Question

Let f(x)and$\mathit{g}\left(\alpha \right)$be a pair of Fourier transforms. Show that$\frac{\mathbf{\text{df}}}{\mathbf{\text{dx}}}$and$\mathit{i}\mathit{\alpha }\mathit{g}\left(\alpha \right)$are a pair of Fourier transforms. Hint: Differentiate the first integral in (12.2)under the integral sign with respect to x. Use to show that

${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathit{\alpha }{\left|g\left(\alpha \right)\right|}^{\mathbf{2}}\mathit{d}\mathit{\alpha }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }\mathbf{i}}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\stackrel{\mathbf{-}}{\mathbf{f}}\left(x\right)\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathit{f}\left(x\right)\mathit{d}\mathit{x}$.

Short answer

The Derivative of with respect to is $\frac{\text{df}}{\text{dx}}={\int }_{-\infty }^{\infty }g\left(\alpha \right)i\alpha e^{i\alpha x}d\alpha$${\int }_{-\infty }^{\infty }\alpha {\left|g\left(\alpha \right)\right|}^{2}d\alpha =\frac{1}{2\mathrm{\pi i}}{\int }_{-\infty }^{\infty }\stackrel{-}{f}\left(x\right)\frac{d}{dx}f\left(x\right)dx$ is verified.

Step-by-step solution

Given information

We have given f(x) and $g\left(\alpha \right)$be a pair of Fourier transforms.

Definition of Inverse Fourier transformation

For a given function u(x) define its Inverse Fourier transformation U as

$\mathbf{\text{U}}\left(\text{k}\right){\mathbf{\text{=f}}}^{\mathbf{-}\mathbf{1}}\left\{\text{u}\left(\text{x}\right)\right\}\mathbf{\text{=}}\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{\pi }}}{\mathbf{\text{∫}}}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}{\mathbf{\text{e}}}^{\mathbf{i}\mathbf{k}\mathbf{x}}\mathbf{\text{u}}\left(x\right)\mathbf{\text{dx}}$

Evaluate dfdx

By the definition of the inverse Fourier transformation

$\text{f}\left(\text{x}\right){\text{=∫}}_{-\infty }^{\infty }\text{g}\left(\alpha \right){\text{e}}^{ikx}\text{dα}$

Differentiate it with respect to x

$\frac{df}{dx}={\int }_{-\infty }^{\infty }g\left(\alpha \right)i\alpha e^{i\alpha x}d\alpha$

Show that ∫-∞∞α|gα|2dα=12πi∫-∞∞f-(x)ddxf(x)dx

To show this, let’s start with equation 12.23

${\int }_{-\infty }^{\infty }\stackrel{-}{g_1}\left(\alpha \right)\times g_2\left(\alpha \right)d\alpha =\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\stackrel{-}{f_1}\left(x\right)\times f_2\left(x\right)dx$

Let $\stackrel{-}{g_1}\left(\alpha \right)=g,g_2\left(\alpha \right)=i\alpha g,\stackrel{-}{f_1}\left(x\right)=\stackrel{-}{f}$and$f_2\left(x\right)=\frac{df}{dx}$

Then result follows

${\int }_{-\infty }^{\infty }\alpha {\left|g\left(\alpha \right)\right|}^{2}d\alpha =\frac{1}{2\pi i}{\int }_{-\infty }^{\infty }\stackrel{-}{f}\left(x\right)\frac{d}{dx}f\left(x\right)dx$