A $\zeta(2n+1)$ series

Let $\zeta$ denote the Riemann Zeta function. Evaluate the series:

$$\sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{(n+1)(2n+1)}$$

(Serafim Tsipelis, Anastasios Kotronis)

Solution

 In what follows $\zeta^* (n)$ stands for $\zeta^*(n)=\left\{\begin{matrix}
\zeta(n) &,  & n \neq 1 \\
 \gamma&,  &n=1
\end{matrix}\right.$. Then it is quite known that

$$\sum_{n=1}^{\infty} \zeta^*(n) x^n = - x \psi(1-x)$$

It follows naturally then that $\displaystyle \sum_{n=1}^{\infty} \frac{\zeta^*(n)}{n}x^n = \log \Gamma(1-x)$. Setting in the last equation $x \mapsto -x$ and substracting the two series we finally get that:

$$ \sum_{n=0}^{\infty} \frac{\zeta^*(2n+1)}{2n+1}x^{2n+1} = \frac{1}{2} \log \left ( \frac{\Gamma(1-x)}{\Gamma(1+x)} \right)$$ 

We have thus established the equation

$$ \gamma x + \sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{2n+1} x^{2n+1} = \frac{1}{2} \log \left ( \frac{\Gamma(1-x)}{\Gamma(1+x)} \right ) \tag{1} \label{*} $$

Integrating \eqref{*} we have that:

\begin{align*}
\frac{1}{2}\gamma + \frac{1}{2}\sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{(2n+1) (n+1)} &=\frac{1}{2} \int_{0}^{1} \log \left ( \frac{\Gamma(1-x)}{\Gamma(x+1)} \right ) \, {\rm d}x\\
 &= \frac{1}{2} \int_{0}^{1} \left [ \log \Gamma(1-x) - \log (1+x) \right ] \, {\rm d}x
\end{align*}

So it remains to evaluate those log Gamma integrals. For the first one:

\begin{align*}
\int_{0}^{1} \log \Gamma(1-x) \, {\rm d}x &\overset{u=1-x}{=\! =\! =\! =\!} \int_{0}^{1}\log \Gamma (x) \, {\rm d}x \\
 &= \frac{1}{2} \left ( \int_{0}^{1} \log \Gamma(1-x) \, {\rm d}+\int_0^1 \log \Gamma(x) \, {\rm d}x \right ) \\
 &= \frac{1}{2} \int_{0}^{1} \log \Gamma(1-x) \Gamma (x) \, {\rm d}x\\
 &= \frac{1}{2} \int_{0}^{1} \log \pi \csc \pi x \, {\rm d}x  \\
 &= \frac{\log \pi}{2} - \frac{1}{2}\int_{0}^{1} \log \sin \pi x \, {\rm d}x \\
 &= \frac{\log \pi}{2} +\frac{\log 2}{2} \\
 &= \frac{\log 2\pi}{2}
\end{align*}

and for the second one , in more general it holds that:

$$\int_0^1 \log \Gamma(x+a) \;{\rm d}x= \frac{\log 2\pi}{2} + \alpha \log a -1 , \;\; a \geq 0$$

since if we define the function $\displaystyle f(a)= \int_{0}^{1} \log \Gamma (x+a) \, {\rm d}x$ and differentiate with respect to $a$ we get that:

 \begin{align*}
f'(a) &=\frac{\mathrm{d} }{\mathrm{d} a} \int_{0}^{1}\log \Gamma(x+a) \, {\rm d}x\\
 &= \int_{0}^{1}\frac{\partial }{\partial a} \log \Gamma(x+a) \, {\rm d}x\\
 &=\int_{0}^{1} \frac{\Gamma'(x+a)}{\Gamma(x+a)} \, {\rm d}x  \\
 &= \int_{0}^{1}\left [ \log \Gamma(x+a) \right ]' \, {\rm d}x\\
 &=  \left [ \log \Gamma(x+a) \right ]_0^1 \\
 &= \log \Gamma(1+a) - \log \Gamma (a) \\
 &= \log a \Gamma (a) - \log \Gamma (a) \\
 &= \log a + \log \Gamma (a) - \log \Gamma (a) \\
 &= \log a
\end{align*}

Thus $f(a)=a\log a - a +c$. But $f$ is continuous at $x_0=0$ thus $\lim \limits_{a \rightarrow 0^+} f(x) = f(0)$. But $f(0)$ is the previous integral and evaluates to $\frac{\log 2\pi}{2}$.

Hence

$$f(a)=\frac{\log 2\pi}{2} +a\log a - a $$

And thus our series evaluates to

$$\sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{(n+1)(2n+1)}= 1- \gamma$$