Question

Find the fourier transform of$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{/}\mathbf{\left(}\mathbf{2}{\mathbf{\sigma }}^{\mathbf{2}}\mathbf{\right)}}$. Hint: Complete the square in the xterms in the exponent and make the change of variable$\mathit{y}\mathbf{=}\mathit{x}\mathbf{+}{\mathit{\sigma }}^{\mathbf{2}}\mathit{i}\mathit{\alpha }$ .Use tables or computer to evaluate the definite integral.

Short answer

The fourier transform of $f\left(x\right)=e^{-x^2/\left(2{\sigma }^2\right)}$ is$g\left(\alpha \right)=\frac{\sigma }{\sqrt{2\pi }}{e}^{-{\sigma }^2{\alpha }^2/2}$.

Step-by-step solution

Given Information.

The given function is$f\left(x\right)=e^{-x^2/\left(2{\sigma }^2\right)}$.

Definition of fourier transform

The Fourier transform is a mathematical technique for expressing a function as the summation of sines and cosines functions.

Step 3: To find the fourier transform of the given function

The fourier transform is given below.

$g\left(\alpha \right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{e}^{-x^2/\left(2{\sigma }^2\right)}{e}^{-i\alpha x}dx$

Put $u=\frac{x}{\sigma \sqrt{2}}$

Differentiate with respect to x.

$du=\frac{dx}{\sigma \sqrt{2}}$

$\begin{array}{l}g\left(\alpha \right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{e}^{-u^2-i\alpha \sqrt{2}\sigma u}\sqrt{2}\sigma du\\ g\left(\alpha \right)=\frac{\sigma }{\pi \sqrt{2}}\underset{-\infty }{\overset{\infty }{\int }}{e}^{-u^2-i\alpha \sqrt{2}\sigma u}du\\ g\left(\alpha \right)=\frac{\sigma }{\pi \sqrt{2}}\underset{-\infty }{\overset{\infty }{\int }}{e}^{-{(u+i\alpha \sigma /\left(\sqrt{2}\right))}^2-{\sigma }^2{\alpha }^2/2}du\end{array}$

Simplify further

$g\left(\alpha \right)=\frac{\sigma }{\pi \sqrt{2}}{e}^{-{\sigma }^2{\alpha }^2/2}\underset{-\infty }{\overset{\infty }{\int }}{e}^{-{(u+i\alpha \sigma /\left(\sqrt{2}\right))}^2}d(u+i\sigma \alpha /\sqrt{2})$

Put$u+i\sigma \alpha /\sqrt{2}=\eta$

$g\left(\alpha \right)=\frac{\sigma }{\pi \sqrt{2}}{e}^{-{\sigma }^2{\alpha }^2/2}\underset{-\infty }{\overset{\infty }{\int }}{e}^{-{\eta }^2}d\eta$

But the integral $\underset{-\infty }{\overset{\infty }{\int }}{e}^{-{\eta }^2}d\eta$is Euler-poisson integral and its value is$\sqrt{\pi }$.

$g\left(\alpha \right)=\frac{\sigma }{\sqrt{2\pi }}{e}^{-{\sigma }^2{\alpha }^2/2}$

The fourier transform of $f\left(x\right)=e^{-x^2/\left(2{\sigma }^2\right)}$is$g\left(\alpha \right)=\frac{\sigma }{\sqrt{2\pi }}{e}^{-{\sigma }^2{\alpha }^2/2}$.