Many of us have seen the evaluation of the integral
$$\int^{\infty}_0 \frac{dx}{x^p(1+x)}\, dx \,\,\, 0<\Re(p)<1$$
It can be solved using contour integration or beta function .
I thought of how to solve the integral
$$\int^{\infty}_0 \frac{\log(1+x)}{x^p(1+x)}\, dx \,\,\, 0<\Re(p)<1$$
It can be solved using real methods as follows
consider the following integral
$$\int^{\infty}_0 x^{-p}(1+x)^{s-1} dx= \frac{\Gamma(1-p)\Gamma(p-s)}{\Gamma(1-s)}$$
Differentiating with respect to $s$ we get
$$\int^{\infty}_0 x^{-p}(1+x)^{s-1}\log(1+x) dx=\frac{\Gamma(1-p)\Gamma(p-s)}{\Gamma(1-s)} \left(\psi_0 (1-s)- \psi_0(p-s)\right)$$
at $s =0$ we get
$$\int^{\infty}_0 x^{-p}\frac{\log(1+x)}{1+x} dx=\frac{\pi}{\sin(\pi p)} \left(\psi_0 (1)- \psi_0(p)\right)$$
where i used the reflection formula .
Statement of question
How to solve the following integral using contour integration
$$\int^{\infty}_0 \frac{\log(1+x)}{x^p(1+x)}\, dx \,\,\, 0<\Re(p)<1$$
I thought we can use the following contour
So the function
$$F(z) = \frac{e^{-p \log(z)}\log(1+z)}{(1+z)} $$
is analytic in and on the contour by choosing the branch cut of $e^{-p \log(z)}$ as $0\leq \text{Arg}(z)<2\pi$ and the branch cut of $\log(1+z)$ as $0\leq \text{Arg}(z+1)<2\pi$ so the function $F(z)$ is analytic everywhere except at $z\geq -1$ . I am finding difficulty finding the integral on the branch point $z=-1$ it seems there is a contribution of the branch point and the pole .
Please don't make any substitutions or simplifications for the integral. Feel free to use another contour if my choice was wrong .
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\ln\pars{1 + x} \over x^{p}\pars{1 + x}}\,\dd x \,\right\vert_{\,0\ <\ \Re\pars{p}\ <\ 2}}:\ {\Large ?}}$.
\begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\ln\pars{1 + x} \over x^{p}\pars{1 + x}}\,\dd x \,\right\vert_{\,0\ <\ \Re\pars{p}\ <\ 2}} = \int_{0}^{\infty}x^{-p} \,\bracks{-\sum_{k = 0}^{\infty}H_{k}\pars{-x}^{k}}\,\dd x \\[5mm] = &\ -\int_{0}^{\infty}x^{\pars{\color{red}{1 - p}} - 1} \,\bracks{\sum_{k = 0}^{\infty} \color{red}{H_{k}\,\Gamma\pars{1 + k}}{\pars{-x}^{k} \over k!}} \,\dd x \\[5mm] = &\ -\Gamma\pars{1 - p}\bracks{H_{p - 1}\,\,\Gamma\pars{p}} = \bbx{-\pi\,{H_{p - 1} \over \sin\pars{\pi p}}} \\ & \end{align}