TITLE: From the series $\sum_{n=1}^{+\infty}\left(H_n-\ln n-\gamma-\frac1{2n}\right)$ to $\zeta(\frac12+it)$.
QUESTION [24 upvotes]: Here is a pretty series 

$$
\displaystyle \sum_{n=1}^{+ \infty} \left(H_{n}-\ln n-\gamma -\frac{1}{2n}\right)=\frac{1}{2} \left(1-\ln (2\pi)+\gamma\right) \tag{*}
$$

where $H_{n}:=\sum_{1}^{n} \frac{1}{k}$ are the harmonic numbers and $\gamma := \lim\limits_{n \to \infty} (H_n- \ln n)$ is the Euler constant.
$$ $$

Now just introduce a parameter in the general term of the series and you get a link with... the Riemann $\zeta$ function on the critical line!

Q 1. What proof would you give for (*)? 
Q 2. What elements would you give to get the link with $\zeta\left(\frac{1}{2}+it\right)$?

REPLY [3 votes]: Some generalizations:
$$
\begin{aligned}
&\boxed{\begin{aligned}
&\sum_{n = 1}^{\infty} n
\left ( H_n-\gamma-\ln n-\frac{1}{2n}+\frac{1}{12n^2}  \right ) = \frac{5}{24}+\frac{\gamma}{12}-\ln A\\
&\sum_{n = 1}^{\infty} n^2
\left ( H_n-\gamma-\ln n-\frac{1}{2n}+\frac{1}{12n^2}  \right ) = \frac{1}{24}-\frac{\zeta(3)}{4\pi^2}\\
&\sum_{n=1}^{\infty}(-1)^{n-1}n^4\left ( H_n-\gamma
-\ln n-\frac{1}{2n} +\frac{1}{12n^2}-\frac{1}{120n^4}   \right )
=\frac{59}{240}-\frac{93\zeta(5)}{4\pi^4} \\
&\sum_{n=1}^{\infty} (-1)^{n-1}n^4
\left(H_n^{(2)}-\frac{\pi^2}{6}+\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{6n^3}    \right)
=\frac{\ln2}{2}-\frac{1}{3}
\end{aligned}}\\ \\
&\boxed{\begin{aligned}
&\sum_{n=1}^{\infty} (-1)^{n-1}\left ( H^{(3)}_n-\zeta(3) \right )
= -\frac{1}{8}\zeta(3)\\
&\sum_{n=1}^{\infty} (-1)^{n-1}n\left ( H^{(3)}_n-\zeta(3) \right )
= \frac{\pi^2}{24} -\frac{7}{16}\zeta(3)\\
&\sum_{n=1}^{\infty} (-1)^{n-1}n^2\left ( H^{(3)}_n-\zeta(3) +\frac{1}{2n^2} \right )
= \frac{\ln2}{2} -\frac{\pi^2}{24}+\frac{1}{4}\\
&\sum_{n=1}^{\infty} (-1)^{n-1}n^3\left ( H^{(3)}_n-\zeta(3) +\frac{1}{2n^2} -\frac{1}{2n^3} \right)
=\frac{7}{32}\zeta(3)-\frac{3}{4}\ln2 +\frac{1}{8}\\
&\sum_{n=1}^{\infty} (-1)^{n-1}n^4\left ( H^{(3)}_n-\zeta(3) +\frac{1}{2n^2} -\frac{1}{2n^3}+\frac{1}{4n^4} \right)
=\frac{\pi^2}{24}-\frac{9}{24}\\
&\sum_{n=1}^{\infty} (-1)^{n-1}n^5\left ( H^{(3)}_n-\zeta(3) +\frac{1}{2n^2} -\frac{1}{2n^3}+\frac{1}{4n^4} \right)
=-\frac{7}{16}\zeta(3)+\frac{5}{4}\ln2-\frac{5}{16}
\end{aligned} }\\ \\
&\boxed{\begin{aligned}
&\sum_{n=1}^{\infty}(-1)^{n-1}\left ( H^{(4)}_n -\frac{\pi^4}{90} \right )
= -\frac{\pi^4}{1440}\\
&\sum_{n=1}^{\infty}(-1)^{n-1}n\left ( H^{(4)}_n -\frac{\pi^4}{90} \right )
=\frac{3}{8}\zeta(3) -\frac{\pi^4}{192}\\
&\sum_{n=1}^{\infty}(-1)^{n-1}n^2\left ( H^{(4)}_n -\frac{\pi^4}{90} \right )
=-\frac{9}{24}\zeta(3)+\frac{\pi^2}{24}\\
&\sum_{n = 1}^{\infty}(-1)^{n-1}n^3\left ( H^{(4)}_n -\frac{\pi^4}{90}+\frac{1}{3n^3}  \right ) 
= -\frac{\pi^2}{16}+\frac{\pi^4}{384}+\frac{\ln2}{2} +\frac{1}{6}
\end{aligned}}
\end{aligned}
$$

Start from the following integral
$$\int_{0}^{\infty} x^{s-1}
\left ( \psi^{(0)}(1+x)-\ln x\right ) \text{d}x
=-\frac{\pi\zeta(1-s)}{\sin\pi s}$$
Where $0<s<1$.
And
$$
\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} 
\frac{-\pi\zeta(1-s)}{\sin(\pi s)}z^{-s}\text{d}s
=\psi^{(0)}(1+z)-\ln z
$$
Where $0<c<1$.
Thus we have
$$\begin{aligned}
&\int_{-\infty}^{\infty} \frac{\zeta\left ( \frac{1}{2}+it  \right ) }{\cosh(\pi t)}\text{d}t  = 2\gamma-2\\
&\int_{-\infty}^{\infty} \frac{\zeta\left ( -\frac{1}{2}-it  \right ) \zeta\left ( \frac{3}{2}+it  \right )}{\cosh(\pi t)}\text{d}t
=\gamma-\ln(2\pi)+1
\end{aligned}$$