A Parseval-like theorem for Mellin transforms

Question

A particular case of Parseval's theorem for Fourier transforms says that if $f$ is square integrable on $\mathbb{R}$, then

$$ \int_{-\infty}^{\infty} |f(t)|^{2} \ dt = \int_{-\infty}^{\infty} |\hat{f} (\omega)|^{2} d \ \omega .$$

I recall coming across a similar theorem for Mellin transforms that states under certain conditions,

$$ \int_{0}^{\infty} \frac{|f(x)|^{2}}{x} \ dx = \frac{1}{2 \pi}\int_{-\infty}^{\infty} |F(it)|^{2} \ d t$$

where $F(s)$ is the Mellin transform of $f(t)$.

Using this theorem we can evaluate an integral like $ \displaystyle \int_{-\infty}^{\infty} \Gamma(a+it) \Gamma(a-it) \ dt$ fairly easily.

But I can't find much information about this theorem on the internet.

Is this somehow just a corollary of the other theorem?

Answer 1

If we substitute $x = e^u$ and write $g(u) = f(e^u)$, on the one hand, we have

$$\int_0^\infty \lvert f(x)\rvert^2\, \frac{dx}{x} = \int_{-\infty}^\infty \lvert f(e^u)\rvert^2\,du = \int_{-\infty}^\infty \lvert g(u)\rvert^2\,du.$$

On the other hand,

$$\begin{align} F(it) &= \int_0^\infty x^{-it} f(x)\,\frac{dx}{x}\\ &= \int_{-\infty}^\infty e^{-iut} f(e^u)\,du\\ &= \int_{-\infty}^\infty g(u)e^{-iut}\,du\\ &= \sqrt{2\pi}\cdot \mathscr{F}[g](t), \end{align}$$

for the $\mathscr{F}[h](\omega) = \frac{1}{\sqrt{2\pi}}\int h(u)e^{-iu\omega}\,du$ variant of the Fourier transform. Since that variant is an isometry of $L^2(\mathbb{R})$ (which is Parseval's theorem), together we obtain

$$\frac{1}{2\pi}\int_{-\infty}^\infty \lvert F(it)\rvert^2\,dt = \lVert \mathscr{F}[g]\rVert_{L^2}^2 = \lVert g\rVert_{L^2}^2 = \int_0^\infty \lvert f(x)\rvert^2\,\frac{dx}{x}$$

under the hypothesis that $g$ is square integrable. Hence

Is this somehow just a corollary of the other theorem?

can be answered with yes.