Fourier Transform Examples

Find the Fourier transform of exp $ \left(-a x^{2}\right) $.

By fourier transform formula we have,

$ F(k)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-i k x-a x^{2}} d x $
$ =\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp \left[-a\left(x+\frac{i k}{2 a}\right)^{2}-\frac{k^{2}}{4 a}\right] d x $
$ =\frac{1}{\sqrt{2 \pi}} \exp \left(-k^{2} / 4 a\right) \int_{-\infty}^{\infty} e^{-a y^{2}} d y $
$ =\frac{1}{\sqrt{2 a}} \exp \left(-\frac{k^{2}}{4 a}\right) $

Find the Fourier transform of a below non periodic function

$ f(x)=\left\{\begin{array}{ll}{1,} & {-1\lt x\lt 1} \\ {0,} & {|x|\gt 1}\end{array}\right. $

The above function is not a periodic function.
A non periodic function cannot be represented as fourier series.But can be represented as Fourier integral.

Then,using Fourier integral formula we get,

$ F(k)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} f(x) e^{-i k x} d x $ $ =\frac{1}{2 \pi} \int_{-1}^{1} e^{-i k x} d x $
$ =\frac{1}{2 \pi}\left.\frac{e^{-i k x}}{-i k}\right|_{-1} ^{1} $$ =\frac{1}{\pi k} \frac{e^{-i k}-e^{i k}}{-2 i}$$=\frac{\sin k}{\pi k} $

This is the Fourier transform of above function.

We can find Fourier integral representation of above function using fourier inverse transform.

$ f(x)=\int_{-\infty}^{\infty} \frac{\sin k}{\pi k} e^{i k x} d x $
$ =\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\sin k(\cos k x+i \sin k x}{k} d k $
$ =\frac{2}{\pi} \int_{0}^{\infty} \frac{\sin k \cos k x}{k} d k $

This is the fourier integral representation of our non periodic function.