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Denote by $\zeta$ the Riemann zeta function. I just learnt from this question $L_2$ bounds for tails of $\zeta(s)$ on a vertical line

that $\int_{T}^{\infty} \frac{\zeta(1/2 + it)}{1/4 + t^2}\mathrm{d}t=O(T^{-1}\log T)$ as $T\rightarrow +\infty$. But is the numerical value for the related integral

$$\int_{-\infty}^{\infty} \frac{\zeta(1/2 + it)}{1/4 + t^2} \mathrm{d}t $$ known?

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    $\begingroup$ Why should this integral converge? $\endgroup$ Oct 27, 2018 at 19:44
  • $\begingroup$ $\frac{\zeta(s)}{s} = \frac{1}{s-1}-\int_1^\infty (x-\lfloor x \rfloor) x^{-s-1}dx$ and $\frac{\zeta(s+1)}{s} = \frac{1}{s^2}-\frac{\gamma}{s} -\int_0^\infty (\sum_{n \le x}\frac{1}{n} - \log(x)+\gamma) x^{-s-1}dx$ thus $F(t)=\zeta(1/2+it)(\frac{1}{1/2+it}+\frac{1}{1-(1/2+it)})-\frac{1}{1/2+it-1}+\frac{1}{(1/2+it-1)^2}-\frac{\gamma}{1/2+it-1}$ is the Fourier transform of $f(u)=(e^u-\lfloor e^u \rfloor) e^{-u/2}-(\sum_{n \le e^u} \frac{1}{n}+u-\gamma)e^{u/2}$ whence $\lim_{T \to \infty}\int_{-T}^T F(t)dt = \lim_{u \to 0} \frac{f(u)+f(-u)}{2}$ $\endgroup$
    – reuns
    Oct 27, 2018 at 21:08
  • $\begingroup$ @reuns, thanks. So it seems to me that $\lim_{u\rightarrow 0} \frac{f(u)+f(-u)}{2}=0$, isn't it ? $\endgroup$
    – acc10
    Oct 27, 2018 at 21:21
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    $\begingroup$ @StanleyYaoXiao, i think the integral should converge since $\zeta(1/2 + it)=o(t)$. $\endgroup$
    – acc10
    Oct 27, 2018 at 21:23
  • $\begingroup$ No because $f(u)$ is supported on $u > 0$ (the $u < 0$ part is in the poles I substracted from $\frac{\zeta(1/2+it)}{t^2+1/4}$) $\endgroup$
    – reuns
    Oct 27, 2018 at 21:24

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For $\Re(s) > 1$, $$\frac{\zeta(s)}{s} = \int_1^\infty (\sum_{1 \le n \le x} 1) x^{-s-1}dx, \qquad \frac{\zeta(s)}{s-1} = \int_1^\infty (x\sum_{1 \le n \le x} \frac1n) x^{-s-1}dx$$ Thus for $\Re(s) \in (0,1)$ $$\frac{\zeta(s)}{s}+\frac{\zeta(s)}{1-s}= \int_0^\infty ((\sum_{1 \le n \le x} 1-x)-x(\sum_{1 \le n \le x} \frac1n-\log x-\gamma)) x^{-s-1}dx$$

So that $\frac{\zeta(1/2+it)}{1/4+t^2}$ is the Fourier transform of $$g(u) = ((\sum_{1 \le n \le e^u} 1)-e^u-e^u(\sum_{1 \le n \le e^u} \frac1n) + e^u u+e^u\gamma)e^{-u/2}$$ and $$\int_{-\infty}^\infty \frac{\zeta(1/2+it)}{1/4+t^2} dt = 2\pi\lim_{u \to 0} \frac{g(u)+g(-u)}{2}=2\pi( \gamma-1)$$

The integral converges absolutely because $\zeta(1/2+it) = O(t^c),c < 1$, and for the same reason you can shift the contour to $1/2+i(-\infty,+\infty)$ to $+\infty+i(-\infty,+\infty)$) and evaluate the integral as $-2 \pi \text{Res}(\frac{\zeta(s)}{s(1-s)},1)$.

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