# Equivalence of two ways of expressing Fourier transform

One way of expressing Fourier transform would be $(\mathcal{F}f)(t)=\int_{-\infty}^\infty f(x)\, e^{-itx}\,dx$ and its inversion $f(x)=\frac{1}{2\pi}\int_{-\infty}^\infty (\mathcal{F}f)(t)\, e^{itx}\,dt$. The other way would be $\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)\ e^{- 2\pi i x \xi}\,dx$ and its inversion being $f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi)\ e^{2 \pi i \xi x}\,d\xi$.

So, how does one prove that two ways of expressing Fourier transform are in fact equal?

Also, for square-integrable functions, satisfying $\int_{-\infty}^\infty\left|f(x)\right|^2\,dx<\infty$, Fourier transform is written as $\widehat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\,e^{-inx}\,dx.$ (with inversion being $\sum_{n=-\infty}^{\infty} \widehat{f}(n)\,e^{inx}=f(x)$), and this at first sight sems to be a little different from how Fourier transform is written in other cases. Why does Fourier transform look like this in this case?

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It doesn't make sense to say that they're equal. The word you want is "equivalent." – Qiaochu Yuan Oct 14 '12 at 2:08

Your two definitions from the first paragraph differ only in the scaling of the $t$-parameter. You have $(\mathcal{F}f)(2\pi t) = \hat{f}(t)$. Conversely, the reverse transform differ by only a change of parameter as well. If you set $t=2\pi\xi$ in the first reverse transformation formula, and then replace $(\mathcal{F}f)(2\pi t)$ by $\hat{f}(\xi)$ you get exactly the second. Note that the $\frac{1}{2\pi}$ cancels out, since the change of parameter adds another factor $2\pi$.
Your second paragraph is plain wrong. That formula is for computing the fourier series coefficients of a periodic function on $[-\pi,\pi]$. The inverse of that is $f(x)=\sum_{i=0}^{\infty}\hat{f}(n)e^{inx}$.