Question: How do you integrate$$\mathcal{I}=\int\limits_0^{\infty}dx\,\frac {x^a\log x}{x^b+1}$$Where $b>a+1$.
I attempted to integrate the function using contour integration by considering the function$$f(z)=\frac {z^a\log z}{z^b+1}$$in a contour sector $C$. The contour has a larger radius of $R$ and a smaller circular detour of radius $r$ because $\log 0$ isn't well defined. Therefore, we can set up$$\oint\limits_{C}dz\, f(z)=\int\limits_r^Rdx\, f(x)+\int\limits_{\gamma_R}dz\, f(z)-e^{2\pi i/b}\int\limits_r^Rdz\, f(z)+\int\limits_{\gamma_r}dz\, f(z)$$However, I'm having difficulty computing the residue at $z=e^{\pi i/b}$$$\operatorname{Res}\left(f,e^{\pi i/b}\right)=\lim\limits_{z\to e^{\pi i/b}}\frac {(z-e^{\pi i/b})z^a\log z}{z^b+1}$$Here's my work$$\begin{align*}\operatorname{Res}\left(f,e^{\pi i/b}\right) & =\lim\limits_{z\to e^{\pi i/b}}\frac {1}{bz^{b-1}}\frac {e^{a\pi i/b}\pi i}{b}\\ & =\frac {\pi i e^{a\pi i/b}}{b^2 e^{(b-1)\pi i/b}}=\color{blue}{\frac {\pi i e^{(a-b+1)\pi i/b}}{b^2}}\end{align*}$$But that means$$\oint\limits_{C}dz\, f(z)=-\frac {2\pi^2 e^{(a-b+1)\pi i/b}}{b^2}$$And I get an even uglier answer when I let $r\to 0$ and $R\to\infty$.
Any ideas on what I should do? Should I change my contour? Did I make a minor mistake?