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Let $f(t)$ be a generic integrable function (where $t$ is time). Let $F(\omega)$ be its Fourier transform.

Suppose every continuity and integrability condition is met. It is legit to say $\int_{-\infty}^{\infty} f(t) dt = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} F(\omega) e^{2i\pi\omega t} dt d\omega$ ?

Thanks in advance!

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Yes, under suitable conditions: $$ F(0)=\int_{-\infty}^{\infty}f(t)e^{-2\pi iw0}dt \\ F(0)=\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}F(w)e^{2\pi i w t}dw\right)e^{-i2\pi i t 0} dt $$

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  • $\begingroup$ @LoScrondo : You have to correct $x$ in $F(x)$ to $F(w)$. Otherwise the inner integral doesn't exist. $\endgroup$ – DisintegratingByParts Oct 7 '15 at 17:15
  • $\begingroup$ Another typing error. I really apologize! That's the correct formula: $\int_{-\infty}^{\infty} f(t) dt = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} F(\omega) e^{2i\pi\omega t} d\omega dt$ $\endgroup$ – Lo Scrondo Oct 7 '15 at 17:20

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