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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.