Applying Fourier Inversion Formula. 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 :)
 A: 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 $\int_{-\infty}^\infty f(x)\,\mathrm{d}x=\mathcal{F}(f)(0)$.
