$. Please tell me if I made any mistake on it or not.
(2)My next problem is to find the poles of $\frac{\zeta(s-1)}{\zeta(s)}\frac{x^s}{s}$.
Well I got stuck at this. My instructor is saying it has a pole of order 1 at 2 but I could not understand what he is trying to say, with due respect. May be I heard him wrong or may be I am missing something. Please help me out. I am not getting confidence.
One may recall that the Riemann zeta function admits a pole at $u=1$ and we have
where $\gamma=0.5772...$ is the Euler Mascheroni constant.
Setting $u:=s-1$, which gives $u-1=s-2$ and $u \to 1$ as $s \to 2$, we may then write $$ \begin{align} \frac{\zeta(s-1)}{\zeta(s)}\frac{x^s}{s}&=\frac{\zeta(u)}{\zeta(u+1)}\frac{x^{u+1}}{u+1}\\\\ &=\frac{x^{u+1}}{u+1}\frac{\frac1{u-1}+\gamma+\mathcal{O}(u-1)}{\zeta((u-1)+2)}\\\\ &=\frac{x^{2+(u-1)}}{2+(u-1)}\frac{\frac1{u-1}+\gamma+\mathcal{O}(u-1)}{\zeta(2)+\mathcal{O}(u-1)}\\\\ &=\frac{x^{2+(u-1)}}{2+(u-1)}\frac{\frac1{u-1}+\gamma+\mathcal{O}(u-1)}{\zeta(2)+\mathcal{O}(u-1)}\\\\ &=\frac{x^2}{2\:\zeta(2)}\frac{1}{(u-1)}+\mathcal{O}(1)\\\\ &=\frac{3x^2}{\pi^2}\frac{1}{s-2}+\mathcal{O}(1)\\\\ \end{align} $$ where we have used $\displaystyle \zeta(2)=\frac{\pi^2}6$, and the pole is of order $1$ with a residue equal to $\displaystyle \frac{3x^2}{\pi^2}$.