Toronto Math Forum
APM346-2015S => APM346--Home Assignments => HA6 => Topic started by: Victor Ivrii on March 05, 2015, 08:37:57 AM
-
Find Fourier transforms of
a. $f(x)=\left\{\begin{aligned} & 1&& |x|\le a,\\
& 0 && |x|\ge a;\end{aligned}\right.$
b. $f(x)=\left\{\begin{aligned} & x && |x|\le a,\\
& 0 && |x|\ge a;\end{aligned}\right.$
c. Using (a) calculate $\int_{-\infty}^\infty \frac{\sin (x)}{x}\,dx$.
-
\begin{gathered}
part(a): \hfill \\
\hat f(\omega ) = \frac{1}{{2\pi }}\int\limits_{ - \infty }^\infty {f(x){e^{ - i\omega x}}dx} \hfill \\
= \frac{1}{{2\pi }}\int\limits_{ - a}^a {{e^{ - i\omega x}}dx} \hfill \\
= \frac{1}{{2\pi }}(\frac{{{e^{i\omega a}} - {e^{ - i\omega a}}}}{{i\omega }}) \hfill \\
= \frac{{\sin (\omega a)}}{{\pi \omega }} \hfill \\
\hfill \\
part(b): \hfill \\
\hfill \\
take f(x) = xg(x),where, g(x) = 1 \hfill \\
hence, \hat f(\omega ) = i\frac{{d\hat g(\omega )}}{{d\omega }} = i\frac{{d(\frac{{\sin (\omega a)}}{{\pi \omega }})}}{{d\omega }} = i(\frac{{a\cos (a\omega )}}{{\pi \omega }} - \frac{{\sin (a\omega )}}{{\pi {\omega ^2}}}) \hfill \\
\hfill \\
part(c): \hfill \\
\hfill \\
IFT: \hfill \\
f(x) = \int\limits_{ - \infty }^\infty { \hat f(\omega ){e^{i\omega x}}d\omega = } \int\limits_{ - \infty }^\infty {\frac{{\sin (\omega a)}}{{\pi \omega }}} {e^{i\omega x}}d\omega \hfill \\
thus,\int\limits_{ - \infty }^\infty {\frac{{\sin (xa)}}{x}{e^{i\omega x}}} dx = \pi , |\omega | \leqslant a,if a = 1,and \omega = 0. \hfill \\
\end{gathered} \]
-
In many papers some answers like $=0$ if this and $=\pi$ if that for (c ) was given. It is a very simple question: calculate $I:=\int_{-\infty}^\infty \frac{\sin(x)}{x}\,dx$. We got that
$$
\frac{1}{\pi}\int _{-\infty}^\infty \frac{\sin(ak)}{k }\underbracket{e^{ikx}}\,dk= \left\{\begin{aligned}&1 &&|x|<a\\ &0 &&|x|\ge a\end{aligned}\right.$$
We do not need underlined factor, so we take $x=0$ and arrive to
$$
\int_{-\infty}^\infty \frac{\sin(a k)}{k}\,dk= \pi.
$$
So taking $a=1$ we have $I=\pi$.
Remark. Above integral does not depend on $a\ne 0$, indeed one can see it changing variables $k_{new}=a k$.