error function integration $\int_{0}^{\infty} \frac{x \operatorname{erf}(a x ) }{x^2+y^2} dx $

Question

I'm interested in the following integral, $$ \int_{0}^{\infty} \frac{x \operatorname{erf}(a x ) }{x^2+y^2} dx $$ where, $\operatorname{erf}$ is error function. Does the analytical solution exist to this integral?

Answer 1

Others have said the integral does not converge, so do not treat this as a converged solution. If we treat this formally only, take the Mellin transform of the integrand with respect to $a$ to a new parameter $s$ to get $$ -\frac{x^{1-s} \Gamma \left(\frac{s}{2}+\frac{1}{2}\right)}{\sqrt{\pi } s \left(x^2+y^2\right)}$$ if we integrate this with your range we get $$ \text{ConditionalExpression}\left[-\frac{\sqrt{\pi } \left(\frac{1}{y^2}\right)^{s/2} \csc \left(\frac{\pi s}{2}\right) \Gamma \left(\frac{s+1}{2}\right)}{2 s},(s-2) \arg \left(\frac{1}{y^2}\right)\geq -2 \pi \land \Re(s)<2\land \left(\Re\left(y^2\right)\geq 0\lor y^2\notin \mathbb{R}\right)\right] $$ dropping the conditions (warning this might make everything invalid) gives $$ -\frac{\sqrt{\pi } \left(\frac{1}{y^2}\right)^{s/2} \csc \left(\frac{\pi s}{2}\right) \Gamma \left(\frac{s+1}{2}\right)}{2 s} $$ The inverse Mellin transform of this (from $s$ to $a$) gives an expression with hypergeometric function and digamma function, as well as erfi: $$ \frac{1}{2} \left(2 a^2 y^2 \, _2F_2\left(1,1;\frac{3}{2},2;a^2 y^2\right)-\pi \sqrt{\frac{1}{y^2}} y \;\text{erfi}(a y)+2 \log (a)-\log \left(\frac{1}{y^2}\right)-\psi ^{(0)}\left(\frac{1}{2}\right)\right) $$ which might be a place to look for analytic continuation or formal solutions in terms of series etc.