How can I calculate $\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$ for the logistic distribution? Take the cumulative distribution function (cdf)
$\Phi(z) = \frac{1}{1+\exp(-z)}$
and probability density function (pdf)
$\phi(z)=\Phi(z)(1-\Phi(z))$
of the logistic distribution.
How can one calculate the integral below?
$$\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$$
 A: Too long for a comment....
This question mixes up the words density and distribution: $\Phi(z) = \frac{1}{1+\exp(-z)}$ is the distribution function  of a logistic random variable, not the density as the OP calls it, while $\phi(z) = \Phi(z)(1-\Phi(z))$ is the corresponding density function even though the OP calls it the distribution function.  (OK, I see that the OP has corrected the nomenclature in the question, making the above not inapplicable any more).
Be that as it may, the result
$$\int_{-\infty}^\infty \Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw = \Phi\left(\frac{-a}{\sqrt{b^2+1}}\right)$$
for normal random variables has been proved in two different ways in the answers to this question. My answer to this question explicitly uses the fact that the difference of two independent normal random variables is itself normally distributed. I suspect that a corresponding result does not hold for logistic random variables and so it might not be possible to adapt my method to solve the OP's problem. On the other hand, the answer by whuber is based on solving partial differential equations, and might be more adaptable to what the OP is trying to do with logistic random variables.
