I'm wondering on how one can go about proving that
$$\frac{1}{2\pi \imath} \int_{\left(\mathcal{C}\right)} \frac{\zeta^2(1-n)\,z^{-n}}{2\cos\left(n\pi /2\right)}\,\mathrm{d}n = -\gamma -\frac12 \log z – \frac{1}{4\pi z}+\frac{z}{\pi}\sum_{n= 1}^{+\infty} \frac{\tau(n)}{z^2+n^2}$$
where $\tau(n)$ represents the divisor function, $\gamma$ denotes the Euler-Mascheroni constant, $\zeta$ represents the Riemann zeta function, $1<\mathcal{C}<2$ and $\int_{(\mathcal{C})}$ denotes the line integral $\int_{\mathcal{C}-\imath \infty}^{\mathcal{C} + \imath\infty}$.
So far I tried complex analytic methods (Residue theorem and contour integration) but no progress.
Any help would be highly appreciated.
Best Answer
I'd say shift the contour to the left to reach the region where $\zeta(1-s)^2=\sum_{m\ge 1} \tau(m) m^{s-1}$ converges absolutely and uniformly on the vertical lines.
No problem to do so because $\zeta(s)$ is $O(|s|^r)$ in the vertical strip so the exponential decay of $1/\cos(\pi s/2)$ makes it ok.
This will add two residues at $0$ and $1$.
Then use the absolute/uniform convergence to say that $$\int_{(-1/2)} \frac{\zeta(1-s)^2}{2 \cos(\pi s/2)} z^{-s}ds=\sum_{m\ge 1} \tau(m)\int_{(-1/2)} \frac{m^{s-1}}{2 \cos(\pi s/2)} z^{-s}ds$$ Where the last integral is easily computed with the residue theorem, for $z\in (0,1)$:
$$\int_{(-1/2)} \frac{m^{s-1}}{2 \cos(\pi s/2)} z^{-s}ds = 2i\pi \sum_{k=0}^\infty Res(\frac{m^{s-1}}{2 \cos(\pi s/2)} z^{-s},-2k-1)$$ $$= 2i \sum_{k=0}^\infty \frac{m^{-2k-2}}{(-1)^k} z^{2k+1} = \frac{2i z}{m^2+z^2} $$
You can extend to the remaining $z\in \Bbb{C}-i \Bbb{Z}_{\ge 1}$ by analytic continuation.