Mellin convolution

Remarks

The substitution $y \mapsto x/y$ shows that $f \star g = g \star f$. Assume that $f$ and $g$ are both Mellin convergent at $s$. Then $f \star g$ is defined at almost every $x > 0$, it is Mellin convergent at $s$, and $\mathcal{M}(f \star g)(s) = \mathcal{M}f(s) \, \mathcal{M}g(s)$. The proof is Fubini's theorem and the substitution $x = yu$, which splits the double integral. Example: $f = g = \mathbf{1}_{(0, 1]}$. The integrand $\mathbf{1}_{(0,1]}(x/y) \mathbf{1}_{(0,1]}(y) / y$ is $1/y$ for $x \le y \le 1$ and $0$ otherwise, which is integrable for each $x > 0$. So $(f \star g)(x) = \int_x^1 dy/y = -\log x$ for $0 < x \le 1$ and $0$ for $x > 1$. As a check, $\int_0^1 x^{s-1} (-\log x) \, dx = 1/s^2 = (1/s)^2$ for $\operatorname{Re} s > 0$. Non-example: $f = g = 1$ on $(0, \infty)$; the integrand $1/y$ is not integrable on $(0, \infty)$, so $(f \star g)(x)$ is defined at no $x$. Some authors use the other normalization $\int_0^\infty f(xy) g(y) \, dy$, whose Mellin transform is $\mathcal{M}f(s) \mathcal{M}g(1 - s)$.