In an earlier post, titled A Fun Integral Problem, I gave a calculation for an integral as an infinite sum $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$. I’ve been told that I should generalize it, so I’ll do that here (hence the title). I will also show how one may calculate some of the particular values of the Riemann $\zeta$ function we will be getting from this generalization, thus making good on my promise to give the value of that sum.
Problem. Calculate the integral
\[
\int_{\left[0,1\right]^{k}}\frac{dx_{1}\cdots dx_{k}}{1-x_{1}\cdots x_{k}}.
\]
Stop here if you want to try to use that previous post to solve this generalization. Continue reading for the solution as well as the actual values of the integrals for various values of $k$.
Solution.
As in the original post, we see that we have a function of the form $\frac{1}{1-x_{1}\cdots x_{k}}$, which suggests that we may use the sum $\sum_{n=0}^{\infty}\left(x_{1}\cdots x_{k}\right)^{n}$ since $|x_{i}|<1$ for $i=1,2,\ldots,k$ in almost all of $\left[0,1\right]$. Thus, our integral becomes
\[
\int_{\left[0,1\right]^{k}}\frac{dx_{1}\cdots dx_{k}}{1-x_{1}\cdots x_{k}} = \int_{\left[0,1\right]^{k}}\sum_{n=0}^{\infty}\left(x_{1}\cdots x_{k}\right)^{n} dx_{1}\cdots dx_{k} =
\]
\[
\sum_{n=0}^{\infty}\int_{\left[0,1\right]^{k}}\left(x_{1}\cdots x_{k}\right)^{n} dx_{1}\cdots dx_{k} = \sum_{n=0}^{\infty}\left(\int_{0}^{1}x^{n} dx\right)^{k}=
\]
\[
\sum_{n=0}^{\infty}\left(\frac{1}{n+1}\right)^{k}=\sum_{n=1}^{\infty}\frac{1}{n^{k}}=\zeta\left(k\right).
\]
Notice that we've now introduced the Riemann $\zeta$ function. It's been seen in previous posts, but now we're actually going to calculate some of the values, in particular the values at all positive even integers.
Let's define a sequence, $B_{k}$, which we will call Bernoulli numbers, by
\[
\frac{z}{e^{z}-1}=\sum_{n=0}^{\infty}B_{n}\frac{z^{n}}{n!}.
\]
One important fact is that these Bernoulli numbers are all rational.
Next, let's look at the following power series expansions of $z\cot z$.
First, we have
\[
z\cot z=1-2\sum_{n=1}^{\infty}\frac{z^{2}}{n^{2}\pi^{2}-z^{2}}=
\]
\[
1-2\sum_{n=1}^{\infty}\left(\frac{z}{n\pi}\right)^{2}\cdot\frac{1}{1-\left(\frac{z}{n\pi}\right)^{2}}=1-2\sum_{n=1}^{\infty}\left(\frac{z}{n\pi}\right)^{2}\sum_{k=0}^{\infty}\left(\frac{z}{n\pi}\right)^{2k}=
\]
\[
1-2\sum_{k=0}^{\infty}\left(\sum_{n=1}^{\infty}\frac{1}{n^{2k+2}}\right)\left(\frac{z}{\pi}\right)^{2k+2}=1-2\sum_{k=1}^{\infty}\left(\sum_{n=1}^{\infty}\frac{1}{n^{2k}}\right)\left(\frac{z}{\pi}\right)^{2k}=
\]
\[
1-2\sum_{k=1}^{\infty}\zeta\left(2k\right)\left(\frac{z}{\pi}\right)^{2k}.
\]
On the other hand, we also have
\[
z\cot z=z\frac{\cos z}{\sin z}=iz\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}=iz+\frac{2iz}{e^{2iz}-1}=iz+\sum_{k=0}^{\infty}B_{k}\frac{\left(2iz\right)^{k}}{k!}=
\]
\[
iz + 1 - iz + \sum_{k=2}^{\infty}B_{k}\frac{\left(2iz\right)^{k}}{k!} = 1+\sum_{k=2}^{\infty}B_{k}\frac{\left(2iz\right)^{k}}{k!}.
\]
Noting that $B_{k}=0$ for all odd $k>2$ and comparing the coefficients for $z^{k}$, we get the following formula:
\[
\zeta\left(2k\right)= \left(-1\right)^{k+1}\frac{\left(2\pi\right)^{2k}}{2\left(2k\right)!}B_{2k},\;\; k\geq1.
\]
Here are a few values, up to $2k=12$ to show that sometimes there is a numerator other than just a power of $\pi$:
\[
\zeta\left(2\right)=\frac{\pi^{2}}{6}, \;\;\; \zeta\left(4\right)=\frac{\pi^{4}}{90}, \;\;\; \zeta\left(6\right)=\frac{\pi^{6}}{945},
\]
\[
\zeta\left(8\right)=\frac{\pi^{8}}{9450}, \;\;\; \zeta\left(10\right)=\frac{\pi^{10}}{93555}, \;\;\; \zeta\left(12\right)=\frac{691\pi^{12}}{638512875}.
\]