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Using the formula of Mellin inverse transformation, I want to obtain $f(t)$ directly from Mellin and its inverse transformation.

i.e., \begin{align} f(t) &= \frac{1}{2\pi i} \int_{a-i\infty}^{a+i \infty} \mathcal{M}[f;s] t^{-s} ds \\ &= \frac{1}{2\pi i} \int_{a-i\infty}^{a+i \infty} \left( \int_0^{\infty} f(\tau) \tau^{s-1} d\tau \right) t^{-s} ds \end{align} After implementing the R.H.S, I have no idea how this becomes $f(t)$.

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    $\begingroup$ Are you familiar with the Fourier inversion formula? It's basically the same thing $\endgroup$
    – D_S
    Dec 29, 2021 at 14:31
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    $\begingroup$ Just perform a change of variables on Fourier inversion formula and eureka $\endgroup$
    – TravorLZH
    Dec 29, 2021 at 14:38
  • $\begingroup$ @TravorLZH, WOW it is really surprising!; without changing variables into the form of Fourier inversion formula, it seems it is not possible to deal with obtaining $f(t)$, but after changing variable it really is! $\endgroup$
    – phy_math
    Dec 29, 2021 at 16:14
  • $\begingroup$ It involves a Dirac Delta integral representation. $\endgroup$ Jan 31, 2022 at 19:39

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This can be derived from Fourier inversion theorem i.e., $f= \int_{-\infty}^{\infty} \mathcal{F}[f;\beta] e^{2\pi i \beta x} d\beta$. : Note that \begin{align} & f(x) = \int_{-\infty}^{\infty} \mathcal{F} [f;\beta] e^{2 \pi i \beta x} d \beta \quad \Rightarrow \quad f(e^{-x}) e^{-ax} = \int_{-\infty}^{\infty} F(s) e^{2\pi i\beta x} d \beta \\ &\quad \Rightarrow \quad f(t) = t^{-a} \int_{-\infty}^{\infty} F(s) t^{-2 \pi i \beta} d\beta = \frac{1}{2\pi i} \int_{a-i \infty}^{a+i \infty} F(s) t^{-s} ds \end{align}

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