Is there an analytic solution for the following Gaussian integral?
$$\int_{-\infty}^{\infty} \Phi(\frac{x}{2}+\frac{k}{x}) e^{-\frac{(x-a)^2}{2}}dx$$
with
$a,k$ are both real numbers,
$\Phi$ CDF of the standard Gauss distribution, that is, $\Phi(x):=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{t^2}{2}}dt$.
I found a similar question here, which solved the integral $\int_{-\infty}^{\infty} \Phi(x) e^{-\frac{(x-a)^2}{2}}dx$.
Is it possible to extend the above solution to my case?
Since $$\Phi(x)=\frac{1}{2} \left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)$$ the problem is $$I= \frac{1}{2}\int_{-\infty}^{+\infty} e^{-\frac{(x-a)^2}{2} } \left(1+\text{erf}\left(\frac{\frac{k}{x}+\frac{x}{2}}{\sqrt{2}}\right)\right)\,dx$$ that is to say $$I=\sqrt{\frac{\pi }{2}}+\frac{1}{2}\int_{-\infty}^{+\infty} e^{-\frac{(x-a)^2}{2} } \,\, \text{erf}\left(\frac{\frac{k}{x}+\frac{x}{2}}{\sqrt{2}}\right)\,dx$$
If we expand the exponential function around $a=0$, we have $$e^{-\frac{(x-a)^2}{2} }=e^{-\frac{x^2}{2}}\sum_{n=0}^\infty P_n(x) \, a^n$$ which means that we face integrals $$J_{2m+1}=\int_{-\infty}^{+\infty} x^{2m+1}\,\,e^{-\frac{x^2}{2} } \,\, \text{erf}\left(\frac{\frac{k}{x}+\frac{x}{2}}{\sqrt{2}}\right)\,dx$$ since, by symmetry, $J_{2m}=0$.
Assuming $k>0$, the only one I have been able to compute is $$J_1=2+2 \left(\frac{1}{\sqrt{5}}-1\right) e^{-\frac{1+\sqrt{5}}{2} k}$$
The coefficient of $x$ is $a\, e^{-\frac{a^2}{2}}$.
This is not much !