I'm trying to compute the contour integral $$\frac{1}{2 \pi i} \int_{c - i \infty}^{c + i \infty} \zeta^2(\omega) \frac{8^\omega}{\omega} \ d \omega$$ where $c > 1$, $\zeta(s)$ is the Riemann zeta function.
Using Perron's Formula and defining $D(x) = \sum_{k \leq x} \sigma_0(n)$, where $\sigma_0$ is the usual divisor counting function, one can show that $$D(x) = \frac{1}{2 \pi i} \int_{c - i \infty}^{c + i \infty} \zeta^2(\omega) \frac{x^\omega}{\omega} \ d \omega.$$ So, to that end one can just compute $D(8)$ and call it a day. However, for my own purposes I want to redefine $D(x)$ by the above integral instead. Hence, why I state the problem for a specific case $x = 8$, for example.
I have made some progress.
By considering a modified Bromwich contour that avoids the branch cut and $z = 0$, lets call it $\mathcal{B}$, we can apply Cauchy's Residue Theorem: $$\oint_{\mathcal{B}} \zeta^2(\omega) \frac{8^\omega}{\omega} d \omega = 2 \pi i \operatorname*{Res}(\zeta^2(\omega) \frac{8^{\omega}}{\omega}; 1) = 8(-1 + 2 \gamma + \ln 8)$$ where $\gamma$ is the Euler-Mascheroni constant. I obtained this by expanding $\zeta^2(\omega) \frac{8^\omega}{\omega}$ into its Laurent series. To get the desired integral we would then need subtract from this value the portions of the contour that are not the vertical line $c - iR$ to $c + iR$, and then take the limit as $R \to \infty$ and $r \to 0$ where $C_r$ is the circle of radius $r$ where the $\mathcal{B}$ dodges the origin.
Feel free to modify this contour in any shape or form, or consider a different positive integer value of $x$.
PS: I’ve had to move this post around between sites. I think here is the most appropriate place, sorry if you’ve seen this before. This is the post’s final resting place.