I had posted this on Stats Stackexchange as I initially thought a probabilistic approach would be best suited. But posting it here for as it has more traffic.
Problem
Evaluate $$I=\int_{-\infty}^\infty \frac{e^{-\frac{1}{2}\left(\frac{x-\mu)}{\sigma} \right)^2}}{\sigma \sqrt{2 \pi}}\frac{1}{1+e^{-x}}\, \mathrm{d}x$$My attempt
Now the first part of the integrand represents the p.d.f of the normal dist and other other the c.d.f. of the standard logistic.So I wrote it as follows. Say $X \sim N(\mu,\sigma^2)$, $Y\sim \mathrm{Logistic}(0,1)$. Let $f_X,F_X$ denote the p.d.f. and c.d.f. of $X$ respectively and similarly $g_Y,G_Y$ for $Y$.
The integral to find is then given as follows, \begin{align*} I&=\int_{-\infty}^\infty f_X(x)G_Y(x)\, \mathrm{d}x\\ &\text{Integrating by parts we get,}\\ &=\underbrace{\left. F_X(x)G_Y(x)\right|_{-\infty}^{\infty}}_{=1}-\int_{-\infty}^\infty g_Y(x)F_X(x)\, \mathrm{d}x\\ &\text{but since we know} \int_{-\infty}^\infty g_Y(x) \mathrm{d}x=1 \text{, we can simplify.}\\ &=\int_{-\infty}^{\infty} g_Y(x)(1-F_X(x))\, \mathrm{d}x\\ &=\int_{-\infty}^{\infty} g_Y(x) \mathbb{P}(X>x)\, \mathrm{d}x\\ &\text{With the law of total probability we get}\\ &=P(X>Y). \end{align*}
Here I'm stuck. Not sure how to simplify it. I read about the convolution of probability distributions and it seems like the initial integral is like the c.d.f. of $X+Y$.
But I don't see how to proceed with that direction either.