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.


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