一道含参三重积分

青青子衿

\[ \color{black}{\int_{0}^{\frac{1}{\alpha}}\int_{0}^{1}\int_{0}^{1}\frac{1}{\left(1+x^{2}+y^{2}+\left(\alpha z\right)^{2}\right)^{2}}\mathrm{d}x\mathrm{d}y\mathrm{d}z=\dfrac{\,\pi^2}{32\alpha}} \]

题目来源：AMM 10771
blog.sina.com.cn/s/blog_4fe74d840102yzai.html

hbghlyj

系统维护中，博文仅作者可见。登陆后可查看本人文章。

😥

hbghlyj

Substituting $\alpha z\mapsto z$, we only need to prove
\[\iiint_{[0,1]^3}\frac{1}{\left(1+x^{2}+y^{2}+z^2\right)^{2}}\mathrm{d}x\mathrm{d}y\mathrm{d}z=\dfrac{\,\pi^2}{32}\]
math.stackexchange.com/questions/2955676

@Z Ahmed 's solution

Let us use $x=r \cos \phi,~ y= r \sin \phi$ and write
$$I=\int_{0}^{1}\int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{(1+x^2+y^2+z^2)^2}= 2\int_{0}^{1} \int_{0}^{\pi/4} \int_{0}^{\sec\phi} \frac{r dr d\phi dz}{(1+r^2+z^2)^2}\tag1$$$$ \Rightarrow I =\int_{0}^{1} \int_{0}^{\pi/4} \left(\frac{1}{1+z^2}-\frac{1}{1+z^2+\sec^2\phi}\right) d\phi dx=\int_{0}^{1} \int_{0}^{1} \frac{1}{(1+z^2)(2+z^2+t^2)} dz~ dt,\tag2$$ $t=\sec\phi$ taken here. Next, interchange $t$ and $z$ to get $$I=\int_{0}^{1} \int_{0}^{1} \frac{dz dt}{(1+t^2)(2+z^2+t^2)}\tag3$$ Adding (2) and (3), we get $$I=\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{1}{2+z^2+t^2} \left( \frac{1}{1+z^2}+\frac{1}{1+t^2} \right)dx~dt=\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{ dz~ dt}{(1+z^2)(1+t^2)}= \frac{1}{2} \left ( \int_{0}^{1}\frac {dz}{1+z^2} \right)^2=\frac{\pi^2}{32}.$$

hbghlyj

It can be written as the divergence of a function of 4 variables:
$$\nabla_{x,y,z,t}\cdot\frac{(x,y,z,-t)}{x^2+y^2+z^2+1}=\frac2{(x^2+y^2+z^2+1)^2}$$

1. Div[{x,y,z,-t}/(x^2+y^2+z^2+1),{x,y,z,t}]//Factor

Copy the Code

@eyeballfrog commented: It can be written as the Laplacian of $-\frac{\tan^{-1}(r)}{2r}$, where $r = \sqrt{x^2+y^2+z^2}$

1. Laplacian[-ArcTan[r]/(2r)/.r->Sqrt[x^2+y^2+z^2],{x,y,z}]//Simplify

Copy the Code

hbghlyj

Let \[I=\iiint_{[0,1]^3}\frac2{\left(1+x^{2}+y^{2}+z^2\right)^{2}}\mathrm{d}x\mathrm{d}y\mathrm{d}z\]we want to prove $I=\frac{\pi^2}{16}$

This integral is equal to the integral over $[0,1]^4$: (since the integrand is independent of $t$.)
\[I=\iint\!\!\!\iint_{[0,1]^4}\frac{2}{\left(x^2+y^2+z^2+1\right)^2}dxdydzdt\]
By calculations in 4#, divergence theorem tells us
\[I=\iiint_{\partial[0,1]^4}\frac{(x,y,z,-t)}{x^2+y^2+z^2+1}\]
Note that $\partial[0,1]^4$ is the union of 8 cubes:
$C_{1,0}=\{0\}\times[0,1]\times[0,1]\times[0,1]$
$C_{1,1}=\{1\}\times[0,1]\times[0,1]\times[0,1]$
$C_{2,0}=[0,1]\times\{0\}\times[0,1]\times[0,1]$
$C_{2,1}=[0,1]\times\{1\}\times[0,1]\times[0,1]$
$C_{3,0}=[0,1]\times[0,1]\times\{0\}\times[0,1]$
$C_{3,1}=[0,1]\times[0,1]\times\{1\}\times[0,1]$
$C_{4,0}=[0,1]\times[0,1]\times[0,1]\times\{0\}$
$C_{4,1}=[0,1]\times[0,1]\times[0,1]\times\{1\}$
Calculate the surface integrals ("flux" through the surface)
\[\iiint_{C_{1,0}}\frac{(0,y,z,-t)}{0^2+y^2+z^2+1}\cdot(-1,0,0,0)=0\]
Similarly, the integral over $C_{2,0},C_{3,0},C_{4,0}$ are zero.
\begin{align*}
\iiint_{C_{1,1}}&\frac{(1,y,z,-t)}{1^2+y^2+z^2+1}\cdot(1,0,0,0)
\\&=\iiint_{[0,1]^3}\frac{1}{y^2+z^2+2}dydzdt
\\&=\iint_{[0,1]^2}\frac{1}{y^2+z^2+2}dydz
\\&=\int_0^1\int_0^1\int_0^{\infty } \exp \left(-t \left(y^2+z^2+2\right)\right) \, dtdydz
\\&=\int_0^{\infty } \exp \left(-2t\right) \left(\int_0^1\exp\left(-ty^2\right)dy\right)\left(\int_0^1\exp\left(-tz^2\right)dz\right)dt
\\\small\text{Using }&\small\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_0^xe^{-t^2} dt
\\&=\frac{\pi}4\int_0^{\infty }{\exp \left(-2t \right)\over t}\operatorname{erf}^2(\sqrt t) dt
\end{align*}By symmetry of $x,y,z$ the integral over $C_{1,1},C_{2,1},C_{3,1}$ should be the same.
\begin{align*}
\iiint_{C_{4,1}}&\frac{(x,y,z,-1)}{x^2+y^2+z^2+1}\cdot(0,0,0,1)\\
&=-\iiint_{[0,1]^3}\frac1{x^2+y^2+z^2+1}dxdydz\\
&=-\int_0^1\int_0^1\int_0^1\int_0^\infty\exp\left(-t(x^2+y^2+z^2+1)\right)\,dtdxdydz
\\&=-\int_0^{\infty } \exp \left(-t\right)\left(\int_0^1\exp\left(-tx^2\right)dx\right)\left(\int_0^1\exp\left(-ty^2\right)dy\right)\left(\int_0^1\exp\left(-tz^2\right)dz\right)dt
\\\small\text{Using }&\small\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_0^xe^{-t^2} dt
\\&=-\frac{\pi^{3/2}}8\int_0^{\infty }{\exp \left(-t \right)\over t^{3/2}}\operatorname{erf}^3(\sqrt t) dt
\end{align*}
Adding up, using this result,
\begin{align*}
I&=3\cdot\frac{\pi}4\int_0^{\infty }{\exp \left(-2t \right)\over t}\operatorname{erf}^2(\sqrt t) dt-\frac{\pi^{3/2}}8\int_0^{\infty }{\exp \left(-t \right)\over t^{3/2}}\operatorname{erf}^3(\sqrt t) dt
\\&=\frac{\pi^2}{16}
\end{align*}