Eq. 2.15.10 in Titchmarsh's book The Theory of the Riemann Zeta-Function comes with no proof:
Let f(x) and $F(s)$ be ralated by $$F(s) = \int_0^{\infty} f(x) x^{s-1} dx, \ \ \ f(x) = \dfrac{1}{2 \pi i} \int_{\sigma- i \infty}^{\sigma+ i \infty} F(s) x^{-s} ds.$$ Then we have, subject to appropriate conditions, $$\dfrac{1}{2 \pi i} \int_{c- i \infty}^{c+ i \infty} F(s) G(w-s) ds = \int_0^{\infty} f(x) g(x) x^{w-1} dx.$$ First thing to do is to compute the integral of $f(x)g(x)x^{w-1}$ and see what happens but I failed to reach a result, especially that I very much am hoping for a rigorous proof. Then searching on web and some books that I had lead me nowhere.
The two questions I have: How the equality is derived and what are the conditions are considered "appropriate"? (perhaps the author is thinking of determining the $c$ based on $\sigma_f$ and $\sigma_g$(?))
Note for Bounty: The post below from "Daigaku no Baku" answers the question partially – to get a complete answer I need to justify the interchange of integrals used in "Daigaku no Baku" 's answer and validity of the real interval for $\max (\alpha_1, 1- \beta_2 ) < c < \max ( \beta_1, 1- \alpha_2)$ mentioned in en.wikipedia.org/wiki/Mellin_transform in the section concerning Parseval's theorem (which is my second question in OP).
Best Answer
$\newcommand{\cat}[1]{\mathcal{#1}}$ Substitute the definition of inverse Mellin transform of $f(x)$ into the integral on the right hand side. Change the order of integration and get $F(s)$ out of the integral that does not depend on $s$. $$\int_{0}^{\infty}f(x)g(x)x^{w-1}dx=\frac{1}{2\pi i}\int_{0}^{\infty}\int_{\sigma-i\infty}^{\sigma+i\infty}F(s)x^{-s}g(x)x^{w-1} \ dx \ ds=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}F(s)\int_{0}^{\infty}x^{w-s-1}g(x)dxds$$ $$\int_{0}^{\infty}x^{w-s-1}g(x)dx=\int_{0}^{\infty}x^{-s}x^{w-1}g(x)dx=G(w-s)$$ Here we used the property of Mellin transform $\cat{M}(x^\nu f(x))(s)=\cat{M}(f(x))(s+\nu)$ $$=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i \infty}F(s)G(w-s)ds$$ The conditions for applicability can be found on the Wikipedia page, in the section concerning Parseval's theorem (which is how this result is called).