Integral involving the CDF of normal distribution I am doing some research and got stuck in solving the following integral (which  I am not sure whether it has a closed form solution or not, I hope it has:))
Here is the integral:
$\int_{-\infty}^{+\infty} e^{-(x-a)^2}N(cx+d)dx$
where
$N(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x} e^{-\frac{z^2}{2}}dz$
any help would greatly be appreciated. (by the way I am not a grad student in math, so please be specific in answers) Thanks a lot. 
 A: In what follows, I am assuming that $c > 0$.
Your $N(x)$ is commonly denoted $\Phi(x)$ and called the cumulative probability distribution function of the standard normal random variable. So, what you have is the evaluation of
$$\int_{-\infty}^\infty e^{-(x-a)^2}\Phi(cx+d)\,\mathrm dx = \sqrt{\pi}\int_{-\infty}^\infty \frac{e^{-(x-a)^2}}{\sqrt{\pi}}\Phi(cx+d)\mathrm dx$$
where the thing multiplying $\Phi(cx+d)$ in the integrand is the probability density function of a normal random variable $X$ with mean $a$ and variance $\frac{1}{2}$.
If $Y$ is another normal random variable with mean $\mu$ and variance $\sigma^2$,
then $$P\{Y \leq x\} = \Phi\left(\frac{x-\mu}{\sigma}\right) = \Phi(cx+d)$$
if we choose $\sigma = \frac{1}{c}$ and $\mu = -\sigma d= -\frac{d}{c}$.
Now, if $X$ and $Y$
are taken to be independent random variables, then the conditional
probability 
$P\{Y \leq X \mid X = x\}$ is just $\Phi(cx+d)$ and the unconditional probability
that $Y$ does not exceed $X$ is, by the law of total probability,
$$\begin{align}
P\{Y \leq X\} &= \int_{-\infty}^\infty P\{Y \leq X \mid X = x\} f_X(x)\,\mathrm dx\\
&= \int_{-\infty}^\infty \frac{e^{-(x-a)^2}}{\sqrt{\pi}}\Phi(cx+d)\mathrm dx
\end{align}$$
which is the integral you want to evaluate (except for the $\sqrt{\pi}$ factor).
But, $P\{Y \leq X\} = P\{Y-X\leq 0\}$ where $Y-X$ is also a normal random
variable with mean $\hat{\mu}= -\frac{d}{c}-a$ and variance 
$\hat{\sigma}^2 = \frac{1}{c}+\frac{1}{2}$
and so 
$$P\{Y \leq X\} = \Phi\left(\frac{0-\hat{\mu}}{\hat{\sigma}}\right)
= \Phi\left(\frac{\frac{d}{c}+a}{\sqrt{\frac{1}{c}+\frac{1}{2}}}\right)$$
giving

$$\int_{-\infty}^\infty e^{-(x-a)^2}\Phi(cx+d)\,\mathrm dx =
\sqrt{\pi}\cdot\Phi\left(\frac{\frac{d}{c}+a}{\sqrt{\frac{1}{c}+\frac{1}{2}}}\right).$$

For $a=d=0$, the argument of $\Phi(\cdot)$ is $0$ and so, 
as pointed out in @wolfie's comment, your integral has
value $\frac{\sqrt{\pi}}{2}$ since $\Phi(0) = \frac{1}{2}$.
Similarly, the value is $\frac{\sqrt{\pi}}{2}$ as long as 
the ratio $\frac{d}{c}$ equals $-a$, and in particular
when $c = \sqrt{2}$ and $d=-a\sqrt{2}$. Note that
the value of $d$ is the negative of what is stated in @Lucas's comment. 
I am  not sure about the value $\sqrt{\pi}/8$ stated
by Lucas for the case $c=\sqrt{2}$, $d=+a\sqrt{2}$.
What if $c < 0$? Well, $\Phi(cx+d)=1-\Phi(-cx-d)$ and so a similar result can
be worked out.
A closely related question came up on stats.SE today and the accepted answer there uses a different method that you
might want to try on your problem.
