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Looking for a way to perform this integral related to the error function. I am thinking an answer in closed form cannot be done, but hoping I missed something.

$$ \int x\operatorname{erf}^{\,3}(x)\,e^{-x^2}dx $$

Edit:

Yuriy : It comes from calculations related to linear moments.
Ives : Thanks, following integration by parts, we have

$\int u\,dv = uv-\int v\,du$

$u=x\,\operatorname{erf}(x)$
$du=\operatorname{erf}(x)+\frac{2x}{\sqrt{\pi}}e^{-x^2}dx$

$v=\frac{\pi}{6}\operatorname{erf}^3(x)$
$dv=\operatorname{erf}^2(x)\,e^{-x^2}dx$

$\int x\operatorname{erf}^{\,3}(x)\,e^{-x^2}\,dx = \frac{\sqrt{\pi}}{6}\,x\,e^{-x^2}\operatorname{erf}^{\,3}(x)-\frac{\sqrt{\pi}}{6}\,\int \operatorname{erf}^3(x)[\operatorname{erf}(x)+\frac{2}{\sqrt{\pi}}xe^{-x^2}]dx$
$=\frac{\sqrt{\pi}}{4}x\operatorname{erf}^4(x)-\frac{\sqrt{\pi}}{4}\int \operatorname{erf}^{\,4}(x)dx$

Then the challenge becomes how to solve:

$$ \int \operatorname{erf}^{\,4}(x)\,dx $$

Edit 2:

Able to simplify(?) a bit more:

$\int \operatorname{erf}^{\,4}(x)\,dx = x\operatorname{erf}^{\,4}(x)+\frac{4}{\sqrt{\pi}}e^{-x^2}\operatorname{erf}^{\,3}(x)-\frac{24}{\pi}\int \operatorname{erf}^{\,2}(x)e^{-2x^2}dx$

Now the challenge becomes how to solve:

$$ \int \operatorname{erf}^{\,2}(x)e^{-2x^2}dx $$

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  • 1
    $\begingroup$ Integration by parts can reduce the integrand to $\text{erf }^4(x)$, indeed not manageable. $\endgroup$ – Yves Daoust Jan 10 at 11:22
  • $\begingroup$ Where did you find this monster? $\endgroup$ – Yuriy S Jan 10 at 11:30
  • $\begingroup$ Let $u=\erf(x)$ and all is good. $\endgroup$ – B. Goddard Jan 10 at 12:06
  • $\begingroup$ @B.Goddard: can you elaborate ? $\endgroup$ – Yves Daoust Jan 10 at 12:40
  • $\begingroup$ If that $x$ wasn't there it'd be a lot easier. $\endgroup$ – Michael Seifert Jan 10 at 21:13

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