Talk:Mellin transform

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[edit] Relationship with other transforms

The definition of the two-sided Laplace transform is

\left\{\mathcal{B} f\right\}(s) = \varphi(s) = \int_{-\infty}^{\infty} e^{-sx} f(x) dx

Under the substitution x=-\ln\, t, t \in [0,\infty] this transforms into

\left\{\mathcal{B} f\right\}(s) = \varphi(s) = \int_{0}^{\infty} t^{s} f(-\ln\,t) {dt \over t} \ .

So the relationship between Mellin transform and two-sided Laplace transform should be

\left\{\mathcal{B} f\right\}(s)=\left\{\mathcal{M} f(-\ln\,x)\right\}(s)

rather than the stated

\left\{\mathcal{B} f\right\}(s)=\left\{\mathcal{M} f(e^{-x})\right\}(s) \ ?

--212.18.24.11 14:58, 22 August 2005 (UTC)

Hello my question is ..let's suppose we know the Mellin inverse transform of F(s) then..how could we calculate the Mellin inverse transform of F(as) when a is a real number, i believe that if g(x) has F(s) as its Mellin transform then g(x1 / a)x1 / a − 1 is the inverse of F(as) but i'm not pretty sure --Karl-H 14:20, 1 October 2006 (UTC)

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