Question

Determine the Fourier transform $\mathit{A}\mathbf{\left(}\mathit{k}\mathbf{\right)}$of the oscillatory function$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ ) and interpret the result. (The identity $\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}{\mathbf{k}}_{\mathbf{0}}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{e}}^{\mathbf{+}\mathbf{i}{\mathbf{k}}_{\mathbf{0}}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{i}{\mathbf{k}}_{\mathbf{0}}}\mathbf{)}$may be useful.)

Short answer

The total function is symmetric around the$y$ -axis, nevertheless, its oscillatory, however, it decays as$x\to \pm \infty$. The function is also peaked around $k=k_0,k=-k_0$ and the width of this peak is inversely proportional to$L$ or the box length (precisely the width is $\frac{4\pi }{L}$).

$A\left(k\right)=\frac{\text{L}}{\mathbf{4}\pi }\frac{\sin \left(({\mathbf{k}}_0+\mathbf{k})\frac{\text{L}}{2}\right)}{({\mathbf{k}}_0+\mathbf{k})\frac{\text{L}}{2}}+\frac{\text{L}}{\mathbf{4}\pi }\frac{\sin \left(({\mathbf{k}}_0-\mathbf{k})\frac{\text{L}}{2}\right)}{({\mathbf{k}}_0-\mathbf{k})\frac{\text{L}}{2}}$

Step-by-step solution

The Fourier transform

The generalization of the Fourier series is known as Fourier transform and it can also refer to both the frequency domain representation and the mathematical function used. The Fourier transform facilitates the application of the Fourier series to non-periodic functions, allowing every function to be viewed as a sum of simple sinusoids.

The equation of the Fourier transform as,

$\mathit{A}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}{\int }_{-\infty }^{+\infty }\psi \mathbf{\left(}\mathit{x}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{k}\mathbf{x}}\mathit{d}\mathit{x}$

Trigonometry relation, $\frac{{\mathbf{e}}^{\mathbf{i}\mathbf{x}}\mathbf{-}\mathbf{e}\mathbf{x}}{\mathbf{2}}\mathbf{=}\mathit{i}\mathit{s}\mathit{i}\mathit{n}\mathit{x}$

Substitute the given function using equation of Fourier transform

Substitute the given function $f\left(x\right)$in the equation for the Fourier transform with proper limits from$-\frac{L}{2}$ to $+\frac{L}{2}$

$\begin{array}{l}\because A\left(k\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\psi \left(x\right)e^{-ikx}dx\\ A\left(k\right)=\frac{1}{2\pi }{\int }_{-\frac{L}{2}}^{\frac{L}{2}}\cos \left(k_{0}x\right)e^{-ikx}dx\\ A\left(k\right)=\frac{1}{2\pi }{\int }_{-\frac{L}{2}}^{\frac{L}{2}}\left(\frac{e^{i{k}_{0}x}+e^{-i{k}_{0}x}}{2}\right)e^{ikx}dx\\ A\left(k\right)=\frac{1}{4\pi }{\int }_{-\frac{L}{2}}^{\frac{L}{2}}(e^{i(k_0+k)x}+e^{-i(k_0-k)x})dx\\ A\left(k\right)=\frac{1}{4\pi }\frac{e^{i(k_0+k)\frac{L}{2}}-e^{-i(k_0+k)\frac{L}{2}}}{i(k_0+k)}+\frac{1}{4\pi }\frac{e^{-i(k_0-k)\frac{L}{2}}-e^{i(k_0-k)\frac{L}{2}}}{-i(k_0-k)}\\ A\left(k\right)=\frac{1}{2\pi }\frac{\sin \left((k_0+k)\frac{L}{2}\right)}{(k_0+k)}+\frac{1}{2\pi }\frac{\sin \left((k_0-k)\frac{L}{2}\right)}{(k_0-k)}\\ A\left(k\right)=\frac{L}{4\pi }\frac{\sin \left(({\mathbf{k}}_0+\mathbf{k})\frac{L}{2}\right)}{({\mathbf{k}}_0+\mathbf{k})\frac{L}{2}}+\frac{L}{4\pi }\frac{\sin \left(({\mathbf{k}}_0-\mathbf{k})\frac{L}{2}\right)}{({\mathbf{k}}_0-\mathbf{k})\frac{L}{2}}\end{array}$

Where, we used the sine identity $\sin \left(x\right)=\frac{e^{ix}-e^{-ix}}{2}$in the transition from above equations with $x$being$\left[(k_0-k)\frac{L}{2}\right]$.

However, the two sine functions $\left(\frac{\sin \left(x\right)}{x}\right)$ that we ended up with in above equation are even functions.

Hence, the total function is symmetric around the $y$-axis, nevertheless, its oscillatory, however, it decays as $x\to \pm \infty$. The function is also peaked around $k=k_0,k=-k_0$ and the width of this peak is inversely proportional to or the box length (precisely the width is $\frac{4\pi }{L}$).

The function is shown below, it has been normalized, with $f\left(k\right)$given by the general expression$A\left(k\right)=\frac{\text{L}}{\mathbf{4}\pi }\frac{\sin \left(({\mathbf{k}}_0+\mathbf{k})\frac{\text{L}}{2}\right)}{({\mathbf{k}}_0+\mathbf{k})\frac{\text{L}}{2}}+\frac{\text{L}}{\mathbf{4}\pi }\frac{\sin \left(({\mathbf{k}}_0-\mathbf{k})\frac{\text{L}}{2}\right)}{({\mathbf{k}}_0-\mathbf{k})\frac{\text{L}}{2}}$