This used to be part of a longer question that I posted earlier but since that question seemed to long I decided to split it up.
Given the function
$$f(x) = \begin{cases} \frac{1}{2}, & \text{if $\lvert x\rvert \le
1$} \\ 0, & \text{else} \end{cases}$$I calculated the fourier transform $\hat{f}(x)=\frac{1}{\sqrt{2
\pi}}\int_{-\infty}^{+\infty}f(x)e^{ikx}dx$ to be:$$\hat{f}(k)=\frac{1}{\sqrt{2\pi}k}\sin k$$
I want to show that the inverse transform $\frac{1}{\sqrt{2
\pi}}\int_{-\infty}^{+\infty}\hat{f}(k)e^{-ikx}dx$ yields the function
$f(x)$ I began with.
My attempt:
$$f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}k}\sin k \cdot e^{-ikx}dk \\ \iff f(x)=\frac{1}{2\pi}\int\frac{1}{k}\sin{(k)} \space e^{-ikx}dk$$
We are given a hint: "Re-express $\frac{1}{k}$ as a derivative in front of the integral". I have no clue how to use this though.
Can someone help me solve the inverse transform integral?
Best Answer
Perhaps what the hint means is $$ f(x)=\frac{-1}{2\pi\mathrm{i}}\frac{\partial}{\partial x}\int_{-\infty}^\infty dk\frac{\sin k}{k^2}e^{-\mathrm{i}kx}=\frac{1}{2\pi}\frac{\partial}{\partial x}\int_{-\infty}^\infty dk\frac{\sin k}{k^2}\sin(kx)\ , $$ where I use the fact that $e^{-\mathrm{i}kx}=\cos (kx)-\mathrm{i}\sin (k x)$ and the integrand with $\cos$ is odd (and thus gives a zero integral). The remaining integrand is now even, so $$ f(x)=\frac{2}{2\pi}\frac{\partial}{\partial x}\int_{0}^\infty dk\frac{\sin k}{k^2}\sin(kx)\ . $$ The integral can be evaluated to be $\pi x/2$ for $|x|<1$, and a constant for $|x|>1$. Therefore, by taking the derivative w.r.t. $x$, you get back what you started with.