Using Perron's formula, I am able to show that
$$ \psi(x)={1\over2\pi i}\int_{a-i\infty}^{a+i\infty}\left[-{\zeta'(s)\over\zeta(s)}\right]{x^s\over s}\mathrm ds $$
where $\psi(x)$ is the Chebyshev's psi function. Now, I aim to find an explicit formula for $\psi(x)$ by evaluating it using this contour:
Writing it mathematically, we have
$$ {1\over2\pi i}\oint_{\gamma+\Gamma}\left[-{\zeta'(s)\over\zeta(s)}\right]{x^s\over s}\mathrm ds={1\over2\pi i}\int_\Gamma\left[-{\zeta'(s)\over\zeta(s)}\right]{x^s\over s}\mathrm ds+{1\over2\pi i}\int_{a-iT}^{a+iT}\left[-{\zeta'(s)\over\zeta(s)}\right]{x^s\over s}\mathrm ds $$
where the left hand side, using argument principle, is
$$ {1\over2\pi i}\oint_{\gamma+\Gamma}\left[-{\zeta'(s)\over\zeta(s)}\right]{x^s\over s}\mathrm ds=x-{\zeta'(0)\over\zeta(0)}-\sum_{\Re(\rho)>0\atop\zeta(\rho)=0}{x^\rho\over\rho}-\frac12\ln(1-x^{-2}) $$
which conforms to the actual explicit formula for $\psi(x)$, indicating that
$$ \lim_{T\to\infty}\int_\Gamma\left[-{\zeta'(z)\over\zeta(z)}\right]{x^z\over z}\mathrm dz=0 $$
Plugging the parametrization for the $\Gamma$ contour gives
$$ \Gamma: z=a+Te^{it},\mathrm dz=iTe^{it}\mathrm dt,t:\pi/2\to3\pi/2 \\ \begin{aligned} \lim_{T\to\infty}\left|\int_\Gamma{\zeta'(z)\over\zeta(z)}{x^z\over z}\mathrm dz\right| &=\lim_{T\to\infty}\left|\int_{\pi/2}^{3\pi/2}{\zeta'(a+Te^{it})\over\zeta(a+Te^{it})}{x^{a+Te^{iT}}\over a+Te^{it}}Te^{it}\mathrm dt\right| \\ &\le\lim_{T\to\infty}x^aT\left|\int_{\pi/2}^{3\pi/2}\left|\zeta'(a+Te^{it})\over\zeta(a+Te^{it})\right|{x^{T\cos t}\over|a-T|}\mathrm dt\right| \\ &=\pi x^a\lim_{T\to\infty}{T\over T-a}\color{blue}{\left|\zeta'(a+Te^{i\eta})\over\zeta(a+Te^{i\eta})\right|}{x^{T\cos\eta}} \end{aligned} $$
Since $\eta$ is chosen between $\frac\pi2$ and $\frac{3\pi}2$, $e^{T\cos\eta}\to0$, but in order to prove that the entire integral goes to zero as $T\to\infty$, I have to establish an upper bound for the blue term. Could anybody give me a hand on this?