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?
Best Answer
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.