# 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.

Now we consider the following function $\displaystyle f(x)=\begin{cases}1-|x|, & |x| \leqslant 1,\\ 0, & |x|>1.\end{cases}$

The Fourier transform of $f$ can be estimated as follows $\displaystyle\begin{gathered} \mathcal{F}[f](y) = \int_{ - 1}^1 {\left( {1 - |x|} \right){e^{ - 2\pi ixy}}dx} \hfill \\ \qquad= \int_{ - 1}^0 {\left( {1 + x} \right){e^{ - 2\pi ixy}}dx} + \int_0^1 {\left( {1 - x} \right){e^{ - 2\pi ixy}}dx} \hfill \\ \qquad= \frac{1}{{ - 2\pi iy}}\left[ {\int_{ - 1}^0 {\left( {1 + x} \right)d({e^{ - 2\pi ixy}})} + \int_0^1 {\left( {1 - x} \right)d({e^{ - 2\pi ixy}})} } \right] \hfill \\ \qquad= \frac{1}{{ - 2\pi iy}}\left[ {\left( {1 + x} \right){e^{ - 2\pi ixy}}\bigg|_{ - 1}^0 + \left( {1 - x} \right){e^{ - 2\pi ixy}}\bigg|_0^1 - \int_{ - 1}^0 {{e^{ - 2\pi ixy}}dx} + \int_0^1 {{e^{ - 2\pi ixy}}dx} } \right] \hfill \\ \qquad= \frac{1}{{2\pi iy}}\left[ {\int_{ - 1}^0 {{e^{ - 2\pi ixy}}dx} - \int_0^1 {{e^{ - 2\pi ixy}}dx} } \right] \hfill \\ \qquad= \frac{1}{{2\pi iy}}\left[ {\frac{{1 - {e^{2\pi ixy}}}}{{ - 2\pi iy}} - \frac{{{e^{ - 2\pi ixy}} - 1}}{{ - 2\pi iy}}} \right] \hfill \\ \qquad= \frac{1}{{2{{(\pi y)}^2}}}\left[ {1 - \frac{{{e^{ - 2\pi iy}} + {e^{2\pi iy}}}}{2}} \right] \hfill \\ \qquad= \frac{1}{{2{{(\pi y)}^2}}}\left[ {1 - \cos (2\pi y)} \right] \hfill \\ \qquad= {\left[ {\frac{{\sin (\pi y)}}{{\pi y}}} \right]^2}. \hfill \\ \end{gathered}$

Thus $\displaystyle\int_{ - \infty }^\infty {{{\left[ {\frac{{\sin (\pi y)}}{{\pi y}}} \right]}^2}{e^{ - 2\pi iyz}}dy} = \mathcal{F}\left[ {\mathcal{F}[f]} \right](z) = f( - z)$.

Now by scaling $\displaystyle\begin{gathered} \int_{ - \infty }^\infty {{{\left[ {\frac{{\sin y}}{y}} \right]}^2}{e^{iyz}}dy} = \int_{ - \infty }^\infty {{{\left[ {\frac{{\sin (\pi y)}}{{\pi y}}} \right]}^2}{e^{i(\pi y)z}}d(\pi y)} \hfill \\ \qquad\qquad= \pi \int_{ - \infty }^\infty {{{\left[ {\frac{{\sin (\pi y)}}{{\pi y}}} \right]}^2}{e^{ - 2\pi iy\frac{z}{{ - 2}}}}dy} \hfill \\ \qquad\qquad= \pi f\left( {\frac{z}{2}} \right). \hfill \\ \end{gathered}$

Thus, this is about to say that $\displaystyle\int_{ - \infty }^\infty {{{\left( {\frac{{\sin x}}{x}} \right)}^2}{e^{itx}}dx} = \begin{cases} \pi (1 - \frac{|t|}{2}),&|t| \leqslant 2,\\ 0, &|t| > 2.\end{cases}$

Similarly, the following integral $\displaystyle\int_{ - \infty }^\infty {\frac{{\sin x}}{x}{e^{itx}}dx}$

can be easily computed via the choice of characteristic function $\chi_{[-1,1]}$ with the fact that $\displaystyle\mathcal{F}[{\chi _{[ - 1,1]}}](y) = \frac{{\sin (2\pi y)}}{{\pi y}}$.

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