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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.)

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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

1 Answers 1

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I don't see what you're doing. Are you trying to show that:

$\int_{\mathbb R^2} f(x) g(x) \, dx = \int_{\mathbb R^2} \widehat{f(\xi)} \overline{\widehat{g(\xi)}} \, d\xi$? This is usually called Parseval's identity.

You can prove it as follows (even in $\mathbb R^n$), first prove by Fubini (very simple) that:

$\int_{\mathbb R^2} f(x) \widehat{g(x)} \, dx = \int_{\mathbb R^2} \widehat{g(x)} f(x) \, dx$

Then let $g = \overline{\hat{h}}$ and note then that $\hat{g} = \overline{h}$ (by Fourier inversion), then we get

$\int_{\mathbb R^2} f(x) h(x) \, dx = \int_{\mathbb R^2} \widehat{g(x)} \overline{\widehat{h(x)}} \, dx$ (assuming $h$ is real, otherwise we get the conjugate).

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    Yes, that is a typo hat instead of widehat, it is the same, $x = (x_1, x_2)$, it is just a shorthand.2010-09-09