I am trying to calculate the integral of the contour

$$ frac {1} {2 pi i} int_ {c – i infty} ^ {c + i infty} zeta ^ 2 ( omega) frac {8 ^ omega} { omega} omega $$

or $ c> $ 1, $ zeta (s) $ is Riemann's zeta function.

Use Perron's formula and define $ D (x) = sum_ {k leq x} sigma_0 (n) $, or $ sigma_0 $ is the usual function of counting the divisors, we 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 for this purpose, we can just calculate $ D (8) $ and call it a day. However, for my own needs, I want to redefine $ D (x) $ by the full above instead. Therefore, why I state the problem for a specific case $ x = $ 8, for example.

I have made some progress.

Considering a modified Bromwich contour which avoids branch cutting and $ z = 0 $, let's call it $ mathcal {B} $, we can apply the Cauchy residue theorem:

$$ oint _ { mathcal {B}} zeta ^ 2 ( omega) frac {8 ^ omega} { omega} d omega = 2 pi i operatorname * {Res} ( zeta ^ 2 ( omega) frac {8 ^ { omega}} { omega}; 1) = 8 (-1 + 2 gamma + ln 8) $$

or $ gamma $ is the Euler-Mascheroni constant. I got this by extending $ zeta ^ 2 ( omega) frac {8 ^ omega} { omega} $ in his Laurent series. To obtain the desired integral, it would then be necessary to subtract from this value the parts of the contour which are not the vertical line $ c – iR $ at $ c + iR $, subtract them from the residual value obtained, then take the limit as $ R to infty $ and $ r to 0 $ or $ C_r $ is the circle of radius $ r $ where the $ mathcal {B} $ dodges the origin.

Feel free to modify this outline in any shape or form, or consider a positive integer value different from $ x $.