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*