Mellin transform

{c}
{title2=$\mathcal Mf(\sigma+i\tau)=\widehat h_\sigma(-\tau)$}

The <change of variables> $x=e^u$, using the <exponential function>, turns the <Mellin transform> into
$$
\mathcal Mf(s)=\int_{-\infty}^\infty e^{su}f(e^u)\,du.
$$
The factor $dx=e^u du$ changes $x^{s-1}$ to $e^{su}$; missing this factor would shift every exponent by one. With $h_\sigma(u)=e^{\sigma u}f(e^u)$ and the <Fourier transform> convention $\widehat h(\omega)=\int_{\mathbb R}h(u)e^{-i\omega u}\,du$, the identity is
$$
\mathcal Mf(\sigma+i\tau)=\int_{\mathbb R}h_\sigma(u)e^{i\tau u}\,du=\widehat h_\sigma(-\tau).
$$
Choosing $\sigma$ inside the <Mellin fundamental strip> makes $h_\sigma$ a <Lebesgue integrable function>. The minus sign in the <Fourier transform> argument follows from the stated convention. For the <Mellin transform of an interval indicator> with $[b,c]=[1,4]$ and $\sigma=0$, $h_0=\mathbf1_{[0,\log4]}$: the <Fourier transform of an interval indicator> is literally the same computation in a different coordinate.

Splitting the <real line> at $u=0$ also connects the <Mellin transform> to two unilateral <Laplace transforms>. If $k_+(v)=f(e^v)$ and $k_-(v)=f(e^{-v})$ for $v\geq0$, then
$$
\mathcal Mf(s)=\mathcal Lk_+(-s)+\mathcal Lk_-(s)
$$
where both <integrals> converge absolutely. Their <Laplace convergence regions> impose opposite endpoint conditions, explaining the two sides of the <Mellin fundamental strip>. Finally, $f(ae^u)=f(e^{u+\log a})$ shows that a change of scale is a <translation> in $u$. Weighted <Fourier transforms> of this <translation> give exactly the factor $a^{-s}$ in the <scaling property of the Mellin transform>. For a smooth <function> with <compact support> inside $(0,\infty)$, the <Mellin inversion theorem> follows from <Fourier inversion>; the existing formula states the corresponding hypotheses.


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 Incoming links: Mellin transform in logarithmic coordinates