Expected value of applying the sigmoid function to a normal distribution Short version:
I would like to calculate the expected value if you apply the sigmoid function $\frac{1}{1+e^{-x}}$ to a normal distribution with expected value $\mu$ and standard deviation $\sigma$.
If I'm correct this corresponds to the following integral:
$$\int_{-\infty}^\infty \frac{1}{1+e^{-x}} \frac{1}{\sigma\sqrt{2\pi}}\ e^{ -\frac{(x-\mu)^2}{2\sigma^2} } dx$$
However, I can't solve this integral. I've tried manually, with Maple and with Wolfram|Alpha, but didn't get anywhere.
Some background info (why I want to do this):
Sigmoid functions are used in artificial neural networks as an activation function, mapping a value of $(-\infty,\infty)$ to $(0,1)$. Often this value is used directly in further calculations but sometimes (e.g. in RBM's) it's first stochastically rounded to a 0 or a 1, with the probabililty of a 1 being that value. The stochasticity helps the learning, but is sometimes not desired when you finally use the network. Just using the normal non-stochastic methods on a network that you trained stochastically doesn't work though. It changes the expected result, because (in short):
$$\operatorname{E}[S(X)] \neq S(\operatorname{E}[X])$$
for most X. However, if you approximate X as a normal distribution and could somehow calculate this expected value, you could eliminate most of the bias. That's what I'm trying to do.

Apart from the the MacLaurin approximation, the usual way to compute that integral in Statistics is to approximate the sigmoid with a probit function.
More specifically $\mathrm{sigm}(a) \approx \Phi(\lambda a)$ with $\lambda^2=\pi/8$.
Then the result would be:
$$\int \mathrm{sigm}(x) \, N(x \mid \mu,\sigma^2) \, dx
\approx \int \Phi(\lambda x) \, N(x \mid \mu,\sigma^2) \, dx
= \Phi\left(\frac{\mu}{\sqrt{\lambda^{-2} + \sigma^2}}\right).$$
