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I'm trying to express the integral

$$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x_1, x_2) \; g(x_1, x_2) \; \mathrm{d}x_1 \mathrm{d}x_2$$

of two real valued functions $f(x_1,x_2)$ and $g(x_1,x_2)$ in terms of their Fourier transforms $\tilde{f}(\omega_1, \omega_2)$ and $\tilde{g}(\omega_1, \omega_2)$. However, as $f(x_1,x_2)$ and $g(x_1,x_2)$ are real-valued functions and $I$ is not defined as the the product of $f(x_1,x_2)$ and $g(x_1,x_2)^\ast$ I'm not working with the complex conjugate).

$$f(x_1,x_2) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{f}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; \mathrm{d}\omega_1 \mathrm{d}\omega_2$$

and

$$g(x_1,x_2) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{g}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; \mathrm{d}\omega_1 \mathrm{d}\omega_2$$

The integral becomes

$$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \; \left[ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{f}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; \mathrm{d}\omega_1 \mathrm{d}\omega_2 \right] \; \left[ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \tilde{g}(\omega_1, \omega_2) \; e^{2 \pi i ( \;\omega_1 x_1 \;+ \;\omega_2 x_2 \;) } \; \mathrm{d}\omega_1 \mathrm{d}\omega_2 \right] \mathrm{d}x_1 \mathrm{d}x_2$$

This is where it starts to get confusing. I'm not sure if I should write:

$$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \mathrm{d}\omega_1 \; \tilde{f}(\omega_1, \omega_2) \; \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \mathrm{d}\omega_2 \tilde{g}(\omega_1, \omega_2) \; \; \int_{-\infty}^{+\infty} \; e^{2 \pi i x_1 ( \; \omega_1 \;+ \;\omega_2 \;) } \; \mathrm{d}x_1 \; \int_{-\infty}^{+\infty} \; e^{2 \pi i x_2 ( \; \omega_1 \;+ \;\omega_2 \;) } \; \mathrm{d}x_2$$

and extract two one-dimensional dirac deltas, or write:

$$I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \mathrm{d}\omega_1 \; \tilde{f}(\omega_1, \omega_2) \; \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} d\omega_2 \tilde{g}(\omega_1, \omega_2) \; \;\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{2 \pi i x_1 ( \; \omega_1 \;+ \;\omega_2 \;) + x_2 ( \; \omega_1 \;+ \;\omega_2 \;) } \; \mathrm{d}x_1 \; \mathrm{d}x_2$$

with the hope of extracting the two-dimensional dirac delta. (I'm not even sure what the inverse Fourier transform of the two dimensional dirac delta is).

I'd appreciate advise on how to proceed. (Thanks.)

  • 3
    This is Plancherel's theorem (not Parseval's). The one-dimensional proof carries over essentially verbatim to $n$ dimensions. See www.ma.utexas.edu/users/odiaz/notas/fourier.pdf for an account by Oliver Diaz that eschews hand-waving such as the use of the "Dirac function".2010-09-08
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    There is no such thing as the "Dirac function" — the Dirac delta is a distribution (i.e. generalized function, not probability distribution).2010-09-09
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    Thanks Robin. I finally understand the proof. I would have given you credit for the answer if you'd posted a full reply instead of a comment.2010-09-12

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