0
$\begingroup$

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.

$\endgroup$

1 Answer 1

5
$\begingroup$

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: $$ \{(t,y) \in\mathbb R^2 | 0<t, 0<y, t^2+y^2<r^2\} $$ 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}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .