Integration of product of exponential function and natural logarithm I am trying to calculate the following integral:
$$\int_{-1}^{1}\ln(1+\gamma x)\exp\left(-\frac{(z-x)^2}{2}\right)dx$$
with $\gamma\in[-1,1]$, $z\in\mathbb{R}$ and ln being the natural logarithm. I tried a lot of thinks like substitution, integration by parts, used the series expansion of the natural logarithm resp of the exponential function. I also searched for some help in Literatur but couldn't find anything helpful. My question is: Can we solve this integral?
 A: Considering first the restricted problem with $z=0$.
We can find an approximate series for $\gamma\rightarrow 0$. First, expand the logarithm
$$\ln(1+\gamma x)=\gamma x - \frac{(\gamma x)^2}{2}+ ... $$
Now notice that on the integration domain, all terms $x^{odd}$ will integrate to zero. We are left with
$$ I(\gamma)=-\int_{-1}^{1}dx\ \left(  \frac{(\gamma x)^2 }{2}+    \frac{(\gamma x)^4 }{4}+...    \right)e^{-x^2/2}$$
Term by term integration gives
$$ I(\gamma)\sim\sum_{n=1}^\infty A_n \gamma^{2n}, \ \gamma\rightarrow 0$$
With the coefficients
$$ A_n =-\frac{2^{n-1/2}}{n}\left( \Gamma(n+1/2)-\Gamma(n+1/2,1/2) \right)$$
The degree to which this is an improvement depends on how you feel about the incomplete gamma function. By repeated use of the identities
$$\Gamma(s+1,x)=s \Gamma(s,x)+x^s e^{-x} $$
$$ \Gamma(1/2,x)=\sqrt{\pi}  \operatorname{erfc}(\sqrt{x})$$
Every term in the series may be cast into the form of $\operatorname{elementary function} \times \operatorname{erfc}(2^{-1/2}) $. This is an improvement because now all those $\operatorname{erfc}$s appear as constants in the series. The first term alone is a fine approximation
$$ I(\gamma)\sim -\gamma^2 \left(\sqrt{\pi/2}  \left( 1-\operatorname{erfc}(2^{-1/2})\right)-e^{-1/2}     \right),\gamma\rightarrow0$$
More importantly we learn that for small $\gamma$, the integral scales as $-\gamma^2$.
Here is a plot of the first term together with the exact (numerically integrated) result

And the same plot with just three terms

We can also (somewhat) extend this to the case $z\rightarrow 0$.
To avoid issues of non-uniform convergence, I'll consider the integral
$$I(\epsilon)= \int_{-1}^1 \ dx \ln(1+\gamma \epsilon x) \exp \left(-\frac{(x-\epsilon z)^2}{2}\right) $$
This of course is still restricted compared to the original. Expanding the exponential around $\epsilon=0$ will produce a series of 'corrections' to the $\ln$ series. The first nonzero term still occurs at $\epsilon^2$. I find
$$ I(\epsilon)\sim -\epsilon^2 \frac{\gamma(\gamma-2z)}{2\sqrt{e}}\left( \sqrt{2e\pi} \operatorname{erf}(2^{-1/2})-2\right),\epsilon\rightarrow 0$$
We learn the integral scales with $-\gamma(\gamma-2z)$ for small $\epsilon$. Here's a plot with $\gamma=1/2$

As expected, when $z>1$ the approximation is bad: this is when the peak of the Gaussian passes outside of the integral bounds.
For large $z$, I think we should be able to write the exponential in such a way as to use Laplace's method, but I don't see how right now.
A: $$I=\int\log(1+\gamma x)\,\exp\left(-\frac{(z-x)^2}{2}\right)dx$$
Let $x=\sqrt{2} t+z$ to make
$$I=\sqrt{2}\int   \log \big[1+\gamma  \left(\sqrt{2} t+z\right)\big]\,e^{-t^2}\,dt$$ Using Taylor around $t=0$
$$ \log \big[1+\gamma  \left(\sqrt{2} t+z\right)\big]=\log (1+\gamma  z)-\sum_{n=1}^\infty \frac{2^{n/2} \left(-\frac{\gamma }{\gamma  z+1}\right)^n}{n} t^n$$ and
$$J_n=\int t^n\,e^{-t^2}\,dt=-\frac{1}{2} \Gamma \left(\frac{n+1}{2},t^2\right)$$
This would be a nightmare.
