Calculus – Sequences and series – The Riemann zeta function

The Riemann zeta function \(\zeta(s)\) is defined as: \(\displaystyle\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}\) for \(\text{Re}(s)>1\).

We will only consider real values of \(s\); in that case we know that the series is absolutely convergent if \(s > 1\). However, in general it is not very easy to find its sum.

Later we will be able to show that \(\displaystyle\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{1}{6}\pi^2\) and \(\displaystyle\zeta(4)=\sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{1}{90}\pi^4\).

The value of \(\zeta(3)\approx1.202057\) is also known as Apéry's constant.

One proof of the first result might be understandable right now. It is based on the Taylor series of \(\displaystyle\frac{\sin(x)}{x}\) and an infinite product. For the infinite product we need to know that the (only but all) zeros of \(\displaystyle\frac{\sin(x)}{x}\) are \(x=\pm n\pi\) with \(n\in\{1,2,3,\ldots\}\) and that \(\displaystyle\lim\limits_{x\to0}\frac{\sin(x)}{x}=1\). Then we have:

\[\frac{\sin(x)}{x}=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}x^{2k}=\prod_{n=1}^{\infty}\left(1-\frac{x}{n\pi}\right)\left(1+\frac{x}{n\pi}\right) =\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right).\]

Comparing the coefficients of \(x^2\) we obtain that \(-\displaystyle\frac{1}{3!}=-\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2} \;\;\Longrightarrow\;\;\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{1}{6}\pi^2\).

Note that

\[\sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}=\sum_{n=1}^{\infty}\frac{1}{n^s}-\sum_{n=1}^{\infty}\frac{1}{(2n)^s}=\left(1-\frac{1}{s^2}\right)\zeta(s)\]

and

\[\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^s}=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}-\sum_{n=1}^{\infty}\frac{1}{(2n)^s} =\sum_{n=1}^{\infty}\frac{1}{n^s}-2\sum_{n=1}^{\infty}\frac{1}{(2n)^s}=\left(1-\frac{1}{2^{s-1}}\right)\zeta(s).\]

Using integration by parts we find that

\[\int_0^1x^n\ln(x)\,dx=\frac{1}{n+1}x^{n+1}\ln(x)\bigg|_0^1-\frac{1}{n+1}\int_0^1x^{n+1}\cdot\frac{1}{x}\,dx=0-\frac{1}{n+1}\int_0^1x^n\,dx =-\frac{1}{(n+1)^2}x^{n+1}\bigg|_0^1=-\frac{1}{(n+1)^2}\]

for \(n=0,1,2,\ldots\). Now we use the series representation \(\displaystyle\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\) for \(|x| < 1\) to find that

\[\int_0^1\frac{\ln(x)}{1-x}\,dt=\sum_{n=0}^{\infty}\int_0^1x^n\ln(x)\,dx=-\sum_{n=0}^{\infty}\frac{1}{(n+1)^2} =-\sum_{n=1}^{\infty}\frac{1}{n^2}=-\zeta(2)=-\frac{1}{6}\pi^2.\]

Furthermore using integration by parts we find that

\[\int_0^1x^n\left(\ln(x)\right)^2\,dx=\frac{1}{n+1}x^{n+1}\left(\ln(x)\right)^2\bigg|_0^1-\frac{2}{n+1}\int_0^1x^{n+1}\ln(x)\cdot\frac{1}{x}\,dx =0-\frac{2}{n+1}\int_0^1x^n\ln(x)\,dx=\frac{2}{(n+1)^3}\]

for \(n=0,1,2,\ldots\). Again we use the series representation \(\displaystyle\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\) for \(|x| < 1\) to find that

\[\int_0^1\frac{\left(\ln(x)\right)^2}{1-x}\,dx=\sum_{n=0}^{\infty}\int_0^1x^n\left(\ln(x)\right)^2\,dx=2\sum_{n=0}^{\infty}\frac{1}{(n+1)^3} =2\sum_{n=1}^{\infty}\frac{1}{n^3}=2\zeta(3).\]

Similarly, we have

\[\int_0^1\frac{\ln(x)}{1+x}\,dx=\sum_{n=0}^{\infty}(-1)^n\int_0^1x^n\ln(x)\,dx=-\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2} =-\left(1-\frac{1}{2}\right)\zeta(2)=-\frac{1}{12}\pi^2\]

and

\[\int_0^1\frac{\left(\ln(x)\right)^2}{1+x}\,dx=\sum_{n=0}^{\infty}(-1)^n\int_0^1x^n\left(\ln(x)\right)^2\,dx=2\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^3} =2\left(1-\frac{1}{4}\right)\zeta(3)=\frac{3}{2}\zeta(3).\]

Using the Taylor series \(\displaystyle\ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}\) for \(|x| < 1\) we find that

\[\int_0^1\frac{\ln(1-x)}{x}\,dx=-\sum_{n=1}^{\infty}\frac{1}{n}\int_0^1x^{n-1}\,dx=-\sum_{n=1}^{\infty}\frac{1}{n}\cdot\frac{1}{n}x^n\bigg|_0^1 =-\sum_{n=1}^{\infty}\frac{1}{n^2}=-\zeta(2)=-\frac{1}{6}\pi^2,\] \begin{align*} \int_0^1\ln(x)\ln(1-x)\,dx&=-\sum_{n=1}^{\infty}\frac{1}{n}\int_0^1x^n\ln(x)\,dx=\sum_{n=1}^{\infty}\frac{1}{n}\cdot\frac{1}{(n+1)^2} =\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}-\frac{1}{(n+1)^2}\right)\\[2.5mm] &=1-\sum_{n=1}^{\infty}\frac{1}{(n+1)^2}=2-\sum_{n=1}^{\infty}\frac{1}{n^2}=2-\zeta(2)=2-\frac{1}{6}\pi^2 \end{align*}

and

\[\int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx=-\sum_{n=1}^{\infty}\frac{1}{n}\int_0^1x^{n-1}\ln(x)\,dx=\sum_{n=1}^{\infty}\frac{1}{n}\cdot\frac{1}{n^2} =\sum_{n=1}^{\infty}\frac{1}{n^3}=\zeta(3).\]

By symmetry (or a substitution) we also have \(\displaystyle\int_0^1\frac{\ln(x)\ln(1-x)}{1-x}\,dx=\zeta(3)\) and therefore

\[\int_0^1\frac{\ln(x)\ln(1-x)}{x(1-x)}\,dx=\int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx+\int_0^1\frac{\ln(x)\ln(1-x)}{1-x}\,dx=2\zeta(3).\]

Similarly, using \(\displaystyle\ln(1+x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}x^{n+1}\) for \(|x| < 1\) we have

\[\int_0^1\frac{\ln(1+x)}{x}\,dx=\sum_{n=0}^{\infty}\int_0^1\frac{(-1)^n}{n+1}x^n\,dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}=\left(1-\frac{1}{2}\right)\zeta(2)=\frac{1}{12}\pi^2,\] \begin{align*} \int_0^1\ln(x)\ln(1+x)\,dx&=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\int_0^1x^{n+1}\ln(x)\,dx=-\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\cdot\frac{1}{(n+2)^2} =\sum_{n=0}^{\infty}\left(\frac{(-1)^n}{n+2}-\frac{(-1)^n}{n+1}+\frac{(-1)^n}{(n+2)^2}\right)\\[2.5mm] &=1-2\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+2)^2}=1-2\ln(2)+1-\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}\\[2.5mm] &=2-2\ln(2)-\left(1-\frac{1}{2}\right)\zeta(2)=2-2\ln(2)-\frac{1}{12}\pi^2 \end{align*}

and

\[\int_0^1\frac{\ln(x)\ln(1+x)}{x}\,dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\int_0^1x^n\ln(x)\,dx=-\sum_{n=1}^{\infty}\frac{(-1)^n}{n+1}\cdot\frac{1}{(n+1)^2} =-\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^3}=-\left(1-\frac{1}{4}\right)\zeta(3)=-\frac{3}{4}\zeta(3).\]

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Last modified on March 15, 2021