Evaluating $\int_{(c)}\frac{x^{s}}{s^{k+1}}ds$

Question

Here, $(c)$ is the path $c+it$, $c\in\mathbb{R}_{>0}$, $-\infty < t < \infty $ oriented 'upwards', $x$ is a positive real number and $k$ is a positive integer. The answer is given as $0$ for $x\in[0,1]$ and $\frac{2\pi i}{k!}\log(x)^{k}$ for $x\geq 1$. The factor $2\pi i$ suggests to me that I should use Cauchy's residue theorem. Indeed, if I am correct, the residue of the integrand at zero is $\lim\limits_{s\rightarrow0}\frac{1}{k!}\frac{d^{k}}{ds^{k}}x^{s}=\frac{1}{k!}\log(x)^{k}$. Therefore, integrating around a contour enclosing zero would give $\frac{2\pi i}{k!}\log(x)^{k}$. This reasoning appears to be in the right direction, but I don't know how to evaluate it over the given path rather than a closed contour, nor how the integral evaluates in a piecewise fashion.