Is there any analytic solution to the following integral?
$$-\frac{2}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}\ln{(1+e^x)}\exp{\Big\{-\frac{(x-\mu)^2}{2\sigma^2}}\Big\}dx$$
where $\mu\in\mathbb{R}$ and $\sigma\in\mathbb{R}^+$.
I've tried to exploit the Taylor series of the natural logarithm (using $\ln{(1+e^x)}$ from $0$ to $-\infty$ and $x+\ln{(1+e^{-x})}$ from $0$ to $\infty$) to get the following equivalent form:
$$\mu\Big\{-1+erf\Big(\frac{\mu}{\sigma\sqrt{2}}\Big)\Big\}-\frac{2\sigma}{\sqrt{2\pi}}e^{-\frac{\mu^2}{2\sigma^2}}+\frac{2e^{-\frac{\mu^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sum_{l=0}^{\infty}\frac{(-1)^l}{l!}n^l\big(2\sigma^2\big)^{\frac{l+1}{2}}\Gamma\Big(\frac{l+1}{2}\Big)\sum_{0\leq k\leq l,\text{ k ODD}}\binom{l}{k}\Big(\frac{\mu}{n\sigma^2}\Big)^k$$
where $\Gamma(\cdot)$ is the gamma function.
But I don't know how to proceed from that.