Strategy for Improper Integrals Related to the Beta Function 2 I am looking for the solution of the following integral 

$$\int_0^1 y^k \log\left(1+\left(\frac y{1-y}\right)^a\right)dy,\quad a>0 $$

I really appreciate it if any one can help.
 A: Changing variables $y = \frac{t^{1/a}}{1+t^{1/a}}$ the integral becomes, denoting $\alpha=\frac{1}{a}$:
$$
   \mathcal{I}= \alpha \int_0^\infty \log(1+t) \frac{t^{\alpha(k+1)}}{{(1+t^{\alpha})^{k+2}}} \frac{\mathrm{d}t}{t}
$$
the integral can be thought of as Mellin convolution of two functions $\int_0^\infty G_1(t) G_2(t) \frac{\mathrm{d}t}{t}$, where:
$$
    G_1(t) = \log(1+t)\qquad G_2(t) = \alpha \frac{t^{\alpha(k+1)}}{{(1+t^{\alpha})^{k+2}}} 
$$
Use Plancherel theorem the integral $\mathcal{I}$ can be written as inverse Mellin transform of product of Mellin images of $G_1$ and $G_2$:
$$
  \mathcal{I} = \frac{1}{2 \pi i} \int_{\gamma - i\infty}^{\gamma + i \infty} \hat{G}_1(-s) \hat{G}_2(s) \mathrm{d}s
$$
where
$$
    \hat{G}_1(s) = \int_0^\infty t^{s-1} \log(1+t) \,\mathrm{d}t = \frac{\pi}{s} \frac{1}{\sin\left(\pi s\right)} \quad \text{ convergent for } -1 < \Re(s)<0
$$
$$
    \hat{G}_2(s) = \int_0^\infty t^{s-1} \alpha \frac{t^{\alpha(k+1)}}{{(1+t^{\alpha})^{k+2}}} \,\mathrm{d}t = \frac{\Gamma\left(1- a s\right) \Gamma\left(1+k + a s\right)}{\Gamma(k+2)} \,\, \text{ for } -\alpha(k+1) < \Re(s)<\alpha
$$
and $\gamma$ being an arbitrary real constant, subject to $0<\gamma<\min(1,\alpha)$:
$$
    \mathcal{I} = \frac{1}{2 \pi i} \int_{\gamma - i\infty}^{\gamma + i \infty} \frac{\pi}{s} \frac{1}{\sin\left(\pi s\right)}  \frac{\Gamma\left(1- a s\right) \Gamma\left(1+k + a s\right)}{\Gamma(k+2)} \mathrm{d}s 
$$
The above integral is known as Fox H-function, using $\frac{\pi}{s} \frac{1}{\sin\left(\pi s\right)} = \Gamma(1-s)\frac{\left(\Gamma(s)\right)^2}{\Gamma(s+1)}$:
$$
  \mathcal{I} = \frac{1}{\Gamma(k+2)} H_{3,3}^{3,2}\left( 1 \left| \begin{array}{ccc}  \left(0,1\right) & \left(0,a\right) & \left(1,1\right) \\ (0,1) & (0,1) & (k+1, a) \end{array} \right. \right)
$$
For $a=1$ the answer is given as 
$$ 
    \mathcal{I}(k,1) = \frac{H_{k+1}}{k+1}
$$
where $H_k$ is the Harmonic number, and for $a=2$, the answer is in terms of the Meijer's G-function:
$$
 \mathcal{I}(k,2) = \frac{2^k }{\pi  (k+1)!} G_{4,4}^{4,3}\left(1\left|
\begin{array}{c}
 0,0,\frac{1}{2},1 \\
 0,0,\frac{k+1}{2},\frac{k+2}{2} \\
\end{array}
\right.\right)
$$
which likely can be simplified further:
$$
   \mathcal{I}(k,2) = \frac{2}{k+1} H_{k+2} - \frac{4}{k+1} \Im \left( \Phi \left( \sqrt{2} \mathrm{e}^{i \pi/4}, 1, k+3 \right) \right)
$$
where $\Phi$ is the Lerch's transcendent.
