# Ngô Quốc Anh

## August 17, 2010

### Evaluate complex integral via the Fourier transform

Filed under: Giải tích 7 (MA4247) — Tags: — Ngô Quốc Anh @ 5:56

As suggested from this topic, we are interested in evaluating the following complex integral

$\displaystyle G(t)=\mathop {\lim }\limits_{A \to \infty } \int\limits_{ - A}^A {{{\left( {\frac{{\sin x}} {x}} \right)}^2}{e^{itx}}dx}$.

The trick here is to use the Fourier transform. Thanks to ZY for teaching me this interesting technique.

In $\mathbb R$, the Fourier transform of function $f$, denoted by $\mathcal F[f]$, is defined to be

$\displaystyle \mathcal F[f](y) = \int_{ - \infty }^\infty {f(x){e^{ - 2\pi ixy}}dx}$.

If we apply the Fourier transform twice to a function, we get a spatially reversed version of the function. Precisely,

$\displaystyle\begin{gathered} \mathcal{F}\left[ {\mathcal{F}[f]} \right](z) = \int_{ - \infty }^\infty {\mathcal{F}[f](y){e^{ - 2\pi iyz}}dy} \hfill \\ \qquad\qquad= \int_{ - \infty }^\infty {\mathcal{F}[f](y){e^{2\pi iy( - z)}}dy} \hfill \\ \qquad\qquad= {\mathcal{F}^{ - 1}}\left[ {\mathcal{F}[f]} \right]( - z) \hfill \\ \qquad\qquad= f( - z) \hfill \\ \end{gathered}$

where $\mathcal F^{-1}$ denotes the inverse Fourier transform.