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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:

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?

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  • $\begingroup$ Yes, that integral indeed tends to $0$. It is well-known that $$\zeta'(s)/\zeta(s) = O(\log^7 t) $$ when $s = \sigma+it$, $\sigma \geq 1, t>1$. Functional equation of $\zeta$ then says $\zeta'(s)/\zeta(s) = O(\log^7 |t|)$ for all $s\in \Gamma$ with $|t|>1$, $\sigma <0$. $\endgroup$ – pisco Oct 13 at 4:53
  • $\begingroup$ @pisco Is there a proof for that? $\endgroup$ – Travor Liu Oct 13 at 4:57
  • $\begingroup$ An introductory reference would be The theory of the Riemann zeta-function (2nd Edition) by Titchmarsh, chapter $3$ section $6$. The key is the identity $3+4\cos x + \cos 2x = 2(1+\cos x)^2$. The book is also excellent if you are interested in Riemann zeta. $\endgroup$ – pisco Oct 13 at 5:07
  • $\begingroup$ @pisco Thanks for that, and you are correct: it is Riemann zeta function that motivates me to learn all these mathematics! $\endgroup$ – Travor Liu Oct 13 at 5:11
  • $\begingroup$ @pisco I read your source, but I really did not find inequalities for $\zeta'(s)/\zeta(s)$ for $\Re(s)<1$ $\endgroup$ – Travor Liu Oct 13 at 8:41

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