I try to solve a gaussian convolution which turns out to be harder than expected. Here is the isolated difficulty: \begin{align} \int_c^{+\infty}\phi(x-y)\big(1-\Phi(y)\big)dy \end{align} where $\phi$ and $\Phi$ are the pdf and cdf of the standard gaussian function.
My idea was to write explicitly the Heaviside and to use the integral property of convolution. But I don't know whether it is correct in this case:
\begin{align} &\int_c^{+\infty}\phi(x-y)\big(1-\Phi(y)\big)dy \\& =\int_{-\infty}^{+\infty}\phi(x-y)H(y-c)\int_y^{+\infty}\phi(z)dzdy \\ &= \int_{-\infty}^{+\infty}\phi(x-y)H(x-y+c-x)\int_{-\infty}^{-y}\phi(z)dzdy \\ &= \int_{-\infty}^{+\infty}\phi(\bar{x}-\bar{y})\bar{H}_{x-c}(\bar{x}-\bar{y})\int_{-\infty}^{\bar{y}}\phi(z)dzd\bar{y} \\ &= \int_{-\infty}^{\bar{x}}\int_{-\infty}^{+\infty}\phi(z-\bar{y})\bar{H}_{x-c}(z-\bar{y})\phi(\bar{y})d\bar{y} dz=\int_{-\infty}^{\bar{x}}(\phi\bar{H}_{x-c})*\phi(z) dz \end{align} which I can compute. But is the passage between the two last lines correct? Is there another way? Here $\bar{x}=-x$ and for the Heaviside $\bar{H}(t)=H(-t)=1-H(t)$.
Some informations about this puzzle, if anybody ever meet this situation:
This integral is not documented in famous tables of integrals, which is not a good sign. Most probably, this integral can not be expressed in a simpler closed form.
This integral arised in a probability problem. I wanted to find the distribution of a random variable defined as a sum of two other random variables with known distributions. I have ended up in this dead end convolving them without noticing that they where in fact not independent variables.