Strategy for Improper Integrals Related to the Beta Function 2

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.