Evaluating $I(n) = \int^{\infty}_{0} \frac{\ln(x)}{x^n(1+x)}\, dx$ for real $n$ I am not sure how to handle the additional parameter $n$. I first need to find out for which real values of $n$ will the integral converge. Based on intuition and checking with mathematica, I believe it will converge only for $0 < n < 1$, although I am not sure and I would like to see exactly how to determine those values. For this integral, I would normally try using a branch cut, but again I am not sure how to apply it given my additional parameter $n$.
$$I(n) = \int^{\infty}_{0} \dfrac{\ln(x)}{x^n(1+x)}\, dx$$

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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\on{I}\pars{n} & \equiv
\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\ln\pars{x} \over x^{n}\pars{1 + x}}\,\dd x\,\right\vert_{\,\Re\pars{n}\ <\ 1}} =
\left.\partiald{}{\nu}\int_{0}^{\infty}
{x^{\color{red}{\nu - n + 1} - 1} \over
1 + x}\,\dd x\,\right\vert_{\,\nu\ =\ 0}
\end{align}
Note that
$\ds{{1 \over 1 + x} =
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k}}
\,{\pars{-x}^{k} \over k!}}$.
Then,
\begin{align}
\on{I}\pars{n} & \equiv
\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\ln\pars{x} \over x^{n}\pars{1 + x}}\,\dd x\,\right\vert_{\,\Re\pars{n}\ <\ 1}}
\\[5mm] = &\
\left.\partiald{}{\nu}\,
\Gamma\pars{\color{red}{\nu - n + 1}}
\Gamma\pars{1 - \bracks{\color{red}{\nu - n + 1}}}
\,\right\vert_{\,\nu\ =\ 0}\quad
\pars{\substack{\ds{Ramanujan's}\\[0.5mm]\ds{Master}
\\[0.5mm]\ds{Theorem}}}
\\[5mm] = &\
\left.\partiald{}{\nu}\,
{\pi \over \sin\pars{\pi\bracks{\nu - n + 1}}}
\,\right\vert_{\,\nu\ =\ 0} =
\left.-\pi\,\partiald{\csc\pars{\pi\bracks{\nu - n}}}{\nu}
\,\right\vert_{\,\nu\ =\ 0}
\\[5mm] = &\
\left.\pi^{2}\csc\pars{\pi\bracks{\nu - n}}
\cot\pars{\pi\bracks{\nu - n}}
\,\right\vert_{\,\nu\ =\ 0}
\\[5mm] = &\
\bbx{\pi^{2}\csc\pars{n\pi}\cot\pars{n\pi}} \\ &
\end{align}
