Explicit computation of Mellin transformation and its inverse
Solution 1:
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}