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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?

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  • $\begingroup$ This is the fourier transform on the LCA group $(\mathbb R^+, \cdot)$ with haar measure $\frac{\mathrm dx}x$. $\endgroup$
    – AlexR
    Apr 27 '15 at 15:19
  • $\begingroup$ Fubini's Theorem seems to be helpful here. $\endgroup$
    – user1337
    Apr 27 '15 at 15:24
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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.

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  • $\begingroup$ 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 $\endgroup$ Apr 27 '15 at 15:47
  • $\begingroup$ In the inside integral, $y$ is treated as constant, so $du/u = d(\log{u}) = d(\log{x}+\log{y}) = dx/x $. $\endgroup$
    – Chappers
    Apr 27 '15 at 15:50
  • $\begingroup$ hi, finally proved it by setting $ y =1/u$ as a change of variable in teh proof of the first identity thanks anyway $\endgroup$ Apr 27 '15 at 19:36

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