A definite integral with with $\mathrm{e}^{\frac{-1}{(1+x)}}$ in terms of Meijer G-function

Question

I am trying to prove the following definite integral calculated in Mathematica where a Meijer G-function is considered

$$ \int_0^\infty\frac{\log(1+x)}{(1+x)^{\kappa+1}} \ \mathrm{e}^{-\frac{1}{(1+x)}} \ \mathrm{d}x = G^{3,0}_{2,3} \left(1 \middle| \begin{array}{c} 1,1 \\ 0,0,\kappa \\ \end{array} \right) - \Gamma(\kappa) \ \psi^{(0)}(\kappa) $$

with the condition that $(\Re(\kappa)>0)$

This equality came out of Mathematica, in whose syntax the right-hand side reads

MeijerG[{{}, {1, 1}}, {{0, 0, k}, {}}, 
  1/\[Theta]] + (Log[1/\[Theta]] + Log[\[Theta]])*
  Gamma[k, 1/\[Theta]] - 
    Gamma[k]*(Log[\[Theta]] + PolyGamma[0, k])

This question is related to

NB: This question is a particular case of a more general one (A definite integral with with $\mathrm{e}^{\frac{-1}{\theta(1+x)}}$ in terms of Meijer G-function) where $\theta=1$. It is also related to: A definite integral in terms of Meijer G-function to which @Leucippus has given an interesting answer

Answer 1

not an answer of how to compute it. But a comment.

Your integeral is, with change of variables, $$ \int_1^\infty \frac{\log x}{x^{\kappa+1}}\;\exp\left(\frac{-1}{x}\right)\;dx $$ But the integral starting at $0$ is much easier $$ \int_0^\infty \frac{\log x}{x^{\kappa+1}}\;\exp\left(\frac{-1}{x}\right)\;dx = -\psi(\kappa)\Gamma(\kappa) $$ so the Meijer G-function is the definite integral $\int_0^1$. Maple does this not using Meijer G-functions, but \begin{align} &\int_0^1 \frac{\log x}{x^{\kappa+1}}\;\exp\left(\frac{-1}{x}\right)\;dx \\&= {\frac {-{\mbox{$_2$F$_2$}(\kappa,\kappa;\,\kappa+1,\kappa+1;\,-1)} \sin \left( \pi\,\kappa \right) +{\kappa}^{2}\Gamma \left( \kappa \right) \left( \pi\,\cos \left( \pi\,\kappa \right) -\sin \left( \pi \,\kappa \right) \psi \left( 1-\kappa \right) \right) }{\sin \left( \pi\,\kappa \right) {\kappa}^{2}}} \end{align} perhaps useful at least when $\kappa$ is not an integer.