Inverse Mellin Transform $(ix)^{-s}$

Question

Consider the inverse Mellin transform of the function $(ix)^{-s}$ for $x>0$. This is: $$ \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} (i x)^{-s} ds $$ What does this evaluate to? I believe that if you simply had $x^{-s}$ then the result would be $\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} x^{-s} ds = \delta(x-1)$. However what happens in this case where there a factor of $i$?

Answer 1

I don't believe the inverse Mellin transform of $i^{-s}$ converges, but one can approximate the inverse Mellin transform of $i^{-s}$ as

$$f(x)=\mathcal{M}_s^{-1}\left[i^{-s}\right](x)=\underset{T\to\infty}{\text{lim}}\left(\frac{1}{2 \pi i}\int\limits_{c-i T}^{c+i T} i^{-s}\, x^{-s} \, ds\right)$$ $$=\underset{T\to \infty }{\text{lim}}\left(\frac{e^{-\frac{1}{2} \pi (T+i c)} \left(e^{\pi T}-x^{2 i T}\right) x^{-c-i T}}{\pi (\pi -2 i \log (x))}\right)\tag{1}.$$

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Figures (1) to (3) below illustrate $|f(x)|$, $\Re(f(x))$, and $\Im(f(x))$ where formula (1) is evaluated with $T=100$ and $T=101$ in blue and orange respectively where in all cases the value $c=2$ was used to evaluate formula (1). I believe $c=2$ is a valid choice since the function $i^{-s}$ is analytic for $s\in\mathbb{C}$. Note formula (1) for $f(x)$ seems to diverge significantly in magnitude as $T$ increases.

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Figure (1): Illustration of $|f(x)|$ with $T=100$ (blue) and $T=101$ (orange)

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Figure (2): Illustration of $\Re(f(x))$ with $T=100$ (blue) and $T=101$ (orange)

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Figure (3): Illustration of $\Im(f(x))$ with $T=100$ (blue) and $T=101$ (orange)

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I'll also note the frequency of the oscillations in $\Re(f(x))$ and $\Im(f(x))$ also seem to diverge (increase) as $T$ increases, but this is perhaps not as noticeable in Figures (1) to (3) above. This is because $T=100$ and $T=101$ are very close in magnitude and the frequency oscillation doesn't seem to diverge as rapidly as the magnitude as $T$ increases.

Answer 2

I start by saying that, in the developments below

1. I follow the path described by @TymaGaidash in their comment to the OP.
2. I approach the problem by using reference [1], precisely example 11.2.7 in chapter 11 "The Mellin Transform" by Jacqueline Bertrand, Pierre Bertrand and Jean-Philippe Orvalez: however the development is not the same.
3. I use the "function" notation for distributions, i.e. I'll write $f(x)$ instead of $\langle f,\varphi\rangle$ for $\varphi \in C^\infty_0(\Bbb R^n)$ and thus I'll write $$\DeclareMathOperator{\Dd}{d\!} \delta(x-x_0)=\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty} e^{-i\omega(x-x_0)} \Dd\omega\label{1}\tag{1} $$ and moreover, for the change of variable formula in the theory of distributions I refer to this old answer of mine and to the references cited there.

The inverse Mellin transform of $i^{-s}$, $s\in \Bbb C$ is $$ \begin{split} f(x) = \mathscr{M}^{-1} [\varphi(s)]( x) & = \frac{1}{2\pi i}\int\limits_{c-i\infty}^{c+i\infty} \varphi(s)x^{-s} \Dd s =\frac{1}{2\pi i}\int\limits_{c-i\infty}^{c+i\infty} (ix)^{-s} \Dd s = \frac{1}{2\pi i}\int\limits_{-i\infty}^{+i\infty} ({x}{i})^{-s} \Dd s\\ &\text{(the choice $c=0$ is allowed by the holomorphy $(ix)^{-s}$}\\ &\text{in the whole $s$-plane, and thus by putting $s=i\omega$ we get)} \\ & = \frac{1}{2\pi}\int\limits_{-\infty}^{+\infty} e^{-i\omega \ln({x}{i})} \Dd \omega = \frac{1}{2\pi}\int\limits_{-\infty}^{+\infty} e^{-i\omega (\ln x+ \ln i)} \Dd \omega\\ &\text{(and finally, by identity \eqref{1}, and by the standard}\\ &\text{change of variable formula for distributions)}\\ & = \delta (\ln x+\ln i)=\delta \left(\ln({x}{i})\right)= \delta (x+i) \end{split} $$ By using the change of variable formula, it can be easily seen that $\delta(x+i)=\delta(ix-1)$ which is the form considered in the comments to the OP.

Interpretation of the result.

As we have seen, the inverse Mellin transform of $\varphi(s)=i^{-s}$ can be explicitly calculated and has the value already given in the comments above. And as @StevenClark points out in his comment to the OP, the result implies that $i^{-s}$ is not the Mellin transform of any function/distribution supported on the non negative real axis $\Bbb R_{\ge0}=\{x\in \Bbb R : x\ge 0\}$, i.e. $$ 0=\int\limits_{0}^{+\infty} \delta(x+i) x^{s-1} \Dd x \neq \mathscr{M} [f(x)](s) \quad\text{ for any }f\in\mathscr{D}^\prime(\Bbb R_{\ge0}) $$ However note that $\delta(x+i)\in \mathscr{D}^\prime(\Bbb C)$ and this is the key to understand the problem: the standard Mellin transform maps a function/distribution supported on the positive semi-axis to a function/distribution supported in a complex semi-plane. If the support has an empty intersection with $\Bbb R_{\ge0}$, then the transformed function is always $0$.
Nevertheless a similar problem has been solved for the Laplace transform and gave rise to the so called (generalized Borel-)Laplace transform (see [2]): the solution consists precisely in defining the transform on a rotated semi-axis in $\Bbb C$ in order to have a non empty intersection between the support of the transform and the support of the function/distribution to be transformed: in the case at hand, if we define the generalied Mellin transform as $$ \color{blue}{ {^{\pi/2}\!\!\mathscr{M}} [f(x)](s)= \!\!\! \!\!\! \int\limits_{0}^{\quad\infty e^{i\frac{\pi}{2}}}\!\!\! f(x) x^{s-1} \Dd x } $$ where $[0,\infty e^{i\frac{\pi}{2}}[$ is the upper imaginary axis in $\Bbb C$ i.e. $\{z\in \Bbb C: \Re{z}=0 \wedge \Im{z}=\pi/2\}$, we have that $$ \color{green}{ \begin{split} {^{\pi/2}\!\!\mathscr{M}} [\delta(x+i)](s) &= \!\!\! \!\!\! \int\limits_{0}^{\quad \infty e^{i\frac{\pi}{2}}} \!\!\! \delta(x+i) x^{s-1} \Dd x \\ &\text{and thus by putting $\xi=ix$ we get)} \\ & = i^{-1}\int\limits_{0}^{+\infty } \delta(-i\xi+i)(-i\xi)^{s-1} \Dd \xi \\ & = i^{-1} (-i)^{s-1} =i^{-1} \left(\frac{1}{i}\right)^{s-1} = i^{-1} (i)^{1-s} = i^{-s} \end{split}} $$ thus we have reestablished a duality between the (generalized) Mellin transform and its (standard) inverse transform.

In sum it is correct to say that $i^{-s}$ is the Mellin image of the distribution $\delta(x+i)$ provided the generalized Mellin transform is considered instead of the standard one.

References

[1] Alexander D. Poularikas (ed.), The transforms and applications handbook, (English) The Electrical Engineering Handbook Series. Aylesfort: CRC Press. New York, NY: IEEE Press, pp. vii+1103 (1996), ISBN:0-8493-8342-0, MR1407750, Zbl 0851.44001.

[2] Boris Yu. Sternin, Victor Shatalov, Borel-Laplace transform and asymptotic theory. Introduction to resurgent analysis, Boca Raton, FL: CRC Press. pp. 270 (1996), Zbl 0852.34001.