An integral involving a Gaussian and a logarithm. Consider a following integral:
\begin{equation}
{\mathcal I}(A) := \int\limits_{\mathbb R} \log\left(1+ A \xi\right) \cdot \frac{\exp(-\frac{\xi^2}{2})}{\sqrt{2 \pi}} d\xi
\end{equation}
By using the trick $\left. d \xi^a/d a \right|_{a=0} = \log(\xi)$ then by substituting for $\xi^2/2$ and then by expanding the resulting power-law term into a binomial expansion and integrating the result term by term using the definition of the Gamma function I have shown that:
\begin{equation}
{\mathcal I}(A) = \log(\frac{A}{\sqrt{2}})+ \imath \frac{\pi}{2} - \frac{\gamma}{2} + \frac{1}{2 A^2} F_{2,2}\left[\begin{array}{rr} 1 & 1 \\ 2 & 3/2\end{array};-\frac{1}{2 A^2}\right]- \imath \frac{\sqrt{\pi}}{\sqrt{2} A} F_{1,1}\left[\begin{array}{r} \frac{1}{2} \\ \frac{3}{2} \end{array};-\frac{1}{2 A^2}\right]
\end{equation}
The real part of the function in question is plotted below:

It clearly behaves as a parabola ${\mathcal I}(A) \simeq -A^2/2$ for $A \rightarrow 0$ and as a logarithm ${\mathcal I}(A) \simeq \log(A)$ for $A\rightarrow \infty$.
Now the question is how would you calculate the integral in question if the Gaussian was replaced by a Tsallis' distribution $\rho_q(\xi) := 1/C_q \cdot e_q(-1/2 \xi^2)$ where $e_q(\xi) := [1+(1-q) \xi]^{1/(1-q)}$ for $q> 1$ ? Here $C_q:=\left(\sqrt{2\pi}/\sqrt{q-1}\right) \cdot \Gamma\left(\frac{3-q}{2(q-1)}\right)/\Gamma\left(\frac{1}{q-1}\right)$ is a normalization constant.
 A: Let us denote the unknown integral with the Tsallis' distribution ${\mathcal I}_q(A)$. In what follows we assume that $3 > q> 1$.
Note that the Tsallis' distribution has a very nice integral representation. In fact it can be seen as a  Gaussian  whose inverse variance is Gamma distributed. To be precise the following equality holds:
\begin{equation}
\rho_q(\xi) = \frac{1}{\Gamma(\frac{1}{q-1}) C_q} \int\limits_0^\infty \frac{ ds}{s} s^{\frac{1}{q-1}} e^{-s} e^{-s (q-1) \frac{\xi^2}{2}}
\end{equation} 
Now, replacing the Gaussian in the formulation of the question by $\rho_q(\xi)$ and swapping the order of integration we immediately see that:
\begin{equation}
{\mathcal I}_q(A) = \frac{1}{\Gamma(\frac{1}{q-1}-\frac{1}{2})} \int\limits_0^\infty \frac{d s}{s} s^{\frac{1}{q-1}-\frac{1}{2}} e^{-s} {\mathcal I}\left(\frac{A}{\sqrt{s (q-1)}}\right)
\end{equation}
Now, the formula for ${\mathcal I}(A)$ consists of four types of terms, firstly a constant$(A)$, secondly a a $\log(A)$, thirdly a negative even power of $A$ and finally a negative odd power of $A$. Replacing ${\mathcal I}$ by those terms in the above equation, we can clearly see (since the integrals are trivial) that except for the logarithm , the terms obtain a multiplicative factor as a result of the above transformation. The case of the logarithm is slightly more complicated since it involves the di-gamma function $\psi$. Bringing everything together we readily have:
\begin{equation}
{\mathcal I}_q(A) = \log(\frac{A}{\sqrt{2(q-1)}}) - \frac{1}{2} \psi(\frac{1}{q-1}-\frac{1}{2}) + \imath \frac{\pi}{2} - \frac{\gamma}{2}+
\frac{1}{2 A^2}\left(1-\frac{1}{2}(q-1)\right) F_{3,2}\left[\begin{array}{rrr}1 & 1 & \frac{1}{q-1}+\frac{1}{2} \\ 2 & 3/2 \end{array};-\frac{(q-1)}{2 A^2}\right]
-\imath \sqrt{\pi} \frac{\sqrt{q-1}}{\sqrt{2} A} \frac{\Gamma(\frac{1}{q-1})}{\Gamma(\frac{1}{q-1}-\frac{1}{2})} F_{2,1}\left[\begin{array}{rr}1/2 & \frac{1}{q-1}\\3/2\end{array};-\frac{(q-1)}{2 A^2}\right]
\end{equation} 
The quantity above clearly tends to ${\mathcal I}$ as $q\rightarrow 1$, as it should be.
