contour integration of a function with two branch points . 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 .
 A: To evaluate the integral using contour integration, consider the integral
$$\oint_C dz \frac{z^{-p} \log{(1+z)}}{1+z} $$
where $C$ is the following contour:

The magnitude of the integral about the large arc of radius $R$ behaves as $\frac{\log{R}}{R^p}$ as $R \to \infty$ and thus vanishes.  Let the radius of the small circular arcs be $\epsilon$.  The contour integral is then equal to, in this limit,
$$\left (1-e^{-i 2 \pi p} \right ) \int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} - e^{-i \pi p}\int_{\infty}^{1+\epsilon}  dx \frac{x^{-p} [\log{(x-1)}+i \pi]}{1-x} \\ - e^{-i \pi p}\int_{1+\epsilon}^{\infty}  dx \frac{x^{-p} [\log{(x-1)}-i \pi]}{1-x}+i \epsilon \int_{\pi}^{-\pi} d\phi \,e^{i \phi} \frac{(e^{i \pi}+\epsilon e^{i \phi})^{-p} \log{(\epsilon e^{i \phi})}}{\epsilon e^{i \phi}}$$
The first integral represents the integral about the branch cut along the positive real axis, which concerns the $z^{-p}$ term only.  Note that the integral about the origin vanishes as $\epsilon \to 0$.  The second and third integrals represent the integrals along each side of the branch cut on the negative axis. Note that, along this branch cut, the argument of $z^{-p}$ is $-\pi p$ on either side of the branch cut, as the branch cut there concerns the log term only.  The fourth integral is the integral about the branch point $z=-1$.
By Cauchy's theorem, the contour integral is zero.  Thus, we have
$$\left (1-e^{-i 2 \pi p} \right ) \int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = i 2 \pi \, e^{-i \pi p} \int_{1+\epsilon}^{\infty} dx  \frac{x^{-p}}{x-1} + i e^{-i \pi p} \int_{-\pi}^{\pi} d\phi \left ( \log{\epsilon} + i \phi \right )$$
Note that
$$\begin{align}\int_{1+\epsilon}^{\infty} dx  \frac{x^{-p}}{x-1} &= \int_0^{1-\epsilon} dx \frac{x^{p-1}}{1-x} + O(\epsilon)  \end{align} $$
and
$$\begin{align}\int_0^{1-\epsilon} dx \frac{x^{p-1}}{1-x} &= \int_0^{1-\epsilon} dx \, x^p \left (\frac1x + \frac1{1-x} \right ) \\ &= \frac1p (1-\epsilon)^p - \left [\log{(1-x)} x^p \right ]_0^{1-\epsilon} + p \int_0^{1-\epsilon} dx \, x^{p-1} \log{(1-x)}\\ &= \frac1p - \log{\epsilon} + p \int_0^1 dx \, x^{p-1} \log{(1-x)} + O(\epsilon)\\ &= \frac1p - \log{\epsilon} - p \sum_{k=1}^{\infty} \frac1{k (k+p)}+ O(\epsilon) \\ &= \frac1p - \log{\epsilon}-\gamma -\psi(1+p)+ O(\epsilon) \\ &= - \log{\epsilon} - (\gamma + \psi(p))+ O(\epsilon) \end{align} $$
where $\psi$ is the digamma function.  Note that the singular $\log{\epsilon}$ pieces cancel.  The second piece of the second integral on the RHS vanishes as it is an odd function over a symmetric interval.  Thus, we may take the limit as $\epsilon \to 0$ and we get
$$\int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = -\frac{i 2 \pi \, e^{-i \pi p}}{1-e^{-i 2 \pi p}} (\gamma + \psi(p)) = -\frac{\pi}{\sin{\pi p}} (\gamma + \psi(p))$$
Alternatively, we may express this so that it is clear that the integral takes a positive value:

$$\int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = \frac{\pi}{\sin{\pi p}} \left (\frac1p - H_p \right ) $$

where $H_p$ is the analytically continued harmonic number at $p$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\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}
