# How get a good approximation, but simple, of $\int_0^{10}\frac{1}{(1+4y^2)^2}\left(\int_2^y\log \zeta(x+iy)dx\right)dy$?

I would to learn to calculate some approximation this kind of integrals, here there is an example that I don't know if my approach is good. Is a double integral involving the logarithm of the Riemann Zeta function. Thus I need to say (if I am wrong tell me) that I've consider the main branch of the logarithm.

Question. How one can evaluate the integral $$\int_0^{10}\frac{1}{(1+4y^2)^2}\left(\int_2^y\log \zeta(x+iy)dx\right)dy?$$ Many thanks.

Which I am interested is to find a way to evaluate this numerically. I know that $\zeta(s)$ has an Euler product convergent for $\Re s>1$, thus there is a closed-form for $\log \zeta(x+iy)$. Is it a good way? I wrote calculations for the integral $\int_2^y$,

$$\int_2^y\sum_{p \text{ prime}}\sum_{m=1}^\infty \frac{1}{m} p^{-mx}e^{-imy\log p}dx=\sum_{p \text{ prime}}\sum_{m=1}^\infty \frac{1}{m}e^{-imy\log p} (-\frac{e^{-mx\log p}}{m\log p}\Big|_2^y .$$

Now I know that I can evaluate and integrate (I presume that with a numerical method) finishig our result, that is a double series $\sum_{p \text{ prime}}\sum_{m=1}^\infty$, of terms of the shape $$\int_0^{10}\frac{1}{(1+4y^2)^2}\cdot\text{exponentials }dy.$$ After I need to take some terms $m$`s and primes $p$ to get an approximation doing calculations from a finite double series. Another way could be to use approximation from Taylor series of the functions in our integrand. Of course isn't required to use Euler products, if you know another way I am waiting.

What's is a good way to get an approximation?

Calculation. With code

integrate (1+4y^2)^(-2)log(RiemannZeta(x+iy))dxdy, from x=2 to y, from y=0 to 10

I get from Wolfram Alpha this approximation $\approx -0.5088+i0.841053.$

I am asking in my Question, about an approximation, isn't required so good as previous, I am interested in the theoretical discussion to obtain an approximation, not the best one.

• I presume that I need a numerical mehtod for our second integral, but as I am asking I don't know if an approximation likes WA did, or more simple, only is required a simple approximation, is possible using bounds or asymptotic behaviours in our integrands. What is your method to get a simple approximation? Many thanks for your attention. – user243301 Dec 14 '16 at 10:17
• $|\log \zeta(s)-\sum_{p^k < x} \frac{p^{-sk}}{k}| < \int_{x-1}^\infty |t^{-s}|dt = \frac{(x-1)^{1-Re(s)}}{1-Re(s)}$ – reuns Dec 14 '16 at 14:03
• Many thanks for this NICE inequality. I wait the approach of users, but I will think in it @user1952009 – user243301 Dec 14 '16 at 14:08
• Do you understand what I wrote and why it is useful here ? – reuns Dec 14 '16 at 14:20
• I belive that you are saying that with your identity one can get the comparison with my calculations, bounding. @user1952009 – user243301 Dec 14 '16 at 14:23