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I am trying to solve the following integral:

$$ \int_{-\infty}^{\infty} \big(\text{erf}(x)\big) ^{n} \exp\left(-3x^{2}\right) \ \mathrm dx $$

where $\text{erf}$ is the error function and $n$ is an integer.

I was wondering if there is an explicit form.

Thank You.

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    $\begingroup$ Have you tried solving the easiest particular cases, $n=1,2$? $\endgroup$
    – David H
    Commented Oct 3, 2021 at 21:32
  • $\begingroup$ An easy observation: the integral is automatically zero for odd $n$ due to the odd symmetry of the integrand. $\endgroup$
    – David H
    Commented Oct 3, 2021 at 23:03
  • $\begingroup$ Hi. Thank you for your comment. $\endgroup$
    – Seouldddd
    Commented Oct 4, 2021 at 2:42
  • $\begingroup$ @DavidH Your observation is correct for odd $\text{n}$. And for even $\text{n}$ I found that it is possible to evaluate the integral if we calculate the following integration: $$ \int_{-\infty}^{\infty} x^2 \big(\text{erf}(x)\big) ^{n+2} \exp\left(-x^{2}\right) \ \mathrm dx $$ Please let me know if you have any ideas to solve it. Thank you! $\endgroup$
    – Seouldddd
    Commented Oct 4, 2021 at 2:55
  • $\begingroup$ Could you elaborate on how to rewrite my integral as a $\text{n+1}$-dimensional multiple integral? $\endgroup$
    – Seouldddd
    Commented Oct 4, 2021 at 4:41

1 Answer 1

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Too Long For Comment:


Define the function $\mathcal{I}:\mathbb{N}\times\mathbb{R}_{>0}\rightarrow\mathbb{R}$ via the (convergent) improper integral

$$\mathcal{I}{\left(n,a\right)}:=\int_{0}^{\infty}\mathrm{d}x\,\exp{\left(-a^{2}x^{2}\right)}\left[\operatorname{erf}{\left(x\right)}\right]^{n},$$

where the error function is typically defined via the integral representation

$$\operatorname{erf}{\left(x\right)}:=\frac{2}{\sqrt{\pi}}\int_{0}^{x}\mathrm{d}t\,\exp{\left(-t^{2}\right)};~~~\small{x\in\mathbb{R}},$$

or equivalently

$$\operatorname{erf}{\left(x\right)}=\frac{2x}{\sqrt{\pi}}\int_{0}^{1}\mathrm{d}t\,\exp{\left(-x^{2}t^{2}\right)};~~~\small{x\in\mathbb{R}}.$$

Then by induction, powers of the error function can then be expressed as multiple integrals in the following way:

$$\left[\operatorname{erf}{\left(x\right)}\right]^{n}=\frac{2^{n}x^{n}}{\pi^{n/2}}\int_{[0,1]^{n}}\mathrm{d}t_{1}\dots\mathrm{d}t_{n}\,\prod_{k=1}^{n}\exp{\left(-x^{2}t_{k}^{2}\right)};~~~\small{n\in\mathbb{N}\land x\in\mathbb{R}}.$$


Recalling the definition of the gamma function as

$$\Gamma{\left(z\right)}:=\int_{0}^{\infty}\mathrm{d}t\,t^{z-1}\exp{\left(-t\right)};~~~\small{z>0},$$

we can show that for any $n\in\mathbb{N}\land x\in\mathbb{R}$,

$$\begin{align} \mathcal{I}{\left(n,a\right)} &=\int_{0}^{\infty}\mathrm{d}x\,\exp{\left(-a^{2}x^{2}\right)}\left[\operatorname{erf}{\left(x\right)}\right]^{n}\\ &=\int_{0}^{\infty}\mathrm{d}x\,\exp{\left(-a^{2}x^{2}\right)}\frac{2^{n}x^{n}}{\pi^{n/2}}\int_{[0,1]^{n}}\mathrm{d}t_{1}\dots\mathrm{d}t_{n}\,\prod_{k=1}^{n}\exp{\left(-x^{2}t_{k}^{2}\right)}\\ &=\frac{2^{n}}{\pi^{n/2}}\int_{0}^{\infty}\mathrm{d}x\int_{[0,1]^{n}}\mathrm{d}t_{1}\dots\mathrm{d}t_{n}\,x^{n}\exp{\left(-a^{2}x^{2}\right)}\prod_{k=1}^{n}\exp{\left(-x^{2}t_{k}^{2}\right)}\\ &=\frac{2^{n}}{\pi^{n/2}}\int_{[0,1]^{n}}\mathrm{d}t_{1}\dots\mathrm{d}t_{n}\int_{0}^{\infty}\mathrm{d}x\,x^{n}\exp{\left(-a^{2}x^{2}\right)}\prod_{k=1}^{n}\exp{\left(-x^{2}t_{k}^{2}\right)}\\ &=\frac{2^{n}}{\pi^{n/2}}\int_{[0,1]^{n}}\mathrm{d}t_{1}\dots\mathrm{d}t_{n}\int_{0}^{\infty}\mathrm{d}x\,x^{n}e^{-\left(a^{2}+\sum_{k=1}^{n}t_{k}^{2}\right)x^{2}}\\ &=\frac{2^{n-1}}{\pi^{n/2}}\int_{[0,1]^{n}}\mathrm{d}t_{1}\dots\mathrm{d}t_{n}\int_{0}^{\infty}\mathrm{d}u\,u^{(n-1)/2}e^{-\left(a^{2}+\sum_{k=1}^{n}t_{k}^{2}\right)u};~~~\small{\left[x=\sqrt{u}\right]}\\ &=\frac{2^{n-1}}{\pi^{n/2}}\int_{[0,1]^{n}}\mathrm{d}t_{1}\dots\mathrm{d}t_{n}\,\frac{1}{\left(a^{2}+\sum_{k=1}^{n}t_{k}^{2}\right)^{(n+1)/2}}\int_{0}^{\infty}\mathrm{d}t\,t^{(n-1)/2}e^{-t};~~~\small{\left[u=\frac{t}{a^{2}+\sum_{k=1}^{n}t_{k}^{2}}\right]}\\ &=\frac{2^{n-1}}{\pi^{n/2}}\int_{[0,1]^{n}}\mathrm{d}t_{1}\dots\mathrm{d}t_{n}\,\frac{1}{\left(a^{2}+\sum_{k=1}^{n}t_{k}^{2}\right)^{(n+1)/2}}\,\Gamma{\left(\frac{n+1}{2}\right)}\\ &=\frac{2^{n-1}\Gamma{\left(\frac{n+1}{2}\right)}}{\pi^{n/2}}\int_{[0,1]^{n}}\mathrm{d}t_{1}\dots\mathrm{d}t_{n}\,\frac{1}{\left(a^{2}+\sum_{k=1}^{n}t_{k}^{2}\right)^{(n+1)/2}}.\\ \end{align}$$


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  • $\begingroup$ The $n$-dimensional integral in the last line above is quite similar to certain families of box integrals studied by Bailey: osti.gov/servlets/purl/964379 . It’s doubtful that a closed form exists for general $n$. $\endgroup$
    – David H
    Commented Oct 4, 2021 at 7:00
  • $\begingroup$ Thank you for your explanation. I will review the referred paper. $\endgroup$
    – Seouldddd
    Commented Oct 5, 2021 at 6:18

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