I have question about fourier inversion formula.
If we use the fourier inversion formula, f and $\hat f$ satisfy the decay condition or moderate decreasing.
In Exercise 8, Ch4,
Suppose $\hat f$ has compact support contained in [-M,M] and let $f(z)=\sum_{n=0}^{\infty} a_n z^n$. Show that $a_n=\frac{(2 \pi i)^n}{n!}\int_{-M}^{M}\hat f(\xi)\xi^nd\xi$.
I thought this problem can be solved by using the fourier inversion formula. However, I don't know why it can be applied the fourier inversion formula. Any help is appreciated.. !!!
$$f\left( x \right)=\sum\limits_{{}}^{{}}{{{a}_{n}}{{z}^{n}}}=\int\limits_{-\infty }^{\infty }{{{e}^{2\pi ikx}}\hat{f}\left( k \right)dk}$$ Hence $$\oint{\frac{1}{{{z}^{m+1}}}\sum\limits_{{}}^{{}}{{{a}_{n}}{{z}^{n}}}dz}=\oint{\frac{1}{{{z}^{m+1}}}dz}\int\limits_{-\infty }^{\infty }{{{e}^{2\pi ikz}}\hat{f}\left( k \right)dk}$$ Where the contour is a circle surrounding z=0. Hence $$2\pi i{{a}_{m}}=\oint{\frac{{{e}^{2\pi ikz}}}{{{z}^{m+1}}}dz}\int\limits_{-\infty }^{\infty }{\hat{f}\left( k \right)dk}$$ Using the taylor series for the exponential we have $\oint{\frac{{{e}^{2\pi ikz}}}{{{z}^{m+1}}}dz}=2\pi i\frac{{{\left( 2\pi i \right)}^{m}}}{m!}$ , hence $${{a}_{m}}=\frac{{{\left( 2\pi i \right)}^{m}}}{m!}\int\limits_{-\infty }^{\infty }{\hat{f}\left( k \right){{k}^{m}}dk}$$