# Mellin convolution and Mellin transform

How can I prove that the Mellin transform of the function defined by

$$\int_{0}^{\infty}K(xy)f(y)dy$$

is equal to the product $K(s)F(1-s)$

and that the Mellin transform of $$\int_{0}^{\infty}K(x/y)f(y)dy/y$$

is just the product of $K(s)F(s)$

where $K(s)=\int_{0}^{\infty}t^{s-1}k(t)$ and $F(s)=\int_{0}^{\infty}t^{s-1}f(t)$

I know this can be proven from the Fourier convolution theorem but what change of variable should I make?

• This is the fourier transform on the LCA group $(\mathbb R^+, \cdot)$ with haar measure $\frac{\mathrm dx}x$. Apr 27 '15 at 15:19
• Fubini's Theorem seems to be helpful here. Apr 27 '15 at 15:24

Your first function's Mellin transform is $$\int_0^{\infty} x^{s-1} \left( \int_0^{\infty} K(xy) f(y) \, dy \right) \, dx$$ Interchange the order of integration by Fubini's theorem to obtain $$\int_0^{\infty} f(y) \left( \int_0^{\infty} x^{s-1} K(xy) \, dx \right) \, dy$$ Now change variables in the inside integral, to $u=xy$, $du/u = dx/x$, which gives $$\int_0^{\infty} f(y) \left( \int_0^{\infty} y^{-s} u^{s-1} K(u) \, du \right) \, dy = \left( \int_0^{\infty} y^{(1-s)-1} f(y) \, dy \right) \left( \int_0^{\infty} u^{s-1} K(u) \, du \right) = F(1-s)K(s),$$ as required. Your second one is done in exactly the same way.
• thanks but i don'te get it all :) $u=xy$ but how i do compute the differential $du$ i forgot how is it done sorry Apr 27 '15 at 15:47
• In the inside integral, $y$ is treated as constant, so $du/u = d(\log{u}) = d(\log{x}+\log{y}) = dx/x$. Apr 27 '15 at 15:50
• hi, finally proved it by setting $y =1/u$ as a change of variable in teh proof of the first identity thanks anyway Apr 27 '15 at 19:36