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.

  • $\begingroup$ I know that the $\frac{1}{2}$ doesn't make sense concerning the sequence of zeros of the cosine function. My main purpose is to know how calculate or get an approximation of this integral. $\endgroup$ – user243301 Dec 14 '16 at 10:44
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    $\begingroup$ Usually we evaluate those integrals using the residue theorem. But i have not seen the case with modulus of a function inside the integrand. If your function has not modules then you can take a closed contour , say a rectangle with sides $(1/2-iT,1/2+iT), (1/2+iT,U+iT),(U+iT,-U-iT),(-U-iT,-1/2-iT)$ the singularities of your function are all inside this rectangle and you calculate the integral using the residue theorem. $\endgroup$ – Beslikas Thanos Dec 14 '16 at 12:26
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    $\begingroup$ If you want just an estimate for this you can take the same contour , or a semicircle with diameter given by the limits of integration $(1/2-iT,1/2+iT)$and you can apply the estimation lemma and then take $T \to \infty$ $\endgroup$ – Beslikas Thanos Dec 14 '16 at 12:30
  • $\begingroup$ Many thanks for your attention @BeslikasThanos, now I read your remarks, and thanks also to Cettt $\endgroup$ – user243301 Dec 14 '16 at 12:31
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    $\begingroup$ You can find these kind of integral in Perron's formula for Dirichlet series. A good book to get a glimpse of Dirichlet series is Titchmarsh's The theory of Functions. But only to start with ! $\endgroup$ – Beslikas Thanos Dec 14 '16 at 12:33

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.

  • $\begingroup$ Thanks to you and users in comments, I am reading your answer. $\endgroup$ – user243301 Dec 14 '16 at 13:02

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