# integral expression involving $\Xi(t)$

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.

• 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) Oct 6 '19 at 23:08
• 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 Oct 6 '19 at 23:10
• 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) Oct 6 '19 at 23:10
• Could you please elaborate on the residue of the pole? Oct 6 '19 at 23:13
• What is your $\Xi$ function here? Oct 6 '19 at 23:17

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 $$\psi(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\pi^{-s}\Gamma(s)\zeta(2s)x^{-s}ds \quad\quad(c\gt \frac{1}{2}).$$