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I am having trouble with calculating the following integral: $$ \int_{0}^{\infty} \ln{(1 + \alpha x)\, G^{k,0}_{k,k}\left[e^{-x}\left|^{(a_k)}_{(b_k)} \right. \right]} \, dx, $$ where $\alpha > 0$ and $G^{m,n}_{p,q}[\cdot | \cdot]$ is the Meijer-G function. Any ideas or references would be very helpful. Thank you!

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I would like to solve the following special case. Let $\alpha,k>0$ and $a_n = a$, $b_n = a-1$ for all $n=1,\dots,k$, where $a>1$.

We will use the following Meijer G identity. For all $|z|<1$ we have: $$ G_{m,m}^{m,0}\left(z\,\middle|\begin{array}c a,\dots,a\\a-1,\dots,a-1\end{array}\right) = \frac{(-1)^{m+1} z^{a-1} \ln^{m-1}(z)}{(m-1)!}. $$

So using this we have $$\begin{align} \int_{0}^{\infty} \ln(1 + \alpha x)\,G_{k,k}^{k,0}\left(e^{-x}\,\middle|\begin{array}c a,\dots,a\\a-1,\dots,a-1\end{array}\right)\,dx \\ = \frac{1}{(k-1)!}\int_0^\infty x^{k-1}\left(e^{-x}\right)^{a-1}\ln(1+\alpha x)\,dx.\end{align} $$ This integral has the following Meijer G representation: $$ \frac{1}{(k-1)!}\int_0^\infty x^{k-1}\left(e^{-x}\right)^{a-1}\ln(1+\alpha x)\,dx = \frac{1}{\alpha^{k}(k-1)!}G_{2,3}^{3,1}\left(\frac{a-1}{\alpha}\middle| \begin{array}{c} -k,1-k \\ 0,-k,-k \\\end{array}\right). $$ Specially for $k=1$ we have: $$ \int_0^\infty \left(e^{-x}\right)^{a-1}\ln(1+\alpha x)\,dx = \frac{1}{1-a}\exp\left(\frac{a-1}{\alpha}\right)\operatorname{Ei}\left(\frac{1-a}{\alpha}\right), $$ where $\operatorname{Ei}$ is the exponential integral. You could solve the $k=1$ case more generally using this identity.

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