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The Mellin inverse is given by

$$ \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}dsF(s)x^{-s} $$

Is it possible to compute the inverse transform of:

$$F(s)=a^{-s} \zeta (s) $$ where: $a>0$

Mathematica gave me solution:

$$\mathcal{M}_s^{-1}\left[a^{-s} \zeta (s)\right](x)=\frac{1}{a x} ? $$ for: $\ln (a)>0\land 0<x<1$

,but if I

$$\mathcal{M}_x\left[\frac{1-\theta (x-1)}{a x}\right](s)=\frac{1}{a (-1+s)}+\frac{2 \pi \delta (i (-1+s))}{a}$$ I not get $a^{-s} \zeta (s) $ ?

Using functional equation formula for Zeta and Borel Regularization for the series I have:

$$\mathcal{M}_s^{-1}\left[a^{-s} \zeta (s)\right](x)=\mathcal{M}_s^{-1}\left[\frac{a^{-s} \zeta (1-s)}{2 (2 \pi )^{-s} \cos \left(\frac{\pi s}{2}\right) \Gamma (s)}\right](x)=\mathcal{M}_s^{-1}\left[\frac{a^{-s} \sum _{k=1}^{\infty } \frac{1}{k^{1-s}}}{2 (2 \pi )^{-s} \cos \left(\frac{\pi s}{2}\right) \Gamma (s)}\right](x)=\sum _{k=1}^{\infty } \mathcal{M}_s^{-1}\left[\frac{a^{-s}}{\left(2 (2 \pi )^{-s} \cos \left(\frac{\pi s}{2}\right) \Gamma (s)\right) k^{1-s}}\right](x)=\sum _{k=1}^{\infty } \frac{2 \cos \left(\frac{2 k \pi }{a x}\right)}{a x}=-\frac{1}{a x}$$ My and Mathematica solution are correct ?

I tried find solution in Tables 1 2 3 of Mellin Transforms ,but I didn't find my example.

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Since

$$\zeta(s)=s\, \mathcal{M}_x[f(x)](-s)=s \int\limits_0^\infty f(x)\, x^{-s-1}\, dx=\sum\limits_{n=1}^\infty \frac{1}{n^s}\,,\quad\Re(s)>1\tag{1}$$

where

$$f(x)=\sum\limits_{n=1}^x 1=\sum\limits_{n=1}^x \theta(x-n)=\sum\limits_{n=1}^\infty \theta(x-n)\tag{2},$$

perhaps you should instead consider the inverse Mellin transform

$$\mathcal{M}_s^{-1}\left[\frac{G(s)}{s}\right]\left(\frac{1}{x}\right)=\sum\limits_{n=1}^\infty \theta(x-a n)\tag{3}$$

where

$$G(s)=a^{-s}\, \sum\limits_{n=1}^\infty \frac{1}{n^s}=a^{-s}\, \zeta(s)\,,\quad\Re(s)>1\tag{4}.$$


In other words

$$s\, \mathcal{M}_x[g(x)](-s)=s \int\limits_0^\infty g(x)\, x^{-s-1}\, dx=a^{-s}\, \zeta(s)\, ,\quad\Re(s)>1\tag{5}$$

where

$$g(x)=\sum\limits_{n=1}^\infty \theta(x-a n)\tag{6}.$$


But if your really want to consider

$$\mathcal{M}_s^{-1}\left[a^{-s}\, \zeta(s)\right](x)\tag{7}$$

then

$$\frac{2}{a x} \sum\limits_{k=1}^{\infty} \cos\left(\frac{2 k \pi}{a x}\right)\ne-\frac{1}{a x}\tag{8}$$

since

$$\mathcal{M}_x\left[-\frac{1}{a x}\right](s)=-\frac{2 \pi}{a}\, \delta(i (s-1))\ne a^{-s}\, \zeta(s)\tag{9}.$$


Rather I believe

$$h(x)=\frac{2}{a x} \sum\limits_{n=1}^{\infty} \cos\left(\frac{2 n \pi}{a x}\right)=\frac{1}{a x} \sum\limits_{n=1}^\infty \delta\left(\frac{1}{a x}-n\right),\quad x>0\tag{10}$$

and note the term-wise Mellin transform

$$\mathcal{M}_x\left[\frac{1}{a x}\, \delta\left(\frac{1}{a x}-n\right)\right](s)=\int\limits_0^{\infty} \frac{1}{a x}\, \delta \left(\frac{1}{a x}-n\right) \, x^{s-1} \, dx=(a n)^{-s}\tag{11}$$

is consistent with

$$\mathcal{M}_x\left[h(x)\right](s)=\int\limits_0^{\infty} h(x)\, x^{s-1} \, dx=a^{-s} \sum\limits_{n=1}^\infty n^{-s}=a^{-s}\, \zeta(s)\,,\quad\Re(s)>1\tag{12}.$$

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