Use the fourier inversion formula to evaluate $\int_{-\infty}^{\infty} \frac{\sin(x)}{x}$.
I know that $\mathcal{F}(\frac{\sin(x)}{x})=c* \mathbb{1}_{[-1,1]}$.
Now the fourier inversion formula states that:
$\mathcal{F}^{-1}(\mathcal{F}(f))=f$ which implies that:
$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f}(\xi)\exp(-ix.\xi)d\xi$.
$\therefore$ I get the following: $f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathbb{1}_{[-1,1]} e^{-ix.\xi}d\xi$?
Have I approached the problem correctly?
Really appreciate the help! Thanks 🙂
Best Answer
There are several normalizations for the Fourier Transform. I find the easiest to remember $$ \mathcal{F}(f)(\xi)=\int_{-\infty}^\infty f(x)\,e^{-2\pi ix\xi}\,\mathrm{d}x\tag1 $$ This normalization gives $$ \mathcal{F}(\mathcal{F}(f))(x)=f(-x)\tag2 $$ Using $(1)$, it is easy to see that $$ \mathcal{F}\!\left[\delta\!\left(x+\tfrac12\right)-\delta\!\left(x-\tfrac12\right)\right]\!(\xi)=e^{\pi i\xi}-e^{-\pi i\xi}\tag3 $$ Integrating $(1)$ by parts yields $$ \mathcal{F}\!\left(f'\right)\!(\xi)=2\pi i\xi\,\mathcal{F}(f)(\xi)\tag4 $$ Therefore, $$\newcommand{\sinc}{\operatorname{sinc}} \begin{align} \mathcal{F}\!\left[-\tfrac12\le x\le\tfrac12\right]\!(\xi) &=\frac{e^{\pi i\xi}-e^{-\pi i\xi}}{2\pi i\xi}\\[3pt] &=\sinc(\pi\xi)\tag5 \end{align} $$ where $[\cdots]$ are Iverson brackets.
Since $\sinc$ is even, Fourier Inversion says $$ \mathcal{F}(\sinc(\pi\xi))(x)=\left[-\tfrac12\le x\le\tfrac12\right]\tag6 $$ The scaling properties of the Fourier transform say $$ \mathcal{F}(\sinc(\xi))(x)=\pi\left[-\tfrac1{2\pi}\le x\le\tfrac1{2\pi}\right]\tag7 $$ Now note that $\int_{-\infty}^\infty f(x)\,\mathrm{d}x=\mathcal{F}(f)(0)$.