I am trying to solve integrals that look like the following:
$\int_{-\infty}^{\infty} \phi(cx+d)^n e^{-(x-a)^2}dx$
where $\phi$ is the cdf of the standard normal distribution. I have no idea how to proceed and it would be nice to know the general way to solve this, since I also have to integrate other terms similar to this.
I don't even know if such terms are exactly solvable, but learning how to estimate them would also be really useful for my research.
Let $c\in {\mathbb R}$ and $d \in {\mathbb R}$. Without the loss of generality we can consider the following integral: \begin{equation} f^{(n)}(c,d):= \int\limits_{{\mathbb R}} [\Phi(c x+ d)]^n e^{-x^2} dx \end{equation} where $\Phi(x) := P(N(0,1)\le x)=1/2(1+erf(x/\sqrt{2}))$.
Now by differentiating with respect to $c$ we get a following identity: \begin{eqnarray} \partial_c f^{(n)}(c,d) &=& -\frac{ n d}{\sqrt{\pi}} \cdot \frac{c}{(c^2+2)^{3/2}} e^{-\frac{d^2}{c^2+2}} \cdot f^{(n-1)}(\frac{\sqrt{2} c}{\sqrt{2+c^2}},\frac{2 d}{c^2+2})+\\ &&\frac{n(n-1)}{2\pi} \cdot \frac{c}{\sqrt{1+c^2} (c^2+2)} \cdot e^{-\frac{d^2}{c^2+1}} \cdot f^{(n-2)}(\frac{c}{\sqrt{1+c^2}},\frac{d}{1+c^2}) \quad (i) \end{eqnarray} subject to conditions $f^{(n)}(0,d) = [\Phi(d)]^n \sqrt{\pi}$ and $f^{(0)}(c,d)=\sqrt{\pi}$.
Likewise by differentiating with respect to $d$ we obtain the following identity: \begin{eqnarray} \partial_d f^{(n)}(c,d) = \frac{n}{\sqrt{\pi}} \cdot \frac{1}{\sqrt{2+c^2}} \cdot e^{-\frac{d^2}{c^2+2}} \cdot f^{(n-1)}(\frac{\sqrt{2} c}{\sqrt{2+c^2}},\frac{2 d}{c^2+2}) \quad (ii) \end{eqnarray} subject to conditions $f^{(n)}(-\infty,d) = 0$ and $f^{(0)}(c,d)=\sqrt{\pi}$.
Now, integrating $(ii)$ over $d$ from minus infinity to $d$ we obtain: \begin{eqnarray} f^{(0)}(c,d)&=& \sqrt{\pi}\\ f^{(1)}(c,d)&=& \frac{\sqrt{\pi}}{2}\left(1+ erf[\frac{d}{\sqrt{c^2+2}}] \right) \\ f^{(2)}(c,d)&=& \sqrt{\pi} \left( \frac{1}{2} (1+erf[\frac{d}{\sqrt{c^2+2}}]) - 2 T(\frac{d \sqrt{2}}{\sqrt{c^2+2}},\frac{1}{\sqrt{1+c^2}})\right) \\ f^{(3)}(c,d)&=& \frac{3}{2} \sqrt{\pi } \left(\frac{1}{2} \left(-4 \text{erf}\left(\frac{d}{\sqrt{c^2+2}}\right) T\left(\frac{\sqrt{2} d}{\sqrt{c^2+1} \sqrt{c^2+2}},\sqrt{\frac{c^2+2}{3 c^2+2}}\right)-4 T\left(\frac{\sqrt{2} d}{\sqrt{c^2+2}},\frac{1}{\sqrt{c^2+1}}\right)+\text{erf}\left(\frac{d}{\sqrt{c^2+2}}\right)+1\right)-4 T\left(-\frac{\sqrt{2} d}{\sqrt{\left(c^2+1\right) \left(c^2+2\right)}},\left\{\sqrt{c^2+1},\sqrt{\frac{c^2+2}{3 c^2+2}}\right\}\right)\right) \end{eqnarray}
where $T(h,a):=\int\limits_h^\infty \phi(\xi) 1/2 erf[a \xi/\sqrt{2}] d\xi$ is the Owen's T function and $T(h,\{a_1,a_2\}):=\int\limits_h^\infty \phi(\xi) \prod\limits_{j=1}^2 (1/2 erf[a_j \xi/\sqrt{2}]) d\xi$ .
Note: If we were to integrate $(i)$ over $c$ from zero to $c$ then of course get the same results as above but in addition get a quite interesting result: \begin{eqnarray} &&\int\limits_{\frac{1}{\sqrt{2}}}^{\sqrt{\frac{1}{2+c^2}}} e^{-d^2 u^2} erf[\frac{d \cdot u^2}{\sqrt{1-u^2}}] du=\\ && % \frac{2 \sqrt{\pi } \left(e^{d^2} T\left(\sqrt{2} d,\sqrt{\frac{1}{c^2+1}}\right)-T\left(\frac{\sqrt{2} d}{\sqrt{c^2+2}},\frac{1}{\sqrt{c^2+1}}\right)\right)}{d}- \frac{2 \sqrt{\pi } \left(e^{d^2} T\left(\sqrt{2} d,1\right)-T(d,1)\right)}{d} \\ && \frac{3}{2\sqrt{\pi}} \int\limits_1^{\frac{1}{\sqrt{1+c^2}}} e^{-d^2 u^2} \frac{erf[\frac{d u^2}{\sqrt{3-u^2}}]}{1+u^2} du + 6 d \int\limits_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2+c^2}}} e^{-d^2 u^2} T(\frac{\sqrt{2} d u^2}{\sqrt{1-u^2}},\frac{1}{\sqrt{3-4 u^2}}) du=\\ && 6 \sqrt{\pi } \left(T\left(-\frac{\sqrt{2} d}{\sqrt{\left(c^2+1\right) \left(c^2+2\right)}},\left\{\sqrt{c^2+1},\sqrt{\frac{c^2+2}{3 c^2+2}}\right\}\right)-T(-d,\{1,1\})\right)+\\ && 3 \sqrt{\pi } \left(\text{erf}\left(\frac{d}{\sqrt{c^2+2}}\right) T\left(\frac{\sqrt{2} d}{\sqrt{c^2+1} \sqrt{c^2+2}},\sqrt{\frac{c^2+2}{3 c^2+2}}\right)-\text{erf}\left(\frac{d}{\sqrt{2}}\right) T(d,1)\right) \end{eqnarray}