Find $\lim_\limits{R\to \infty}{1\over 2\pi i}\int_{{1\over 2}-iR}^{{1\over 2}+iR}{x^s\over s}ds$ $\lim_\limits{R\to \infty}{1\over 2\pi i}\int_{{1\over 2}-iR}^{{1\over 2}+iR}{x^s\over s}ds$ where $x>0$. Split it to cases: $x>1,x=1,0<x<1$. I tried using contour integration but I am still very confused as to how I should approach it. When $x=1$ I get $\ln|s|$. I spot no difference between $x>1$ and $x<1$ when it is presented like this, but I do sense it has something to do with $Log$. I would really appreciate any guiding. (When writing this, no match came up, and looking for it didn't bring me much forward.)

Rewrite as
$$\lim_{R \to \infty} \frac1{i 2 \pi}\int_{\frac12-i R}^{\frac12+i R} ds \frac{e^{s \log{x}}}{s} $$
When $x \gt 1$, $\log{x} \gt 0$ and we may close to the left of $\operatorname{Re}{s}=\frac12$ with a circle of radius $R$ centered at the origin.  The integral about the circle vanishes when $R \to \infty$.  In this case, the closed contour encloses the pole at $s=0$ so that, in the limit, the integral is $1$.  However, when $0 \lt x \lt 1$, $\log{x} \lt 0$ and we must close to the right of the line $\operatorname{Re}{s}=\frac12$ in order for the integral about the circle to vanish as $R \to \infty$.  In this case, there are no poles and the integral is $0$ in this limit.  Thus
$$\lim_{R \to \infty} \frac1{i 2 \pi}\int_{\frac12-i R}^{\frac12+i R} ds \frac{x^s}{s} = \begin{cases}0 & 0 \lt x \lt 1 \\ 1 & x \gt 1 \end{cases}$$
