How to evaluate $\int_0^\infty\operatorname{erfc}^n x\ \mathrm dx$?

Let $\operatorname{erfc}x$ be the complementary error function.

I successfully evaluated these integrals:
$$\int_0^\infty\operatorname{erfc}x\ \mathrm dx=\frac1{\sqrt\pi}\tag1$$
$$\int_0^\infty\operatorname{erfc}^2x\ \mathrm dx=\frac{2-\sqrt2}{\sqrt\pi}\tag2$$

(Both $\operatorname{erfc}x$ and $\operatorname{erfc}^{2}x$ have primitive functions in terms of the error function.)

But I have problems with
$$\int_0^\infty\operatorname{erfc}^3x\ \mathrm dx\tag3$$
and a general case
$$\int_0^\infty\operatorname{erfc}^n x\ \mathrm dx.\tag4$$
Could you suggest an approach to evaluate them as well?

Answer

Here is my trial, which is partially successful but still not fully answering to your question.

Using some coordinate change, I derived that

$$ I_{n} := \int_{0}^{\infty} \mathrm{erfc}^{n}(x) \, dx = \frac{1}{\sqrt{n}} \left( \frac{2}{\sqrt{\pi}} \right)^{n-1} \int_{T^{n-1}} \int_{0}^{\infty} s^{n-1}e^{-(|x|^{2}-1)s^{2}} \mathrm{erfc}(s) \, ds d\sigma_{x}, \tag{*} $$

where $T^{n-1} = \{ x \in [0, \infty)^{n} : x_{1} + \cdots + x_{n} = \sqrt{n} \}$ is an $(n-1)$-simplex and $d\sigma_{x}$ is the surface measure on $T^{n-1}$.


Case $n = 1$. Since $T^{0} = \{1\}$ and $d\sigma_{x} = \delta(x-1)$ is a unit mass located at $x = 1$, the equation $\text{(*)}$ gives no new information.


Case $n = 2.$ It is easy to find that

$$ \int_{0}^{\infty} s e^{-(|x|^{2}-1)s^{2}} \mathrm{erfc}(s) \, ds = \frac{1}{2|x|(|x|+1)}$$

and by suitable parametrization of the line segment $T^{1}$, we get

\begin{align*}
I_{2}
&= \sqrt{\frac{2}{\pi}} \int_{T^{1}} \frac{1}{2|x|(|x|+1)} \, d\sigma_{x} \\
&= \sqrt{\frac{2}{\pi}} \int_{0}^{\sqrt{2}} \frac{\sqrt{2} \, dt}{2\sqrt{t^{2} + (\sqrt{2}-t)^{2}} ( \sqrt{t^{2} + (\sqrt{2}-t)^{2}} + 1)} \\
&= \sqrt{\frac{2}{\pi}} \int_{0}^{\frac{\pi}{4}} \frac{d\theta}{1+\cos\theta}
= \sqrt{\frac{2}{\pi}} \tan\left(\frac{\pi}{8}\right)
= \frac{2 – \sqrt{2}}{\sqrt{\pi}}.
\end{align*}

We can also write

$$ I_{2} = \boxed{ \displaystyle \frac{2}{\sqrt{\pi}} – \frac{2\sqrt{2}}{\pi^{3/2}} \arctan(\infty) }. $$


Case $n = 3$. We have

$$ \int_{0}^{\infty} s^{2}e^{-(|x|^{2}-1)s^{2}} \mathrm{erfc}(s) \, ds = \frac{1}{2\sqrt{\pi}(|x|^{2}-1)} \left( \frac{\arctan\sqrt{|x|^{2}-1}}{\sqrt{|x|^{2}-1}} – \frac{1}{|x|^{2}} \right). $$

Now we know that $T^{2}$ is a regular triangle with side length $\sqrt{6}$. Thus by introducing polar coordinate change, we get

\begin{align*}
I_{3}
&= \frac{2}{\pi^{3/2}\sqrt{3}} \int_{T^{2}} \frac{1}{|x|^{2}-1} \left( \frac{\arctan\sqrt{|x|^{2}-1}}{\sqrt{|x|^{2}-1}} – \frac{1}{|x|^{2}} \right) \, d\sigma_{x} \\
&= \frac{4\sqrt{3}}{\pi^{3/2}} \int_{0}^{\frac{\pi}{3}} \int_{0}^{\frac{\sec\theta}{\sqrt{2}}} \frac{1}{r} \left( \frac{\arctan r}{r} – \frac{1}{r^{2} + 1} \right) \, drd\theta \\
&= \frac{4\sqrt{3}}{\pi^{3/2}} \int_{0}^{\frac{\pi}{3}} \left( 1- \frac{\arctan\left(\sec\theta / \sqrt{2}\right)}{\sec\theta / \sqrt{2}} \right) \, d\theta \\
&= \frac{4\sqrt{3}}{\pi^{3/2}} \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \left( 1 – \frac{\arctan u}{u} \right) \frac{du}{u\sqrt{2u^{2} – 1}}.
\end{align*}

I haven’t tried the last integral, but Mathematica suggests that

$$ \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \left( 1 – \frac{\arctan u}{u} \right) \frac{du}{u\sqrt{2u^{2} – 1}} = \frac{\sqrt{3}\pi}{4} -\sqrt{\frac{3}{2}} \arctan\left( \sqrt{2} \right).$$

This gives

$$ I_{3} = \boxed{ \displaystyle \frac{3}{\sqrt{\pi}} – \frac{6\sqrt{2}}{\pi^{3/2}} \arctan\left( \sqrt{2} \right) }, $$

which can be numerically checked.


I am trying to generalize this calculation for higher dimensions, but it seems somewhat daunting. For example, if $n = 4$ we have to evaluate

$$ \int_{0}^{\infty} \mathrm{erfc}^{4}(x) \, dx = \frac{1}{\pi^{3/2}} \int_{T^{3}} \frac{2|x|+1}{(|x|+1)^{2}|x|^{3}} \, d\sigma_{x}. $$

Nevertheless, using the formula

$$ \int_{T^{n-1}} f(|x|) \, d\sigma_{x} = n! \int_{0}^{\sqrt{\frac{1}{n-1}}} \int_{0}^{\sqrt{\frac{n}{n-2}}s_{1}} \cdots \int_{0}^{\sqrt{\frac{3}{1}}s_{n-2}} f\left( \sqrt{1+s_{1}^{2} + \cdots + s_{n-1}^{2}} \right) \, ds_{n-1} \cdots ds_{1} $$

together with aid of Mathematica, I was able to evaluate $I_{4}$ and it was

$$ I_{4} = \boxed{ \displaystyle \frac{4}{\sqrt{\pi}} – \frac{24\sqrt{2}}{\pi^{3/2}} \arctan \left( \frac{1}{\sqrt{8}} \right) }. $$


These series of observations lead us to believe that $I_{n}$ is of the form

$$ I_{n} = \frac{n}{\sqrt{\pi}} – \frac{n!\sqrt{2}}{\pi^{3/2}} \arctan \alpha_{n} $$

for some reasonably nice $\alpha_{n}$ (with $\alpha_{1} = 0$, $\alpha_{2} = \infty$, $\alpha_{3} = \sqrt{2}$ and $\alpha_{4} = \frac{1}{\sqrt{8}}$), but inverse symbolic calculations show that this seems no longer the case for $n \geq 5$.

Indeed, for $n = 5$ we have

$$ I_{5} = \frac{5}{\sqrt{\pi}} – \frac{80\sqrt{2}}{\pi^{3/2}} \arctan \sqrt{\frac{2}{3}} + \frac{240\sqrt{2}}{\pi^{5/2}} A\left( \sqrt{\frac{5}{3}}, \sqrt{\frac{3}{2}}, \sqrt{\frac{4}{5}} \right), $$

where $A(p, q, r)$ is the generalized Ahmed’s integral. I have no idea whether it will reduce to a familiar closed form or not.

Attribution
Source : Link , Question Author : Hanna K. , Answer Author : Sangchul Lee

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