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The Mellin convolution of two functions, when it exists, is of the form $$ (f \ast_M g)(t) = \int_0^\infty f\left( \frac{t}{\tau} \right) g(\tau) \frac{\mathrm{d}\tau}{\tau} $$ and has the property that $$ \mathscr{M}(f \ast_M g)(s) = \mathscr{M}f \cdot \mathscr{M}g $$ where $\mathscr{M}$ is the Mellin transform and $\cdot$ is pointwise multiplication.

Sometimes, however, it can be useful to evaluate convolution integrals which are identical to the one given above, but integrating from $-\infty$ to $\infty$ rather than from 0 to $\infty$. In other words $$ (f \ast_\widehat{M} g)(t) = \int_{-\infty}^\infty f\left( \frac{t}{\tau} \right) g(\tau) \frac{\mathrm{d}\tau}{\tau} $$ Does there exist a “modified Mellin transform” which has the similar property that $$ \widehat{\,\mathscr{M}}(f \ast_\widehat{M} g)(s) = \widehat{\,\mathscr{M}}f \cdot \widehat{\,\mathscr{M}}g $$ with this “bilateral” Mellin convolution?

To note, I am aware that many authors have defined a bilateral Mellin transform according to different conventions (often as a pair of unilateral Mellin transforms). So I am not asking for an “ad hoc” definition of a bilateral Mellin transform. Rather, what I am asking is, is it possible to somehow work “backward” from the above bilateral convolution definition to obtain a modified Mellin transform, which has the property that the pointwise product in this modified Mellin domain yields the above “bilateral multiplicative Mellin convolution” in the time domain?

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