$ \int_0^{\infty} \, e^{-a^2x^2} \operatorname{erf}(bx) \, dx$

Question

I'd like to know the proof of this formula

$$ \int_0^{\infty} \, e^{-a^2x^2} \operatorname{erf}(bx) \, dx=\frac{1}{a\sqrt{\pi}}\arctan{\frac{b}{a}}$$

where $\operatorname{erf}(x)$ is the error function $\operatorname{erf}(x):=\frac{2}{\sqrt{\pi}}\int_0^{x} e^{-y^2} \, dy$.

Answer 1

Hint: transform the double integral into polar coordinates: $$\frac{2b} {\sqrt\pi}\int_0^\infty e^{-a^2x^2}dx\int_0^x e^{-b^2y^2}dy\\ =\frac{2b} {\sqrt\pi}\int_0^{\pi/4} d\phi\int_0^\infty e^{-r^2 (a^2\cos^2\phi+b^2\sin^2\phi)}rdr =\frac b {\sqrt\pi}\int_0^{\pi/4} \frac {d\phi}{a^2\cos^2\phi+b^2\sin^2\phi} $$ and use $t=\tan \phi $ substitution.

Answer 2

Your idea to use Feynman trick was good $$I(b)=\int_0^\infty e^{-a^2 x^2} \text{erf}(b x)\,dx$$ $$I'(b)=\frac{2}{\sqrt{\pi }}\int_0^\infty x e^{- \left(a^2+b^2\right)x^2}\,dx=\frac{1}{\sqrt{\pi }}\int_0^\infty e^{- \left(a^2+b^2\right)t}\,dt=\frac{1}{\sqrt{\pi } \left(a^2+b^2\right)}$$ $$I(b)=\frac{1}{\sqrt \pi }\int \frac{db}{ \left(a^2+b^2\right)}=\frac{\tan ^{-1}\left(\frac{b}{a}\right)}{\sqrt{\pi } a}+C$$ and, since $I(0)=0$, then the result.