Evaluating $\int_0^\infty\operatorname{erfi}(x)e^{-x^2}\frac{dx}x$

Question

I want to evaluate

$$I=\int_0^\infty\operatorname{erfi}(x)e^{-x^2}\frac{dx}x$$ where $\operatorname{erfi}(x)=\frac{2}{\sqrt\pi}\int_0^x e^{t^2}dt$.

I can only prove this integral converges.
Not-so-useful Attempt
Integrating by parts gives $$I=\int_0^\infty-\frac{1}{\sqrt{\pi}}e^{x^2}\operatorname{Ei}(-x^2)dx$$ Substituting $x^2=t$ gives $$I=\int_0^\infty-\frac{1}{2\sqrt{\pi x}}e^{x}\operatorname{Ei}(-x)dx$$ where $\operatorname{Ei}$ is the exponential integral function.

Answer 1

Introduce $$ I(\lambda)=\int_0^\infty\mathrm{erfi}(\lambda x)\frac{\exp(-x^2)}{x}\,\mathrm{d}x $$ for $\lambda\in[0,1]$. Note that $I(\lambda)$ is continuous (and even real-analytic on $(0,1)$) and differentiating with respect to $\lambda$ under the integral sign, we have $$ I'(\lambda)=\frac{2}{\sqrt{\pi}}\int_0^\infty\exp(-(1-\lambda^2) x^2)\,\mathrm{d}x=\frac{1}{\sqrt{1-\lambda^2}} $$ for $\lambda\in (0,1)$. Furthermore, $I(0)=0$ and so $$ I=I(1)=\int_0^1\frac{1}{\sqrt{1-\lambda^2}}\,\mathrm{d}\lambda=\frac{\pi}{2} $$

Answer 2

A change of variable and Fubini's theorem are enough: $$\begin{eqnarray*}\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\int_{0}^{x}e^{-x^2+y^2}\,dy\frac{dx}{x}&\stackrel{y\mapsto x z}{=}&\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\int_{0}^{1}e^{-x^2} e^{x^2 z^2}\,dz\,dx\\&=&\int_{0}^{1}\frac{dz}{\sqrt{1-z^2}}=\frac{\pi}{2}. \end{eqnarray*}$$