I'm trying to prove the following identity

$$\int_0^\infty \frac{\Xi(t)}{t^2 + \frac{1}{4}} \cos(xt) dt = \frac{1}{2} \pi (e^{\frac{1}{2}x} - 2e^{-\frac{1}{2}x} \psi(e^{-2x}))$$


$$\psi(x)=\sum_{n = 1}^{\infty} e^{-n^2 \pi x}$$


$$\Xi(t) = \xi(\frac{1}{2} + it)$$

is the xi function on the critical line.


I only don't understand the following equality

$$-\frac{1}{4 i \sqrt{y}} \int_{\frac{1}{2} - i\infty}^{\frac{1}{2} + i\infty} \Gamma(\frac{1}{2}s) \pi^{-\frac{1}{2}s} \zeta(s) y^s ds = -\frac{\pi}{\sqrt{y}} \psi(\frac{1}{y^2}) + \frac{1}{2} \pi \sqrt{y}.$$

Where does the latter summand $\frac{1}{2} \pi \sqrt{y}$ come from? When I write $(\frac{1}{y})^{-s}$ and substitute with $s = 2w$ I only get the first summand. We know that by the Mellin transform theorem $\psi(x)$ can be recovered by the inverse Mellin transform integral over $\Gamma(s) \pi^{-s} \zeta(2s)$.


This is from Titchmarsh's book "The Theory of the Riemann Zeta-function" p. 35−36.

enter image description here

  • $\begingroup$ haha ! $\theta(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s)$ is the Mellin transform of $\psi(x) = \sum_{n \ge 1} e^{-\pi n^2 x^2}$, but only for $Re(s) > 1$ ! for $Re(s) < 1$ it doesn't converge ($\theta(s)$ has a pole at $s=1$) and substracting it, you can see that on $Re(s) \in (0,1)$ , $\theta(s)$ is the Mellin transform of $\psi(x) - 1$ $\endgroup$ – reuns Jun 23 '16 at 17:27
  • 2
    $\begingroup$ to be clear : for $Re(s) > 1$ : $\int_0^1 x^{s-2} dx = \int_0^\infty \frac{1_{x < 1}}{x} x^{s-1} dx = \frac{1}{s-1}$ so that $\theta(s) - \frac{1}{s-1} = \int_0^\infty (\psi(x) - \frac{1_{x < 1}}{x} ) x^{s-1} dx$ converges for $Re(s) > 0$. $\ \ $ note also that for $Re(s) <1 $ : $\int_1^\infty x^{s-2} dx = \int_0^\infty \frac{1_{x > 1}}{x} x^{s-1} dx = -\frac{1}{s-1}$ i.e. for $Re(s) \in (0,1)$ : $$\theta(s) = \theta(s) - \frac{1}{s-1}+ \frac{1}{s-1} = \int_0^\infty (\psi(x) - \frac{1_{x < 1}}{x} - \frac{1_{x > 1}}{x}) x^{s-1} dx = \int_0^\infty (\psi(x) - \frac{1}{x}) x^{s-1} dx$$ $\endgroup$ – reuns Jun 23 '16 at 17:34
  • $\begingroup$ but if it was $\psi(x) - 1$ we'd get $\frac{1}{2} \pi \frac{1}{\sqrt{y}}$ $\endgroup$ – fje Jun 23 '16 at 17:37
  • 2
    $\begingroup$ no. Let $f(x) = \sum_{n=1}^\infty e^{- \pi n^2 x^2}$ and $\theta(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s)$. for $Re(s) \in (0,1)$ : $$\theta(s) = \int_0^\infty (f(x) - \frac{1}{x}) x^{s-1} dx$$ so $$f(x) - \frac{1}{x} = \frac{1}{2i \pi} \int_{1/2-i \infty}^{1/2+i \infty} \theta(s) x^{-s} ds$$ and $$\frac{1}{2i \pi} \int_{1/2-i \infty}^{1/2+i \infty} \theta(s) y^{s} ds = f(1/y) - y$$ which is what is expected $\endgroup$ – reuns Jun 23 '16 at 18:37
  • 1
    $\begingroup$ I explained this above, by substracting $\displaystyle\frac{1_{x < 1}}{x}$ I substract $\frac{1}{s-1}$, and by substracting $\displaystyle\frac{1_{x > 1}}{x}$ I add it back, so that substracting $\frac{1}{x}$ doesn't change the function but only its domain of convergence $\endgroup$ – reuns Jun 23 '16 at 18:42

I know it's too late to post an answer but maybe it will be useful in the future. 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 $$ \psi(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\pi^{-s}\Gamma(s)\zeta(2s)x^{-s}ds \quad\quad(c\gt \frac{1}{2}). $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.