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?

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    $\begingroup$ For those who may be interested, there is additional reading at this MSE link. One can use / adapt the method presented there to obtain a direct proof of the claim by replacing the Fourier transform integrals with their Mellin transform counterparts. (The statement above seems to have assumed that zero i.e. the imaginary axis lies in the fundamental strip.) $\endgroup$ Mar 7, 2014 at 1:51

1 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.

  • $\begingroup$ Thanks. What you refer to as a variant of the Fourier transform is how I typically define the Fourier transform since that is how it is presented in a PDE textbook that I use often. Is that atypical? $\endgroup$ Mar 7, 2014 at 1:44
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    $\begingroup$ No, it's one of the common definitions, and one of the two I like best. The other is $\int f(t)e^{-2\pi i t\omega}\,dt$, which also is an $L^2$-isometry. The latter plays nicer with convolutions, the former with differentiation. Then there's also the engineering variant, $\int f(t) e^{-it\omega}\,dt$ that puts the $2\pi$ entirely in the inverse transform (and isn't an isometry), and the probability variant $\int f(t) e^{it\omega}\,dt$ which is the engineering variant with the sign in the exponent flipped and called "characteristic function" in probability. There may be yet more variants. $\endgroup$ Mar 7, 2014 at 1:50
  • $\begingroup$ Can the formula be generalized to $$\int_{0}^{\infty} \frac{f(x) \overline{g(x)}}{x} \ dx = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(it) \overline{G(it)} \ dt ?$$ $\endgroup$ Mar 7, 2014 at 2:02
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    $\begingroup$ That depends on whether you call $s \mapsto \int x^{-s}f(x)\,\frac{dx}{x}$ the Mellin transform, or $s \mapsto \int x^s f(x)\,\frac{dx}{x}$, same as with the Fourier transform (probability vs. engineering), it doesn't make a difference for the result. $\endgroup$ Mar 7, 2014 at 16:36
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    $\begingroup$ For the integral $\int\lvert F(it)\rvert^2\,dt$, you get just a flip of sign, $\int\lvert F(-it)\rvert^2\,dt$ has the same value of course. With the other convention for the Mellin transform, we'd get $F(it) = \sqrt{2\pi}\mathscr{F}^{-1}[g](t)$ instead of $\mathscr{F}[g]$, but of course Plancherel's theorem can also be formulated $$\int \mathscr{F}^{-1}[f](t)\overline{\mathscr{F}^{-1}[g](t)}\,dt = \int f(u)\overline{g(u)}\,du.$$ $\endgroup$ Mar 7, 2014 at 18:56

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