integrationpower seriesriemann-zetazeta-functions
I would like to ask if someone would help me solve the following integral.
\begin{equation}
\zeta(3) = \int_0^1 \int_0^1 \int_0^1 \frac{1}{1-xyz}\ dx\,dy\,dz
\end{equation}
It occurred to me that if I transformed the integral into spherical, cylindrical, polar coordinates, or turned the volume created by integrals. Maybe it would help in solving.
Thank you in advance.
Here, I evaluate the forth part of the region so after evaluating the triple integral you should multiply the result to $4$. According to Fig below, we have a rectangle triangular $OCD$. So $\sin(\phi_0)=1/2$ and so $\phi_0=\pi/6$. It means that $\phi\in [\pi/6,\pi/2]$. As you found out the rang of $\theta$ is $[0,\pi/2]$. But about the $\rho$:
You see that our region is starting to shape from the intersection till we meet $z=0$. For finding the range of $\rho$, make an arrow arbitrary to cut the cylinder at $A$ and next cut the sphere at $B$. What is $\rho$ at $A$ and what is at $B$? We know that in a spherical coordinates: $$x=\rho\sin(\phi)\cos(\theta)\\y=\rho\sin(\phi)\sin(\theta)\\z=\rho\cos(\phi)$$ then the equation of our cylinder $x^2+y^2=1$ be converted to $\rho^2\sin^2(\phi)=1$ or $\rho=\frac{1}{\sin(\phi)}$. This is the first coordinate $\rho$ at $A$. Clearly, $\rho$ at $B$ is $2$. Hence you have: $$\int_0^{\pi/2}d\theta\int_{\pi/6}^{\pi/2}\sin(\phi)d\phi\int_{\frac{1}{\sin(\phi)}}^{2}\rho^2d\rho$$.
Look at this plot in $1/8$ of all space, $x,y,z\ge0$:
We have to find two volumes in this part of $\mathbb R^3$. Once, you get the whole volume of the following volume inside the sphere, you can multiply it by $8$ and then subtract it from $4/3\pi(2a)^3$. But about this below shape:
$$V_1: \phi|_0^{\pi/6}, ~~\theta|_0^{\pi/2},~~\rho|_0^{2a}\\ V_2: \phi|_{\pi/6}^{\pi/2}, ~~\theta|_0^{\pi/2},~~\rho|_0^{a\csc(\phi)}$$
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} {\dd x\,\dd y\,\dd z \over 1 - xyz}} = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \bracks{\sum_{n = 0}^{\infty}\pars{xyz}^{n}}\dd x\,\dd y\,\dd z \\[5mm] = &\ \sum_{n = 0}^{\infty}\pars{\int_{0}^{1}x^{n}\,\dd x} \pars{\int_{0}^{1}y^{n}\,\dd y} \pars{\int_{0}^{1}z^{n}\,\dd z} = \sum_{n = 0}^{\infty}{1 \over \pars{n + 1}^{3}} \\[5mm] = &\ \sum_{n = 1}^{\infty}{1 \over n^{3}} = \bbx{\zeta\pars{3}} \end{align}