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This is my question:

How to evaluate or prove this integral: \begin{equation} \iiint_{[0,1]^3} \frac{1}{\left(1+x^2+y^2+z^2\right)^{\frac{5}{2}}}\mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z =\frac{1}{\sqrt 2}\cot^{-1}\left(2\sqrt2 \right) \end{equation}

This question arose from my study of this integral: \begin{equation} I=\int_0^{+\infty}e^{-x} \mathrm{erf}^n \left(\sqrt x \right) \, \mathrm{d}x \end{equation} when $n = 3$.

My Attempt and Conjecture:

\begin{align} I&=\int_0^{+\infty}2ue^{-u^2}\mathrm{erf}^3(u) \, \mathrm{d}u\ \ \ (x=u^2)\\ &=\int_0^{+\infty}2ue^{-u^2}\left(\int_0^u e^{-x^2} \mathrm{d}x \right)^3 \mathrm{d}u\\ &=\int_0^{+\infty}2u^4e^{-u^2}\left(\int_0^1 e^{-u^2y^2} \mathrm{d}y \right)^3 \mathrm{d}u\\ &=2\int_0^1\int_0^1\int_0^1\int_0^{+\infty}u^4e^{-(1+x^2+y^2+z^2)u^2}\mathrm{d}u \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z \\ &=2\cdot\frac{1}{2}\Gamma \left(\frac{5}{2}\right)\int_0^1\int_0^1\int_0^1\frac{1}{\left(1+x^2+y^2+z^2\right)^{\frac{5}{2}}}\mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z. \end{align}

Conjecture:

\begin{align} n=1 &: I_1 = \frac{1\cdot \sqrt 2}{\pi}\cot^{-1}\left(0\cdot \sqrt2 \right)\\ n=2 &: I_2 = \frac{2\cdot\sqrt2}{\pi}\cot^{-1}\left(1\cdot\sqrt2 \right)\\ n=3 &: I_3 = \frac{3\cdot\sqrt2}{\pi}\cot^{-1}\left(2\cdot\sqrt2 \right)\\ &\vdots \\ n=k &: I_k = \frac{k\cdot\sqrt2}{\pi}\cot^{-1}\left((k-1)\cdot\sqrt2 \right)? \end{align} However, when $n=4$, it becomes incorrect, and thus I have failed. Can someone assist me in correctly calculating this problem? $$ \bbox[13px,#90EE90,border:5px solid #111a68]{\int_0^{+\infty}e^{-x} \mathrm{erf}^n \left(\sqrt x \right) \mathrm{d}x= \, ?} $$

Benjamin Wang
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Laura Olatex
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    A slick vector calculus method would be to apply the divergence theorem noting that $$\boldsymbol{\nabla}\cdot\left(\frac{\mathbf{r}}{\left(1+r\right)^{3/2}}\right)=\frac{3}{\left(1+r\right)^{5/2}}.$$ Reducing the triple integral to a double integral over the faces of the cube. What you find is $$\iiint\limits_{[0,1]^3}\frac{\mathrm{d}V}{\left(1+x^2+y^2+z^2\right)^{5/2}}=\iint_{[0,1]^2}\frac{\mathrm{d}u,\mathrm{d}v}{\left(2+u^2+v^2\right)^{3/2}}=\int_{0}^{1}\frac{\mathrm{d}v}{\left(v^2+2\right)\sqrt{v^2+3}}=\frac{1}{\sqrt{2}}\cot^{-1}\left(2\sqrt{2}\right).$$ – KStar Feb 25 '25 at 14:46
  • @KStar why $\iiint \frac{dV}{(\color{red}{1}+x^2+y^2+z^2)^{\frac{5}{2}}}=\iint \frac{dudv}{(\color{red}{2}+u^2+v^2)^{\frac{3}{2}}}$?I know the divergence theorem, but I'm still not quite clear about it here. Could you explain it in more detail? – Laura Olatex Feb 25 '25 at 16:54
  • @LauraOlatex For some reason it is true. Nice work KStar! – Kurt G. Feb 25 '25 at 17:00
  • @LauraOlatex One computes the dot product with the normals of the faces with $x=0,1$, $y=0,1$, $z=0,1$. The cases where $x=y=z=0$ have zero flux, and by symmetry of the cube, the others are equal which cancels with the $3$. When any one of them are $1$, the denominator becomes $2+u+v$, where $u,v$ are the other variables of $x,y,z$ that are not fixed to be $1$. We can push this idea once more to $n=4$: $$\frac{\pi^2}{32}\cdot I_4=\int\limits_{[0,1]^4}\frac{\mathrm{d}V}{\left(1+x^2+y^2+z^2+w^2\right)^3}=\int_{[0,1]^3}\frac{\mathrm{d}y,\mathrm{d}z,\mathrm{d}w}{\left(2+y^2+z^2+w^2\right)^2}.$$ – KStar Feb 25 '25 at 17:53
  • We can perform the integration to arrive at $$I_4=\frac{32}{\pi^2}\int_{0}^{1}\frac{\arctan\left(\frac{1}{\sqrt{3+x^2}}\right)}{\left(2+x^2\right)\sqrt{3+x^2}},\mathrm{d}x$$ which is an Ahmed-type integral that can be evaluated here with $p=\frac{1}{\sqrt{3}}$, $q=\sqrt{3}$, and $r=\frac{1}{\sqrt{2}}$, but the result is significantly more messy. Going beyond $n=4$ seems unfeasible. – KStar Feb 25 '25 at 17:56

2 Answers2

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Alternative to the existing approaches, here is an answer using a identity that is easily handled. We have for $x\in\mathbb{C}$, $$ \operatorname{erf}(\sqrt{x})^2 =1-\frac{2}{\pi} \int_{0}^{1} \frac{e^{-x(1+t)}}{\sqrt{t} \left ( 1+t \right ) } \text{d} t, $$ thus making the integral more tractable, $$ \begin{aligned} &\int_{0}^{\infty}e^{-x} \operatorname{erf}\left ( \sqrt{x} \right ) ^3\text{d}x,\\ =&\int_{0}^{\infty}e^{-x}\operatorname{erf}\left ( \sqrt{x} \right ) \left(1-\frac{2}{\pi} \int_{0}^{1} \frac{e^{-x(1+t)}}{\sqrt{t} \left ( 1+t \right ) } \text{d} t\right)\text{d}x,\\ =&\frac{1}{\sqrt{2} } -\frac{2}{\pi} \int_{0}^{1} \frac{1}{\sqrt{t} (1+t)} \cdot\frac{1}{\left ( 2+t \right )\sqrt{3+t} } \text{d}t,\\ =&\frac{1}{\sqrt{2} } -\frac{2\sqrt{2} }{\pi} \arctan\left ( \frac{\sqrt{2} }{5} \right ) ,\end{aligned} $$ where a simple identity valid for $\Re(\alpha)>0$, $$ \int_{0}^{\infty} e^{-\alpha x}\operatorname{erf}\left ( \sqrt{x} \right ) \text{d}x=\frac{1}{\alpha\sqrt{\alpha+1} } $$ is used. With regard to generalizations, such things probably wouldn't have been evaluated in terms of usual mathematical functions and constants when we are faced with high powers of error function. But we have the following integral, defining its complementary $\operatorname{erfc}\left ( \sqrt{x} \right )=1-\operatorname{erf}\left ( \sqrt{x} \right )$, $$ \int_{0}^{\infty} \operatorname{erfc}\left ( \sqrt{x} \right )^5 \text{d}x=\frac{1}{6\pi^2} \left ( 3\pi^2+\left ( 80\sqrt{3}-60 \right )\pi-120\sqrt{3}\arctan\left ( \sqrt{15} \right ) \right ) $$ exactly by the same strategy while employing the following identity $$ \int_{0}^{\infty}e^{-\alpha x}\operatorname{erfc}\left ( \sqrt{x} \right )^2 \text{d}x =\frac{1}{\alpha} \left ( 1-\frac{4}{\pi} \frac{\arctan\left ( \sqrt{1+\alpha} \right ) }{\sqrt{1+\alpha} } \right ),\qquad \Re(\alpha)>-2 .$$

Setness Ramesory
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4

Convert to spherical coordinates, and use symmetry (see here for details) to reduce the integral over the box to $6$ times the integral over a tetrahedron. Roll a die to pick a subset of the box:

$$\begin{align*} I &= \iiint_{[0,1]^3} \frac{dV}{\left(1+x^2+y^2+z^2\right)^{5/2}} \\ &= 6 \int_0^\tfrac\pi4 \int_{\operatorname{arccot}(\sin\theta)}^\tfrac\pi2 \int_0^{\sec\theta\csc\phi} \frac{\rho^2 \sin\phi}{\left(1+\rho^2\right)^{5/2}} \, d\rho \, d\phi \, d\theta \\ &= 2 \int_0^\tfrac\pi4 \int_{\operatorname{arccot}(\sin\theta)}^\tfrac\pi2 \frac{\sin\phi}{\left(1 + \cos^2\theta\sin^2\phi\right)^{3/2}} \, d\phi \, d\theta \\ &= 2 \int_0^\tfrac\pi4 \int_0^{\sin\theta} \frac{df\,d\theta}{\left(1+f^2+\cos^2\theta\right)^{3/2}} & f = \cot\phi \\ &= \sqrt2 \int_0^\tfrac\pi4 \frac{\sin\theta}{1+\cos^2\theta} \, d\theta \\ &= \sqrt2\left(\frac\pi4 - \arctan\frac1{\sqrt2}\right) = \sqrt2\arctan(3-2\sqrt2) = \boxed{\frac1{\sqrt2} \arctan\frac1{2\sqrt2}} \end{align*}$$

which follow from the identities

$$\begin{align*} \arctan\frac{1-t}{1+t} &= \frac\pi4 - \arctan t \\ \arctan t &= 2 \arctan \frac t{1+\sqrt{1+t^2}} \end{align*}$$

user170231
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