I am trying to find a closed form of the following integral
$$
\int _0^{\infty }\int _0^x\int _0^y\int _0^z \exp \left( -\frac{a x^2}{2}-\frac{b y^2}{2}-\frac{c z^2}{2}-\frac{d w^2}{2} \right) \,\mathrm{d}w\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x
$$
where $a,b,c,d>0$ are some constants.
My idea is to change variable to to polar system by letting
$$
x=\frac{r \cos (\alpha ) \cos (\beta ) \cos (\theta )}{\sqrt{a}}, \quad
y=\frac{r \cos (\alpha ) \cos (\beta ) \sin (\theta )}{\sqrt{a}}
$$
$$
z=\frac{r \sin (\alpha ) \cos (\beta )}{\sqrt{c}}, \quad w=\frac{r \sin (\beta )}{\sqrt{d}}
$$
This reduces the original integral into
$$
\int_0^{\tan ^{-1}\left(\sqrt{\frac{b}{a}}\right)} \frac{\sin (\theta ) \tan ^{-1}\left(\sin (\theta ) \sqrt{\frac{d}{b+c \sin ^2(\theta )}}\right)}{\sqrt{a b d \left(b+c \sin ^2(\theta )\right)}} \, d\theta
$$
But then I get stuck here.
PS: I am interested in this because I found that
$$
\int _0^{\infty }\int _0^x \exp \left(-\frac{a x^2}{2}-\frac{b y^2}{2}\right)\,\mathrm{d}y\,\mathrm{d}x
=
\frac{\tan ^{-1}\left(\sqrt{\frac{b}{a}}\right)}{\sqrt{a b}}
$$
and
$$
\int _0^{\infty }\int _0^x\int _0^y \exp \left( -\frac{a x^2}{2}-\frac{b y^2}{2}-\frac{c z^2}{2} \right) \,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x
=
\frac{\sqrt{\pi/2 } }{\sqrt{a b c}}
\left(\tan ^{-1}\left(\sqrt{\frac{c}{b}}\right)-\tan ^{-1}\left(\sqrt{\frac{a c}{b (a+b+c)}}\right)\right)
$$
So I am trying to generalize this. Maybe this is already known?
Best Answer
Define: \begin{eqnarray} I(a,b,c,d):=\int_0^{\tan ^{-1}\left(\sqrt{\frac{b}{a}}\right)} \frac{\sin (\theta ) \tan ^{-1}\left(\sin (\theta ) \sqrt{\frac{d}{b+c \sin ^2(\theta )}}\right)}{\sqrt{a b d \left(b+c \sin ^2(\theta )\right)}} \, d\theta \quad (i) \end{eqnarray} Then also define:
\begin{eqnarray} {\mathfrak F}^{(A,B)}_{a,b} &:=& \int\limits_A^B \frac{\log(z+a)}{z+b} dz\\ &=& F[B,a,b] - F[A,a,b] + 1_{t^* \in (0,1)} \left( -F[A+(t^*+\epsilon)(B-A),a,b] + F[A+(t^*-\epsilon)(B-A),a,b] \right) \quad (ii) \end{eqnarray} where \begin{eqnarray} t^*:=-\frac{Im[(A+b)(b^*-a^*)]}{Im[(B-A)(b^*-a^*)]} \end{eqnarray} and \begin{equation} F[z,a,b] := \log(z+a) \log\left( \frac{z+b}{b-a}\right) + Li_2\left( \frac{z+a}{a-b}\right) \end{equation} for $a$,$b$,$A$,$B$ being complex.
Then we have: \begin{eqnarray} I(a,b,c,d)&=& \frac{1}{\sqrt{a}} \int\limits_0^{\sqrt{\frac{d}{a+b+c}}} \frac{u \tan ^{-1}(u)}{\left(d-c u^2\right) \sqrt{d-u^2 (b+c)}} du\\ &=& \frac{\sqrt{d}}{\sqrt{a} (b+c)}\int\limits_0^{\sin ^{-1}\left(\sqrt{\frac{b+c}{a+b+c}}\right)} \frac{\sin (\phi ) \tan ^{-1}\left(\sin (\phi ) \sqrt{\frac{d}{b+c}}\right)}{d-\frac{c d \sin ^2(\phi )}{b+c}} d\phi\\ &=&-\frac{2 i}{\sqrt{a} \sqrt{d}} \int\limits_0^{\frac{\sqrt{\frac{b+c}{a+b+c}}}{\sqrt{\frac{a}{a+b+c}}+1}} \frac{t}{b \left(t^2+1\right)^2+c \left(t^2-1\right)^2} \log \left(\frac{2 i t \sqrt{\frac{d}{b+c}}+t^2+1}{-2 i t \sqrt{\frac{d}{b+c}}+t^2+1}\right) dt\\ &=&\frac{1}{4} \frac{1}{\sqrt{a b c d}} \sum\limits_{\xi=1}^4 \sum\limits_{\eta=1}^4 (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\left\lfloor \frac{\xi -1}{2}\right\rfloor } \int\limits_0^{\frac{\sqrt{\frac{b+c}{a+b+c}}}{\sqrt{\frac{a}{a+b+c}}+1}} \frac{\log \left(i (-1)^{\left\lfloor \frac{\xi -1}{2}\right\rfloor } \sqrt{\frac{d}{b+c}}+i (-1)^{\xi -1} \sqrt{\frac{b+c+d}{b+c}}+t\right)}{t-i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\eta +1} e^{i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor } \tan ^{-1}\left(\frac{\sqrt{c}}{\sqrt{b}}\right)}} dt \\ &=&\frac{1}{4} \frac{1}{\sqrt{a b c d}} \sum\limits_{\xi=1}^4 \sum\limits_{\eta=1}^4 (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\left\lfloor \frac{\xi -1}{2}\right\rfloor } % {\mathfrak F}^{(0,\frac{\sqrt{\frac{b+c}{a+b+c}}}{\sqrt{\frac{a}{a+b+c}}+1})}_{i (-1)^{\left\lfloor \frac{\xi -1}{2}\right\rfloor } \sqrt{\frac{d}{b+c}}+i (-1)^{\xi -1} \sqrt{\frac{b+c+d}{b+c}},-i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\eta +1} e^{i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor } \tan ^{-1}\left(\frac{\sqrt{c}}{\sqrt{b}}\right)}} \end{eqnarray}
In the top line we substituted for $u=\sin(\theta) \sqrt{d/(b+c \sin(\theta)^2)}$. In the second line we substituted $u = \sqrt{d/(c+b)} \sin(\phi)$. In the third line we substituted $t=\tan(\phi/2)$. In the forth line we used partial fraction decomposition and properties of the logarithm. Finally in the fifth line we used the anti-derivative defined in $(ii)$.
Update: As a sanity check look at the case $a=b=c=d=1$. Then define: \begin{eqnarray} M1&:=&\left( \begin{array}{cccc} -1+\sqrt{3} & \sqrt{2} & \sqrt{2} & -1+\sqrt{3} \\ 1 & \sqrt{2-\sqrt{3}} & \sqrt{2-\sqrt{3}} & 1 \\ \sqrt{2-\sqrt{3}} & 1 & 1 & \sqrt{2-\sqrt{3}} \\ \sqrt{2} & -1+\sqrt{3} & -1+\sqrt{3} & \sqrt{2} \\ \end{array} \right)\\ M2&:=&\left( \begin{array}{cccc} \frac{1}{\sqrt{2}} & \frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{2} \left(-1+\sqrt{3}\right) & \frac{1}{2} \left(-1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} \\ \frac{1}{2} \left(-1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{2} \left(-1+\sqrt{3}\right) \\ \frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{2} \left(1+\sqrt{3}\right) \\ \end{array} \right)\\ A1&:=&\left( \begin{array}{cccc} -\frac{\pi }{6} & \frac{\pi }{12} & -\frac{\pi }{4} & 0 \\ \frac{5 \pi }{6} & \frac{\pi }{12} & \frac{5 \pi }{12} & -\frac{\pi }{3} \\ -\frac{5 \pi }{12} & \frac{\pi }{3} & -\frac{5 \pi }{6} & -\frac{\pi }{12} \\ \frac{\pi }{4} & 0 & \frac{\pi }{6} & -\frac{\pi }{12} \\ \end{array} \right)\\ A2&:=&\left( \begin{array}{cccc} -\frac{\pi }{12} & \frac{\pi }{6} & -\frac{\pi }{6} & \frac{\pi }{12} \\ \frac{7 \pi }{12} & -\frac{\pi }{6} & \frac{\pi }{6} & -\frac{7 \pi }{12} \\ -\frac{\pi }{6} & \frac{7 \pi }{12} & -\frac{7 \pi }{12} & \frac{\pi }{6} \\ \frac{\pi }{6} & -\frac{\pi }{12} & \frac{\pi }{12} & -\frac{\pi }{6} \\ \end{array} \right) \end{eqnarray} and we have \begin{eqnarray} I(1,1,1,1)=\frac{1}{4} \sum\limits_{\xi=1}^4 \sum\limits_{\eta=1}^4 (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\left\lfloor \frac{\xi -1}{2}\right\rfloor } \left( Li_2(M1_{\xi,\eta}\exp(\imath A1_{\xi,\eta}))- Li_2(M2_{\xi,\eta}\exp(\imath A2_{\xi,\eta})) \right) \end{eqnarray} We have checked numerically that this quantity above coincides with $\pi^2/96$ to one hundred digits. It would be interesting to prove this analytically.