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 $n$.
$$I(n) = \int^{\infty}_{0} \dfrac{\ln(x)}{x^n(1+x)}\, dx$$
 A: Let $y = \frac{x}{1+x}$, then
$$dy = \frac{dx}{(1+x)^2},\;1 - y = \frac{1}{1+x}\;\text{ and }\;x = \frac{y}{1-y}$$
For $0 < n < 1$, we have
$$\begin{align}
I_n = & \int_0^\infty \frac{\log x}{x^n(1+x)} dx\\
= & -\frac{\partial}{\partial n}\left[\int_0^\infty \frac{1}{x^n(1+x)} dx\right]\\
= & -\frac{\partial}{\partial n}\left[\int_0^\infty \left(\frac{1+x}{x}\right)^n\left(\frac{1}{1+x}\right)^{n-1} \frac{dx}{(1+x)^2}\right]\\
= & -\frac{\partial}{\partial n}\left[\int_0^1 y^{-n}(1-y)^{n-1} dy\right]\\
= & -\frac{\partial}{\partial n}\left[\frac{\Gamma(n)\Gamma(1-n)}{\Gamma(1)}\right]
= -\frac{\partial}{\partial n}\left[\frac{\pi}{\sin(\pi n)}\right]
=  \frac{\pi^2 \cos(\pi n)}{\sin(\pi n)^2}
\end{align}
$$
A: I think the OP wanted to see some complex analysis.  Here goes:
Consider
$$\oint_C dz \frac{z^{-n} \log{z}}{1+z}$$
where $C$ is a keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$.  Thus the contour integral has four pieces:
$$\int_0^R dx \frac{x^{-n} \log{x}}{1+x} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} R^{-n} e^{-i n \theta} \frac{\log{(R e^{i \theta})}}{1+R e^{i \theta}} \\ + e^{-i 2 \pi n} \int_R^0 dx \, x^{-n} \frac{\log{x}+i 2 \pi}{1+x} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \epsilon^{-n} e^{-i n \phi} \frac{\log{(\epsilon e^{i \phi})}}{1+\epsilon e^{i \phi}}$$
As $R \to\infty$, the second integral vanishes only when $n \gt 0$.  As $\epsilon \to 0$, the fourth integral vanishes only when $n \lt 1$.  Thus we restrict $n \in (0,1)$, and the integral is equal to
$$\left (1-e^{-i 2 \pi n} \right )\int_0^{\infty} dx \frac{x^{-n} \log{x}}{1+x} - i 2 \pi \, e^{-i 2 \pi n} \int_0^{\infty} dx \frac{x^{-n} }{1+x}$$
The contour integral is equal to, by the residue theorem, $i 2 \pi$ times the residue at the pole $z=-1 = e^{i \pi}$, or
$$i 2 \pi e^{-i n \pi} (i \pi) = -2 \pi^2 (\cos{n \pi} - i \sin{n \pi})$$
We find the integral sought after by equating real and imaginary parts.  Let $A$ be the first integral (with the log) and $B$ the second (without the log).  Then
$$(1-\cos{2 \pi n}) A - (2 \pi \sin{2 \pi n}) B = -2 \pi^2 \cos{\pi n}$$
$$(\sin{2 \pi n}) A - (2 \pi \cos{2 \pi n}) B = 2 \pi^2 \sin{\pi n}$$
Eliminating the $B$ pieces by multiplying by $\cos{2 \pi n}$ in the first equation and $-\sin{2 \pi n}$ in the second, we get
$$(\cos{2 \pi n}-1) A = -2 \pi^2 \cos{\pi n}$$
or
$$A = \int_0^{\infty} dx \frac{x^{-n} \log{x}}{1+x} = \pi^2 \frac{\cos{\pi n}}{\sin^2{\pi n}}$$
As a bonus, you can show that
$$B = \int_0^{\infty} dx \frac{x^{-n} }{1+x} = \frac{\pi}{\sin{\pi n}}$$
A: Note that if $$f(a) = \int_0^{\infty} \dfrac{dx}{x^a(1+x)}$$ then $I(n) = -f'(n)$. And from here, we have
$$f(a) = \pi \csc(\pi a) \implies I(a) = \pi \csc(\pi a) \cot(\pi a)$$
A: Firstly, to determine the range of $n$ for which the integral converges, just look at the singularity at $0.$ To integrate, and put the pesky $n$ where it is easier to deal with, make the substitution $u=\log x.$ This will transform your integral to $\int_{-\infty}^{\infty} u \exp(-(n-1) u)/(1+\exp u) du,$ which should be easy by standard contour integration methods. To check your computation, the answer is $\pi^2 \cot(n \pi) \csc(n \pi),$ provided $0<n<1.$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$\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}
