# Gaussian Expectation of 2 Erfs

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

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

• 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. Commented Feb 19, 2020 at 18:55
• 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? Commented Feb 19, 2020 at 19:45
• 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 $\frac{u}{\sqrt 2}=x$. Commented Feb 19, 2020 at 20:05
• Also the idea should be the same as here, just interchange the order of integrals. Commented Feb 19, 2020 at 20:09

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 $$\int_{-\infty}^{\infty} exp{(-\frac{u^2}{d})}~ erf{(au)}~erf{(bu)}~du$$ $$=\int_{-\infty}^{\infty} exp{(-x^2)}~ erf{(a\sqrt{d} x)}~erf{(b\sqrt{d} x)}~\sqrt{d} dx$$
$$=\frac{2 \sqrt{d}}{\sqrt{\pi}} ~ atan\big(\frac{ab\sqrt{d}\sqrt{d}}{\sqrt{a^2d+b^2d+1}}\big)$$ $$=\frac{2 \sqrt{d}}{\sqrt{\pi}} ~ atan\big(\frac{ab\sqrt{d}}{\sqrt{a^2+b^2+\frac{1}{d}}}\big).$$