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For the Standard Gaussian Expectation

$$\int_{-\infty}^{\infty} exp{(-u^2)}~ erf{(au)}~erf{(bu)}~du,$$

where $erf(x)$ is the cumulative densitiy function of a normal, I have found investigate the formulae

$$\frac{2}{\sqrt{\pi}} ~ atan\big(\frac{ab}{\sqrt{a^2+b^2+1}}\big)$$ (UPDATE: square-root was missing!)

described here. Can anybody confirm this result?

Anyway, I was wondering, how the formulae changes when instead a scaled Gaussian $exp{(-\frac{u^2}{d})}$ is used instead. That is, how can I compute

$$\int_{-\infty}^{\infty} exp{(-\frac{u^2}{d})}~ erf{(au)}~erf{(bu)}~du?$$

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  • $\begingroup$ Please bring some more context. It's quite obvious that the result is incorrect, take for example $a=b=1$ then the integral is easy to compute $\int_{-\infty}^\infty e^{-x^2} \operatorname{erf}^2 x dx=\frac{2}{\sqrt \pi} \frac{\operatorname{erf}^3 x}{3}\bigg|_{-\infty}^\infty=\frac{\sqrt \pi}{3}$. The formula that you stated there doesn't give this. $\endgroup$
    – Zacky
    Commented Feb 19, 2020 at 18:55
  • $\begingroup$ Ups, there was an error, I missed the square-root in the denominator. I changed it and now your example should work. What do you mean with more context? $\endgroup$
    – Delgado
    Commented Feb 19, 2020 at 19:45
  • $\begingroup$ Yes, now it should work. See here about context. Like, what have you tried? Or how did you encounter this integral? Btw the two integrals are the same, after you substitute $\frac{u}{\sqrt 2}=x$. $\endgroup$
    – Zacky
    Commented Feb 19, 2020 at 20:05
  • $\begingroup$ Also the idea should be the same as here, just interchange the order of integrals. $\endgroup$
    – Zacky
    Commented Feb 19, 2020 at 20:09

1 Answer 1

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Tanks for the hint. It was actually not that hard. I used $$\frac{u^2}{d}=x^2$$ leading to $$u = \sqrt{d} x$$ and $$du = \sqrt{d} dx.$$

Plugging-in yields $$\int_{-\infty}^{\infty} exp{(-\frac{u^2}{d})}~ erf{(au)}~erf{(bu)}~du$$ $$=\int_{-\infty}^{\infty} exp{(-x^2)}~ erf{(a\sqrt{d} x)}~erf{(b\sqrt{d} x)}~\sqrt{d} dx$$

$$=\frac{2 \sqrt{d}}{\sqrt{\pi}} ~ atan\big(\frac{ab\sqrt{d}\sqrt{d}}{\sqrt{a^2d+b^2d+1}}\big)$$ $$=\frac{2 \sqrt{d}}{\sqrt{\pi}} ~ atan\big(\frac{ab\sqrt{d}}{\sqrt{a^2+b^2+\frac{1}{d}}}\big).$$

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