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Story

I'm trying to prove the following identity

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

where

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

and

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

is the xi function on the critical line.

Problem

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 $\Gamma(s) \pi^{-s} \zeta(2s)$.

Source

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

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

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