Integral power of error function with exponential

Question

I'm trying to solve the integrals below:
$$ \int_{-\infty}^{\infty}dx \hspace{5pt}\mathrm{erf}\left(x b\right)^2\exp\left(-x^2a\right) ,$$

with $a,b>0$. Unfortunately, I could not find this integral type in Korotkov's Integrals Related to the Error Function book. Would anyone have any ideas? Or is there no closed solution at all?

Answer 1

Note that your integral can be reduced to the evaluation of $$ f \colon (0,\infty) \to (0,\infty), \, f(r) = \int \limits_0^\infty \operatorname{erf}^2(r t) \mathrm{e}^{-t^2} \, \mathrm{d} t \, , $$ since we can exploit the symmetry of the integrand and the substitution $x = t/\sqrt{a}$ to show that it equals $2 f(b/\sqrt{a})/\sqrt{a}$. Using the representation $$ \operatorname{erf}(s) = \frac{2}{\sqrt{\pi}} \int \limits_0^1 s \, \mathrm{e}^{-s^2 u^2} \, \mathrm{d}{u} $$ we can write \begin{align} f(r) &= \frac{4 r^2}{\pi} \int \limits_0^1 \int \limits_0^1 \int \limits_0^\infty t^2 \mathrm{e}^{-[1 + r^2 (u^2 + v^2)]t^2} \, \mathrm{d} t \, \mathrm{d} u \, \mathrm{d} v = \frac{r^2}{\sqrt{\pi}} \int \limits_0^1 \int \limits_0^1 \frac{\mathrm{d} u \, \mathrm{d} v}{[1 + r^2(u^2+v^2)]^{3/2}} \\ &= \frac{r^2}{\sqrt{\pi}} \int \limits_0^1 \frac{\mathrm{d} v}{(1+r^2 v^2) \sqrt{1+r^2 + r^2 v^2}} \, . \end{align} Now we let $ w = r^2 v/\sqrt{1 + r^2 + r^2 v^2}$ and find $$ f(r) = \frac{1}{\sqrt{\pi}} \int \limits_0^{r^2/\sqrt{1+2r^2}} \frac{\mathrm{d} w}{1+w^2} = \frac{1}{\sqrt{\pi}} \arctan \left(\frac{r^2}{\sqrt{1+2r^2}} \right) \, .$$

Answer 2

If $b=a^2$ then Maple can compute the integral as $\frac{\sqrt{\pi}}{3|a|}$ but it fails to compute it otherwise, which is a good indicator that it's not computable...

Answer 3

Using $x=\frac{t}{\sqrt{a}}$ and $c=\frac{b}{\sqrt{a}}$, we have $$I=\frac{1}{\sqrt{a}}\int_{-\infty}^{+\infty} e^{-t^2} \,\text{erf}(c t)^2\,dt$$

Expanded as series around $c=0$ $$[\text{erf}(c t)]^2=\frac 1 \pi \sum_{n=0}^\infty d_{2n}\, t^{2n}\,c^{2n}$$ where, with $(d_0=0,d_1=0,d_2=4,d_3=0)$, $$d_{2n}=-2\,\frac{ (n-2) (3 n-8) \, d_{n-2}+4 (n-4) \,d_{n-4}}{n(n-1)(n-2) }$$ and $$\int_{-\infty}^{+\infty}e^{-t^2}\, t^{2 n}\,dt=\Gamma \left(n+\frac{1}{2}\right)$$ This gives $$I=\frac{1}{\pi\sqrt{a}}\sum_{n=0}^\infty d_{2n}\,\Gamma \left(n+\frac{1}{2}\right)\,c^{2n}$$ which, fortunately, correspond to the series expansion of $$\frac{2}{\sqrt{\pi }}\tan ^{-1}\left(\frac{c^2}{\sqrt{2 c^2+1}}\right)$$ already given by @ComplexYetTrivial in his/her elegant answer (which I would be unable to find from the infinite series).