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I have two doubts in the answer for evaluating the following integral:

$$\int_0^1\int_0^1\int_0^1(1+u^2+v^2+w^2)^{-2}du~dv~dw$$

Solution: call this integral as $I$. By symmetry we may compute it over the domain $\{(u,v,w):0\leq v\leq u\leq 1\}$ and then double the result.
Substitute $u=r\cos(\theta)$, $v=r\sin(\theta)$, $w=\tan(\phi)$. Now the limits of integration become $0\leq \theta,\phi\leq \pi/4$ and $0\leq r\leq \sec(\theta)$. Then finally we have:
$$I=2\int_0^{\pi/4}\int_0^{\pi/4}\int_0^{\sec(\theta)}\frac{r\sec^2(\phi)}{(r^2+\sec^2(\phi))^2}dr~d\theta ~d\phi.$$

My doubts are:

1.) I don't get how by symmetry are we computing $I$ on $\{(u,v,w):0\leq v\leq u\leq 1\}$ .
2.) I don't understand why we substitute $u=r\cos(\theta)$, $v=r\sin(\theta)$, $w=\tan(\phi)$?

Kindly help me with above doubts. Thanks in advance.

Integrand
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coool
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2 Answers2

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This integral is a real beaut. I'll try to answer your questions, explain the substitutions, and then finally reduce the integral to a (very difficult) known definite integral.

  1. "I don't get how by symmetry are we computing $I$ on $\{(u,v,w) : 0\le v\le u \le 1\}$."

A picture or two is worth a thousand words: setting $w=0$ (this doesn't change things), compare the whole and half ranges.

integral_1

integral_2

  1. "I don't understand why we substitute $u=r\cos(\theta),$ $v=r\sin(\theta)$, $w=\tan(\phi)$."

Basically, we want to exploit the radial symmetry of the integrand. After using the symmetry trick, the region of integration is a triangular prism. You have correctly computed the new limits of integration and the Jacobian determinant, so we're ready to go.

Start with the new integral $$ I = \int _{0}^{\pi/4}\int_{0}^{\pi/4} \int _0^{\sec(\theta)} \frac{2r \sec^2(\phi)}{(r^2+\sec^2(\phi))^2}\,\mathrm{d}r \mathrm{d}\theta \mathrm{d}\phi $$Integrating with respect to $r$ doesn't pose too much trouble: $$ I = \left. \int _{0}^{\pi/4}\int_{0}^{\pi/4} \frac{-\sec ^2(\phi )}{r^2+\sec ^2(\phi )}\right|_{0}^{\sec(\theta)}\,\mathrm{d}\theta \mathrm{d}\phi $$ $$ = \int _{0}^{\pi/4}\int_{0}^{\pi/4} \frac{1}{1+\cos ^2(\theta ) \sec ^2(\phi )}\,\mathrm{d}\theta \mathrm{d}\phi $$This is basically an arctan integral, we just have to scale it appropriately. Using $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x} \frac{\arctan\left(\frac{\tan (x)}{\sqrt{1+M^2}}\right)}{\sqrt{1+M^2}}=\frac{1}{1+M^2\cos^2(x)}}$, we have $$ I = \left. \int _{0}^{\pi/4}\frac{\arctan\left(\frac{\tan (\theta )}{\sqrt{1+\sec ^2(\phi )}}\right)}{\sqrt{1+\sec ^2(\phi )}} \right|_{0}^{\pi/4}\,\mathrm{d}\phi $$ $$ = \int _{0}^{\pi/4}\frac{\arctan\left(\frac{1}{\sqrt{1+\sec ^2(\phi )}}\right)}{\sqrt{1+\sec ^2(\phi )}} \,\mathrm{d}\phi $$Let $\sec(\phi)=\sqrt{y^2+1}$. Then $\sec(\phi)\tan(\phi)\mathrm{d}\phi = \frac{y}{\sqrt{y^2+1}}\mathrm{d}y$, or $\mathrm{d}\phi=\frac{1}{1+y^2} \mathrm{d}y$: $$ = \int _{0}^{1}\frac{\arctan\left(\frac{1}{\sqrt{2+y^2}}\right)}{\sqrt{2+y^2}(1+y^2)} \,\mathrm{d}y $$Use the reciprocal relation of arctan: $$ = \int _{0}^{1}\frac{\pi/2-\arctan\left({\sqrt{2+y^2}}\right)}{\sqrt{2+y^2}(1+y^2)} \,\mathrm{d}y $$ $$ = \underbrace{\frac{\pi}{2}\int _{0}^{1}\frac{1}{\sqrt{2+y^2}(1+y^2)} \,\mathrm{d}y}_{I_1}-\underbrace{\int _{0}^{1}\frac{\arctan\left({\sqrt{2+y^2}}\right)}{\sqrt{2+y^2}(1+y^2)} \,\mathrm{d}y}_{I_2} $$For the first integral, we have $$ I_1=\frac{\pi}{2}\int _{0}^{1}\frac{(2+y^2)}{(2+y^2)^{3/2}(1+y^2)} \,\mathrm{d}y $$ $$ =\frac{\pi}{4}\int _{0}^{1}\frac{(2+y^2)}{(1+y^2)}\cdot \frac{2}{(2+y^2)^{3/2}} \,\mathrm{d}y $$ $$ =\frac{\pi}{2}\int _{0}^{1}\frac{1}{2}\cdot\frac{1}{1+\left(\frac{y}{\sqrt{2+y^2}}\right)^2}\cdot \frac{2}{(2+y^2)^{3/2}} \,\mathrm{d}y $$ $$ =\left.\frac{\pi}{2}\arctan\left(\frac{y}{\sqrt{2+y^2}}\right)\right|_0^1 = \frac{\pi^2}{12} $$$I_2$ is much more difficult to evaluate; fortunately, it is known as Ahmed's Integral and a very nice calculation giving the value of $\displaystyle{\frac{5\pi^2}{96}}$ can be found here. Putting it together, we have $\displaystyle{I = \frac{\pi^2}{32}}$.

sai-kartik
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Integrand
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \pars{1 + u^{2} + v^{2} + w^{2}}^{-2}\,\,\dd u\,\dd v\,\dd w}:\ {\Large ?}}$.

With $\ds{\large\vec{r} \equiv u\,\hat{u} + v\,\hat{v} +w\,\hat{w}}$: \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \pars{1 + u^{2} + v^{2} + w^{2}}^{-2}\,\,\dd u\,\dd v\,\dd w} = \iiint_{\large\pars{0,1}^{3}}{\dd^{3}\vec{r} \over \pars{1 + r^{2}}^{2}} \\[5mm] = &\ \iiint_{\large\pars{0,1}^{3}}\bracks{% -\,{1 \over 4\pi}\,\nabla^{2}\Phi\pars{r}}{\dd^{3}\vec{r}} \\[2mm] &\ \mbox{where}\ \Phi\pars{r} = 4\pi\int_{0}^{\infty} {1 \over \pars{1 + r\,'^{2}}^{2}}\,{r\, '^{2}\,\dd r\, ' \over \max\braces{r,r'}} = 2\pi\,{\arctan\pars{r} \over r} \end{align}

It turns out -via Gauss-Ostrogradsky's Divergence Theorem- that \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \pars{1 + u^{2} + v^{2} + w^{2}}^{-2}\,\,\dd u\,\dd v\,\dd w} \\[5mm] = & -\,{1 \over 4\pi}\iint_{S} \nabla\Phi\pars{r}\cdot\dd\vec{S}\quad \pars{\begin{array}{l} \ds{S}\ \mbox{is the}\ \ds{\pars{0,1}^{3}}\ \mbox{box surface} \end{array}} \end{align} By symmetry considerations:

  • The contribution, to the surface integration, from the sides that intersects the origin of coordinates $\underline{vanish\ out}$.
  • Each of the $\underline{\color{red}{three}}$ remaining sides yields the same contribution to the surface integration.

Then, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \pars{1 + u^{2} + v^{2} + w^{2}}^{-2}\,\,\dd u\,\dd v\,\dd w} \\[5mm] = &\ \color{red}{3}\braces{-\,{1 \over 4\pi}\iint_{\pars{0,1}^{2}} \bracks{\partiald{\Phi\pars{r}}{z}}_{\ z\ = 1}\dd x\,\dd y} \\[5mm] = &\ -\,{3 \over 2}\iint_{\pars{0,1}^{2}} \braces{\partiald{}{r}\bracks{{\arctan\pars{r} \over r}} \, {1 \over r}}_{\ z\ =\ 1}\dd x\,\dd y \\[5mm] = &\ 3\int_{1}^{\infty}\bracks{\int_{0}^{1}\int_{0}^{1} {\dd x\,\dd y \over \pars{x^{2} + y^{2} + t^{2} + 1}^{2}}}\dd t \\[5mm] = &\ 3\int_{1}^{\infty}{\mrm{arccot}\pars{\root{2 + t^{2}}} \over \pars{1 + t^{2}}\root{2 + t^{2}}}\,\dd t \\[5mm] = &\ {3\pi \over 2}\ \underbrace{\int_{1}^{\infty} {\dd t \over \pars{1 + t^{2}}\root{2 + t^{2}}}} _{\ds{\pi \over 12}}\ -\ 3\ \underbrace{\int_{1}^{\infty}{\arctan\pars{\root{2 + t^{2}}} \over \pars{1 + t^{2}}\root{2 + t^{2}}}\,\dd t} _{\ds{\pi^{2} \over 32}} \\[5mm] = &\ \bbx{\large{\pi^{2} \over 32}} \approx 0.3084 \\ & \end{align}

Felix Marin
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