# difficult integral involving exponentials and error functions

I'm trying to prove that for any $$r \geq 0$$, we have: $$\int_0^r \frac{x\,e^{x^2} \mathrm{erf}(x)}{\sqrt{r^2-x^2}}\,\mathrm{d}x=\frac{\sqrt{\pi }}{2} \left(e^{r^2}-1\right)$$

Where $$\mathrm{erf}(x) := \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\mathrm{d}t$$ is the standard error function. Tried a bunch of things; mostly integration by parts and trig subsitutions, but no luck.

First, substitute $$y := \sqrt{r^2-x^2}$$: $$I(r) := \int_0^r \frac{x\,e^{x^2} \mathrm{erf}(x)}{\sqrt{r^2-x^2}}\,\mathrm{d}x = -\int_r^0 e^{r^2-y^2} \mathrm{erf}\left(\sqrt{r^2-y^2}\right)\,\mathrm{d}y$$ Next, using the definition of $$\mathrm{erf}$$: $$I(r) = \frac{2}{\sqrt\pi} e^{r^2} \int_0^r \int_0^{\sqrt{r^2-y^2}} e^{-y^2} e^{-t^2}\,\mathrm{d}t\,\mathrm{d}y$$ This is a double integral over the following domain: $$\{(t,y) \in\mathbb R^2 | 0 That is a quarter circle with radius $$r$$. Using polar coordinates: \begin{align} I(r) &= \frac{2}{\sqrt\pi} e^{r^2} \int_0^r \int_0^{\sqrt{r^2-y^2}} e^{-(t^2+y^2)}\,\mathrm{d}t\,\mathrm{d}y \\ &= \frac{2}{\sqrt\pi} e^{r^2} \int_0^r \int_0^{\pi/2} e^{-\rho^2} \rho\,\mathrm{d}\varphi \,\mathrm{d}\rho \\ &= \sqrt\pi e^{r^2} \int_0^r e^{-\rho^2} \rho\,\mathrm{d}\rho \end{align} Lastly, substitute $$u := \rho^2$$: \begin{align} I(r) &= \frac{\sqrt\pi}{2} e^{r^2} \int_0^{r^2} e^{-u} \,\mathrm{d}u\\ &= \frac{\sqrt\pi}{2} e^{r^2} \left(1-e^{-r^2}\right)\\ &= \frac{\sqrt\pi}{2} \left(e^{r^2}-1\right) \end{align}