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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

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?

Mingzhou Liu
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    Yes: $\sqrt{\pi/2}$ independently of $k$ – Sal Oct 07 '23 at 04:26
  • @Sal. You are right but why is it independent of $k$ ? Cheers :-) – Claude Leibovici Oct 07 '23 at 04:41
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    @ClaudeLeibovici Hi Claude! I write $\Phi(x)=\frac{1}{2}+g(x)$ where $g(x)$ is an odd function. Then because $\frac{x}{2}+\frac{k}{x}$ is also odd, the only contribution to the original integral from $\Phi$ is the constant $1/2$ – Sal Oct 07 '23 at 04:51
  • @sal. This is what you could have written in the first comment (even if it is obvious). Cheers :-) – Claude Leibovici Oct 07 '23 at 04:56
  • @Sal Apologize for the typo. It should be $e^{-\frac{(x-a)^2}{2}}$ with $a$ some real number. I have edited the problem. Any idea? – Mingzhou Liu Oct 07 '23 at 07:04
  • @MingzhouLiu in that case I doubt there is a 'nice' exact answer – Sal Oct 09 '23 at 06:13

1 Answers1

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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 !

Claude Leibovici
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