A definite integral with with $\mathrm{e}^{\frac{-1}{\theta(1+x)}}$ in terms of Meijer G-function I have solved the following definite Integral using Mathematica. However, as I am not familiar with Meijer G-functions, it is not trivial for me to use relevant functional identities involving Meijer G-functions in order to prove
$$
\int_0^\infty\frac{\log(1+x)}{\theta^{\kappa} (1+x)^{\kappa+1}} \ \mathrm{e}^{-\frac{1}{\theta(1+x)}} \ \mathrm{d}x = \Gamma\left( \kappa,\frac{1}{\theta} \right) \left( \log(1/\theta) + \log(\theta) \right) + G^{3,0}_{2,3} \left(\frac{1}{\theta} \middle|
\begin{array}{c}
 1,1 \\
 0,0,\kappa \\
\end{array} \right) - \Gamma(\kappa) \left( \log(\theta)+\psi^{(0)}(\kappa) \right)
$$
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 related to:
A definite integral in terms of Meijer G-function
to which @Leucippus has given an interesting answer
 A: Let $\alpha = -1/\theta$. Show that
$$ \int_{\mathbb R^+} \frac {\ln(x + 1)} {(x + 1)^p} dx =
-\frac d {dp} \int_{\mathbb R^+} \frac {1} {(x + 1)^p} dx =
\frac 1 {(p - 1)^2}, \quad \operatorname {Re} p > 1$$
and that
$$\int_{\mathbb R^+}
 \frac {\ln(x + 1)} {(x + 1)^{\kappa + 1}} e^{\alpha/(x + 1)} dx =
\sum_{j \geq 0} \int_{\mathbb R^+}
 \frac {\ln(x + 1)} {(x + 1)^{\kappa + 1}}
 \frac {\alpha^j} {j! (x + 1)^j} dx = \\
\sum_{j \geq 0} \frac {\alpha^j} {j! (\kappa + j)^2} =
\kappa^{-2} \hspace {1px} {_2 F_2}(\kappa, \kappa; \kappa + 1, \kappa + 1; \alpha)$$
(the last step is optional). Then apply the residue theorem to the integral representation of the G-function:
$$\operatorname* {Res}_{s = 0}
 \frac {\Gamma^2(s) \Gamma(\kappa + s)} {\Gamma^2(1 + s)} (-\alpha)^{-s} =
\Gamma(\kappa) (\psi(\kappa) - \ln(-\alpha)), \\
\operatorname* {Res}_{s = -\kappa - j}
 \frac {\Gamma^2(s) \Gamma(\kappa + s)} {\Gamma^2(1 + s)} (-\alpha)^{-s} =
\frac {(-\alpha)^\kappa \alpha^j} {j! (\kappa + j)^2},
\quad j \in \mathbb N^0.$$
If we're taking the principal branches of $z^p$ and $\ln z$, then $\ln \theta + \ln(1/\theta) = 0$  and $\theta^p (1/\theta)^p = 1$ for $\theta \not \in (-\infty, 0]$, so the two formulas match for those values of $\theta$.
