Does $\int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty}\frac{\log (\left| \cos(s)\right|)}{ \left| s \right|^2}dt$ converge? Can you evaluate it?

Question

Let $s=\sigma+it$ the complex variable, after I read the statement of an equivalence to an unsolved problem involving the Riemann Zeta function, and since I would like to know more facts from comparisons between $\zeta(s)$ and simple functions I thought this

Question. Evaluate $$\int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty}\frac{\log (\left| \cos(s)\right|)}{ \left| s \right|^2}dt.$$

That is, in our integrand $\log (\left| \cos(s)\right|$ is the natural logarithm of the modulus of the complex cosine $\cos(s)=\cos(\sigma+it)$, and in the denominator $\left| s \right|^2=\left| \sigma+it \right|^2$, from the notation of our complex variable.

I don't know how start to study this. Is well-possed (does converge)? I know easy facts like to calculate the modulus of our cosine. Can you get an approximation or get the exact value of this integral? Many thanks.

Answer 1

Look at the integral

$$\int_{1/2-i T}^{1/2+i T} dt \frac{\log{(|\cos{(1/2+i t)}|)}}{1/4+t^2} $$

$$\cos{(1/2+i t)} = \cos{(1/2)} \cosh{t} - i \sin{(1/2)} \sinh{t}$$ $$|\cos{(1/2+i t)}|^2 = \cos^2{(1/2)} \cosh^2{t} + \sin^2{(1/2)} \sinh^2{t} $$

$$\log{(|\cos{(1/2+i t)}|)} = \frac12 \log{(\cos^2{(1/2)} \cosh^2{t} + \sin^2{(1/2)} \sinh^2{t} )} $$

As $T \to \infty$, each of the cosh and sinh terms look like $\frac14 e^{2 t}$, so the log becomes a linear term in $t$. Thus, the integral diverges logarithmically.