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?