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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?

Random Variable
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    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.) – Marko Riedel Mar 07 '14 at 01:51

1 Answers1

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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.

Daniel Fischer
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  • 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? – Random Variable Mar 07 '14 at 01:44
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    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. – Daniel Fischer Mar 07 '14 at 01:50
  • Can the formula be generalized to $$\int_{0}^{\infty} \frac{f(x) \overline{g(x)}}{x} \ dx = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(it) \overline{G(it)} \ dt ?$$ – Random Variable Mar 07 '14 at 02:02
  • Yes. You can either repeat the above argument (slightly modified), or use the polarisation formula to deduce it from the above result. – Daniel Fischer Mar 07 '14 at 02:05
  • I just noticed that you have $$F(it) = \int_0^\infty x^{{\color{red}{-}}it} f(x),\frac{dx}{x}$$

    Should the negative be in the exponent?

     – Random Variable Mar 07 '14 at 15:57
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    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. – Daniel Fischer Mar 07 '14 at 16:36
  • If we use the engineering variant of the Mellin transform, then we would have to use the probability variant of the Fourier transform to get the result? The switching between variants is a tad confusing. I always thought that the characteristic function of a random variable was similar to but not quite a Fourier transform. – Random Variable Mar 07 '14 at 17:36
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    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 $F(it) = \sqrt{2\pi}\mathscr{F}^{-1}g$ instead of $\mathscr{F}[g]$, but of course Plancherel's theorem can also be formulated $$\int \mathscr{F}^{-1}f\overline{\mathscr{F}^{-1}g},dt = \int f(u)\overline{g(u)},du.$$ – Daniel Fischer Mar 07 '14 at 18:56
  • For Fourier transforms, is $\int_{-\infty}^{\infty} f(x) \hat{g}(x) \ dx = \int_{-\infty}^{\infty} \hat{f}(x) g(x) \ dx$ a valid identity? It seems like it follows from simply expressing $\hat{g}(x)$ as an integral and then switching the order of the integration (when it's justified). And it's almost like a variant of Parseval's theorem. But I can't find it stated in any textbook. – Random Variable May 02 '14 at 15:47
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    Yes, it's valid (under suitable conditions). For $L^1$ functions, you get it by changing the order of integration, for $L^2$ functions by continuous extension from $L^1\cap L^2$ (for example), if one is a tempered distribution (whose Fourier transform is again given by a locally integrable function) and the other a Schwartz function, it also holds. Off the top of my head, I can't give a situation where the integrals make sense but the equality doesn't hold. – Daniel Fischer May 02 '14 at 16:15