91Ó°ÊÓ

Given functions are $f\left(x\right)=e^{−\frac{x^2}{2{\mathrm{σ}}^2}}$from Problem 21 and $g\left(\mathrm{Î\pm }\right)=\sqrt{\frac{h}{2\mathrm{Ï€}}}\mathrm{Ï•}\left(p\right)$, and substitution $\mathrm{Î\pm }=\frac{2\mathrm{Ï€}p}{h}$.

Here $p$is the new variable, h is a constant.

The Fourier transform of a function is a complex-valued function that symbolises the original function's complex sinusoids. A Fourier transform is a mathematical operation that breaks down geographically or temporally dependent functions into functions with those same dependencies.

Calculate definite integral of square of function$f\left(x\right)=e^{−\frac{x^2}{2{\mathrm{σ}}^2}}$

Find the integral,

$\begin{array}{l}I={∫}_{−\mathrm{∞}}^{\mathrm{∞}}{\left|f\left(x\right)\right|}^{2}dx\\ I={∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−\frac{x^2}{{\mathrm{σ}}^2}}dx\end{array}$

Find the square of the integral,

$\begin{array}{l}{I}^2=\left({∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−\frac{x^2}{{\mathrm{σ}}^2}}dx\right)\left({∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−\frac{y^2}{{\mathrm{σ}}^2}}dy\right)\\ I^2={∫}_{−\mathrm{∞}}^{\mathrm{∞}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−\frac{(x^2+y^2)}{{\mathrm{σ}}^2}}dxdy\\ I^2={∫}_0^{\mathrm{∞}}{∫}_0^{2\mathrm{π}}{e}^{−\frac{r^2}{{\mathrm{σ}}^2}}rdrd\mathrm{θ}\\ I^2=2\mathrm{π}{∫}_0^{\mathrm{∞}}{e}^{−\frac{r^2}{{\mathrm{σ}}^2}}rdr\end{array}$

Solve further,

$\begin{array}{c}{I}^2=2\mathrm{π}{∫}_0^{\mathrm{∞}}{e}^{−\frac{r^2}{{\mathrm{σ}}^2}}d(−\frac{r^2}{{\mathrm{σ}}^2})d(−\frac{{\mathrm{σ}}^2}{2})\\ I^2=−\mathrm{π}{\left[{\mathrm{σ}}^2{e}^{−\frac{r^2}{{\mathrm{σ}}^2}}\right]}_0^{\mathrm{∞}}\\ I^2=\mathrm{π}{\mathrm{σ}}^2\\ I=\sqrt{\mathrm{π}}\mathrm{σ}\end{array}$

Find the normalization factor .

$\begin{array}{l}N=\frac{1}{\sqrt{I}}\\ N=\frac{1}{\sqrt{\mathrm{σ}\sqrt{\mathrm{π}}}}\end{array}$

Define the new function $Nf\left(x\right)=\mathrm{ψ}\left(x\right)$.

$\mathrm{ψ}\left(x\right)=\frac{1}{\sqrt{\mathrm{σ}\sqrt{\mathrm{π}}}}{e}^{−\frac{x^2}{2{\mathrm{σ}}^2}}$ …… (1)

Find the function $\mathrm{Ï•}\left(p\right)$from Problem 35, and use equation (1) to evaluate it.

$\begin{array}{l}\mathrm{ϕ}\left(p\right)=\frac{1}{\sqrt{h}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}\mathrm{ψ}\left(x\right)e^{−\frac{2\mathrm{π}ipx}{h}}dx\\ \mathrm{ϕ}\left(p\right)=\frac{1}{\sqrt{h\mathrm{σ}\sqrt{\mathrm{π}}}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−\frac{x^2}{2{\mathrm{σ}}^2}−\frac{2\mathrm{π}ipx}{h}}dx\end{array}$

Here, let,

$\begin{array}{c}u=\frac{x}{\mathrm{σ}\sqrt{2}}\\ du=\frac{dx}{\mathrm{σ}\sqrt{2}}\\ \mathrm{β}=\frac{2\mathrm{π}p}{h}\end{array}$

Solve further,

$\begin{array}{l}\mathrm{ϕ}\left(p\right)=\sqrt{\frac{2\mathrm{σ}}{\sqrt{\mathrm{π}}h}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−u−i\mathrm{β}\mathrm{σ}\sqrt{2}u}du\\ \mathrm{ϕ}\left(p\right)=\sqrt{\frac{2\mathrm{σ}}{\sqrt{\mathrm{π}}h}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−{(u+i\frac{\mathrm{β}\mathrm{σ}}{\sqrt{2}})}^2−\frac{{\mathrm{β}}^2{\mathrm{σ}}^2}{2}}du\\ \mathrm{ϕ}\left(p\right)=\sqrt{\frac{2\mathrm{σ}}{\sqrt{\mathrm{π}}h}}{e}^{−\frac{{\mathrm{β}}^2{\mathrm{σ}}^2}{2}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−{(u+i\frac{\mathrm{β}\mathrm{σ}}{\sqrt{2}})}^2}du\end{array}$

Here, let,

$\begin{array}{c}z=u+i\frac{\mathrm{β}\mathrm{σ}}{\sqrt{2}}\\ dz=du\end{array}$

Solve further,

$\begin{array}{l}\mathrm{ϕ}\left(p\right)=\sqrt{\frac{2\mathrm{σ}}{\sqrt{\mathrm{π}}h}}{e}^{−\frac{{\mathrm{β}}^2{\mathrm{σ}}^2}{2}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−z^2}dz\\ \mathrm{ϕ}\left(p\right)=\sqrt{\frac{2\mathrm{σ}}{\sqrt{\mathrm{π}}h}}{e}^{−\frac{{\mathrm{β}}^2{\mathrm{σ}}^2}{2}}\left(\sqrt{\mathrm{π}}\right)\\ \mathrm{ϕ}\left(p\right)=\sqrt{\frac{2\mathrm{σ}\sqrt{\mathrm{π}}}{h}}{e}^{−\frac{{\mathrm{β}}^2{\mathrm{σ}}^2}{2}}\\ \mathrm{ϕ}\left(p\right)=\sqrt{\frac{2\mathrm{σ}\sqrt{\mathrm{π}}}{h}}{e}^{−\frac{2{\mathrm{π}}^2{p}^2{\mathrm{σ}}^2}{h^2}}\end{array}$

Check whether the function $\mathrm{Ï•}\left(p\right)$is normalized or not.

${∫}_{−\mathrm{∞}}^{\mathrm{∞}}{\left|\mathrm{ϕ}\left(p\right)\right|}^{2}dp=\frac{2\mathrm{σ}\sqrt{\mathrm{π}}}{h}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−\frac{4{\mathrm{π}}^2{p}^2{\mathrm{σ}}^2}{h^2}}dp$

Here, let,

$\begin{array}{c}t=\frac{2\mathrm{π}p\mathrm{σ}}{h}\\ dt=\frac{2\mathrm{π}\mathrm{σ}}{h}dp\end{array}$

Solve further,

$\begin{array}{c}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}{\left|\mathrm{ϕ}\left(p\right)\right|}^{2}dp=\frac{1}{\sqrt{\mathrm{π}}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}{e}^{−t^2}dt\\ =\frac{1}{\sqrt{\mathrm{π}}}\left(\sqrt{\mathrm{π}}\right)\\ =1\end{array}$

So, from (1) the normalized function is,

$\mathrm{ψ}\left(x\right)=\frac{1}{\sqrt{\mathrm{σ}\sqrt{\mathrm{π}}}}{e}^{−\frac{x^2}{2{\mathrm{σ}}^2}}$

Therefore, normalization of $f\left(x\right)=e^{−\frac{x^2}{2{\mathrm{σ}}^2}}$is $\mathrm{ψ}\left(x\right)=\frac{1}{\sqrt{\mathrm{σ}\sqrt{\mathrm{π}}}}{e}^{−\frac{x^2}{2{\mathrm{σ}}^2}}$, where, $\mathrm{ψ}\left(x\right)=Nf\left(x\right)$and factor of normalization is $N=\frac{1}{\sqrt{\mathrm{σ}\sqrt{\mathrm{π}}}}$, and $\mathrm{ϕ}\left(p\right)=\sqrt{\frac{2\mathrm{σ}\sqrt{\mathrm{π}}}{h}}{e}^{−\frac{2{\mathrm{π}}^2{p}^2{\mathrm{σ}}^2}{h^2}}$for which Parseval’s theorem ${∫}_{−\mathrm{∞}}^{\mathrm{∞}}{\left|\mathrm{ϕ}\left(p\right)\right|}^{2}dp=1$is verified.