I have been trying to calculate the following triple integral:
$$ I(a,b,c) \,=\, \int_{x=0}^{a}\int_{y=0}^{b}\int_{z=0}^{c} \frac{dx\,dy\,dz}{(1+x^{2}+y^{2}+z^{2})^{3}} $$
I can find values numerically for given $a,b,c$ but, since I know that $I(a,b,c)\rightarrow\frac{\pi^{2}}{32}$ as $a,b,c\rightarrow\infty$, I wondered whether the integral has a closed-form solution for arbitrary $a,b,c$ ? I certainly haven't found one and hoped someone might be able to help.
On
Here's a one-dimensional integral with well-studied functions. $$I(a,b,c)=\frac{1}{2}\Big(\frac{\sqrt{\pi}}{2}\Big)^3 \int_0^\infty u^{1/2} e^{-u} \, \text{erf}(a\,\sqrt{u}) \,\text{erf}(b\,\sqrt{u})\, \text{erf}(c\,\sqrt{u})\,du \,.$$ The nice thing is that you get the value of $\pi^2/32$ for $a, b, c \to \infty$ by use of Euler's integral for the gamma function since all the erf's $\to$ 1. The derivation is simple. Use the gamma function representation $$ \frac{1}{2}\int_0^\infty u^2 \,\exp{(-u\,(1+x^2+y^2+z^2))}\,du = \frac{1}{(1+x^2+y^2+z^2)^3}$$ Insert this into the definition of $I$ and interchange the integrals (allowed because the integrands are bounded) $$ I(a,b,c)= \frac{1}{2} \int_0^\infty \! du\, u^{2} e^{-u} \int_0^a e^{-u\,z^2}dz \int_0^b e^{-u\,y^2}dy \int_0^c e^{-u\,x^2}dx $$ Each of the inner integrals have the same form $$ \int_0^c e^{-u\,x^2}dx = \frac{\sqrt{\pi}}{2\sqrt{u}} \text{erf}(c\,\sqrt{u})$$ so the top formula follows.
On
Just using a CAS $$I_1=\int_{z=0}^{c} \frac{dz}{(1+x^{2}+y^{2}+z^{2})^{3}}=\frac{1}{8} \left(\frac{c \left(3 c^2+5 \left(x^2+y^2+1\right)\right)}{\left(x^2+y^2+1\right)^2 \left(c^2+x^2+y^2+1\right)^2}+\frac{3 \tan ^{-1}\left(\frac{c}{\sqrt{x^2+y^2+1}}\right)}{\left(x^2+y^2+1\right)^{5/2}}\right)$$ $$I_2=\int_{y=0}^{b} I_1\,dy$$ gives the nice expression $$I_2=\frac c {8b} \left(\frac{1}{\left(x^2+1\right) \left(x^2+c^2+1\right)}-\frac{1}{\left(x^2+b^2+1\right) \left(x^2+b^2+c^2+1\right)} \right)+$$ $$\frac{\frac{b \left(3x^2+ b^2+3\right) \tan ^{-1}\left(\frac{c}{\sqrt{x^2+b^2+1}}\right)}{\left(x^2+b^2+1\right)^{3/2}}+\frac{c \left(3 x^2+2 c^2+3\right) \tan ^{-1}\left(\frac{b}{\sqrt{x^2+c^2+1}}\right)}{\left(x^2+c^2+1\right)^{3/2}}}{8\left(x^2+1\right)^2}$$ Now, the "easy part" $$I_3=\int_{x=0}^{a} \left(\frac{1}{\left(x^2+1\right) \left(x^2+c^2+1\right)}-\frac{1}{\left(x^2+b^2+1\right) \left(x^2+b^2+c^2+1\right)} \right)\,dx$$ $$I_3=\frac{\tan ^{-1}\left(\frac{a}{\sqrt{b^2+c^2+1}}\right)}{c^2 \sqrt{b^2+c^2+1}}+\frac{\tan ^{-1}(a)}{c^2}-\frac{\tan ^{-1}\left(\frac{a}{\sqrt{c^2+1}}\right)}{c^2 \sqrt{c^2+1}}-\frac{\tan ^{-1}\left(\frac{a}{\sqrt{b^2+1}}\right)}{c^2\sqrt{b^2+1} }$$
I give up for the remaining.
Here's an idea that might work. It should at least get you started. By Gaus' Law, $\int\int\int(\nabla\cdot \vec{a}) dxdydz=\int\int\vec{a}\cdot\hat{n}dA$ and
$$ I(a,b,c) \,=\, \int_{x=0}^{a}\int_{y=0}^{b}\int_{z=0}^{c} \frac{dx\,dy\,dz}{(1+x^{2}+y^{2}+z^{2})^{3}} $$
Is there an $\vec{a}$ whose divergence is the denominator?
Solve for $a_r$ $\frac{1}{r^2}\frac{d}{d r}(r^2a_r)=\frac{1}{(1+r^2)^6}$ where $r^2=x^2+y^2+z^2$
$$a_r= \frac{1}{r^2}\int\frac{r^2dr}{(1+r^2)^6}$$
Then let $\vec{a}=a_r\hat{r}=\frac{a_r}{\sqrt{x^2+y^2+z^2}}(x\hat{i}+y\hat{j}+z\hat{k})$ and $a_r$ is also converted into an expression in cartesian coordinates.
The boundaries of the triple integral correspond to 6 faces of a rectangular prism. Each face has one of the cartesian unit vectors as a normal. Calculate the integral of $\vec{a}\cdot\hat{n}$ on each face where $\hat{n}$ is the outward facing normal.