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I need to find the following limiting value: $$lim_{T\to\infty}\int_{2-iT}^{2+iT} \frac{a^s}{s}ds$$ where $s \in \mathbb{C}$ and $a\in \mathbb{R}$

This came up in the context of trying to understand the distribution of primes with respect to the zeros of the Riemann Zeta Function. You might find some printing mistakes, but you may refer to 'A History of PNT' by LJ Goldstein for the same.

Thank you in advance

user511110
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  • $a \ge 0$ (for other $a$ it diverges). It depends if you prefer the residue theorem or Fourier analysis. Both give the answer $2i\pi \ 1_{a > 1}$ (at the discontinuity point it is the mean value $1/2$) – reuns Jan 16 '20 at 09:31
  • Would you please elaborate, I can't quite see it. – user511110 Jan 16 '20 at 09:34
  • Does this proof mean something to you ($1/s$ is the Laplace transform of $1_{t > 0}$) ? Otherwise do you know what is $Res(a^s/s,0)$ ? – reuns Jan 16 '20 at 09:36
  • I do know that residue is calculated over a simple closed contour – user511110 Jan 16 '20 at 09:45
  • Also, the integral along a contour with $0$ within it will give $Res(a^s/s,0)$=$2\pi i$ – user511110 Jan 16 '20 at 09:51

1 Answers1

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Let $C_T$ be the rectangle $[-T,2]+i[-T,T]$. Then for $a > 1$ $$\lim_{T\to \infty} \frac1{2i\pi}\int_{\partial \ C_T} \frac{a^s}{s}ds=lim_{T\to\infty}\frac1{2i\pi}\int_{2-iT}^{2+iT} \frac{a^s}{s}ds$$ On the other hand $$\frac1{2i\pi}\int_{\partial \ C_T} \frac{a^s}{s}ds=Res(\frac{a^s}{s},0)=1$$ For $a \in (0,1)$ it is the same with the rectangle $R_T=[2,T]+i[-T,T]$, it contains no pole thus $\frac1{2i\pi}\int_{\partial \ R_T} \frac{a^s}{s}ds=0$.

This way we have proven the Mellin inversion theorem for $1_{x > 1}$

reuns
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  • How is the contour chosen for integration? – Fella Jan 16 '20 at 11:46
  • $\partial \ C_T$ is the boundary of a rectangle – reuns Jan 16 '20 at 11:50
  • That is certainly understood, but how are the rectangles chosen with respect to the variation of values of $a$? – Fella Jan 16 '20 at 11:51
  • Was looking for the reason for choosing different contours for these kinds of integrals. Would be obliged if answered. – Fella Jan 16 '20 at 12:11
  • It is because for $a >1$, $a^s$ is rapidly decreasing as $\Re(s)\to - \infty$ whereas for $a \in (0,1)$ it is as $\Re(s)\to \infty$ – reuns Jan 16 '20 at 12:26