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I'm trying to prove that $$ \int_{0}^{\infty}\frac{\Xi(t)}{t^2+\frac{1}{4}}\cos(xt)dt=\frac{\pi}{2}\Big(e^{x/2}-2e^{-x/2}\psi(e^{-2x})\Big), $$ where $\Xi(t)=\xi(1/2+it)$ and $\psi(x)=\sum_{n=1}^{\infty}e^{-n^2\pi x}$. For this purpose I'm following the book "The Theory of the Riemann Zeta-function" by Titchmarsh and I got stuck in the following equality: $$ \frac{-1}{4\sqrt{y}i}\int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty}\Gamma\Big(\frac{s}{2}\Big)\pi^{-s/2}\zeta(s)y^sds=-\frac{\pi}{\sqrt y}\psi\Big(\frac{1}{y^2}\Big)+\frac{\pi}{2}\sqrt y. $$ My attempt: By Mellin inversion we know that $$ \psi(x)=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\pi^{-s}\Gamma(s)\zeta(2s)x^{-s}ds \quad\quad (\sigma\gt 1/2). $$ Now to prove the equality, let $u=s/2$ so that $$ \frac{-1}{4\sqrt{y}i}\int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty}\Gamma\Big(\frac{s}{2}\Big)\pi^{-s/2}\zeta(s)y^sds=\frac{-1}{2\sqrt{y}i}\int_{\frac{1}{4}-i\infty}^{\frac{1}{4}+i\infty}\Gamma(u)\pi^{-u}\zeta(2u)y^udu, $$ and then try to use Mellin inversion to obtain the desired the result. The problem is that I don't know how to change the interval of integration from $1/4-i\infty$ to $1/4+i\infty$, to something of the form $\sigma-i\infty$ to $\sigma + i\infty$ for some $\sigma\gt 1/2$. Also, if I apply Mellin inversion directly I only get the factor $-\frac{\pi}{\sqrt y}\psi(y^{-2})$, so I don't know where the factor $\frac{\pi}{2}\sqrt y$ comes from.

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  • $\begingroup$ You can move the line of integration using the residue theorem, $\Gamma(s)$ is fast decreasing on vertical lines (eg. from the reflection formula) while $\zeta(s)$ is polynomially bounded on vertical strips. To check that you didn't make any mistake : check that the Mellin/Laplace/Fourier transform of the obtained function converges for $s$ in the considered vertical line (you know the asymptotics of $\psi$ from the Poisson summation formula) $\endgroup$
    – reuns
    Oct 6 '19 at 23:08
  • $\begingroup$ That's right, thanks. The problem is that if I apply Mellin inversion after moving the line of integration I only get the first factor, I don't know where the factor $\frac{\pi}{2}\sqrt{y}$ comes from $\endgroup$
    – Sebitas
    Oct 6 '19 at 23:10
  • $\begingroup$ You are supposed to use that $\Xi$ or $\Gamma(s/2)\pi^{-s/2}\zeta(s)$ is invariant under $s\to 1-s$ somewhere. The second factor is the residue at the pole (the term that need to be substracted to change the domain of convergence of the Laplace transform) $\endgroup$
    – reuns
    Oct 6 '19 at 23:10
  • $\begingroup$ Could you please elaborate on the residue of the pole? $\endgroup$
    – Sebitas
    Oct 6 '19 at 23:13
  • $\begingroup$ What is your $\Xi$ function here? $\endgroup$
    – clathratus
    Oct 6 '19 at 23:17
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The idea is to move the line of integration from $1/2-i\infty\to 1/2+i\infty$ to $2-i\infty\to 2+i\infty$ passing through the pole of $\zeta(s)$ at $s=1$. Considering the rectangular contour with vertices at $1/2-iR, 1/2+iR, 2-iR , 2-iR$ and letting $R\to\infty$, by the Residue Theorem it follows that $$ -\frac{1}{4 i \sqrt{y}} \int_{\frac{1}{2} - i\infty}^{\frac{1}{2} + i\infty} \Gamma(\frac{1}{2}s) \pi^{-\frac{1}{2}s} \zeta(s) y^s ds=-\frac{1}{4 i \sqrt{y}} \int_{2 - i\infty}^{2 + i\infty} \Gamma(\frac{1}{2}s) \pi^{-\frac{1}{2}s} \zeta(s) y^s ds + \frac{\pi}{2} \sqrt{y}, $$ where the $\pi/2 \sqrt{y}$ is the residue of the integrand at $s=1$ (The integral along the horizontal lines vanishes). Now use Mellin inversion to recover $\psi$ from $\Gamma$ and $\zeta$, namely $$ \psi(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\pi^{-s}\Gamma(s)\zeta(2s)x^{-s}ds \quad\quad(c\gt \frac{1}{2}). $$

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