# Calculating the Inverse Fourier Transform of $\frac{1}{\sqrt{2\pi}k}\sin k$

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?

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.
• First, thank you very much for your answer. I just a have a few follow up questions. So basically you rewrote the integral: $$\frac{1}{2 \pi} \int_{-\infty}^{+\infty} \frac{1}{k} \sin(k) e^{-ikx} dk$$ to $$\frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{1}{k} \sin(k) (\cos{kx}-i\sin{kx}) dk$$ Since cosine is an odd function it disappears. Why does an odd function equal zero when integrated from negative to positive infinity? We get: $$\frac{1}{2 \pi} \int_{-\infty}^{+\infty} -\frac{1}{k} i\sin(k)\sin(kx)dk$$ I don't get how you rewrite the integral after that. – bluemoon Jan 7 '16 at 21:21
• 'Since cosine is an odd function it disappears' is not quite right. $\cos(k x)$ is an even function in $k$, because $\cos(kx)=\cos((-k)x)$. The "integrand with cos" I was referring to is $\sin(k)\cos(kx)/k^2$, which is odd [note that there is a $k^2$ in the denominator now, there is a typo in your comment]. The integral $\int_{-\infty}^\infty g(k)dk$, with $g(k)=-g(-k)$ [odd], is always zero: just split the integral as $\int_{-\infty}^0 +\int_0^\infty$, change variable $k\to -k$ in the first one and use the oddness property. – Pierpaolo Vivo Jan 8 '16 at 6:13
The inversion integral can be evaluated as a Cauchy principal value integral $$\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\frac{1}{\sqrt{2\pi}}\frac{\sin k}{k}e^{ikx}dk$$ The above will converge in $L^2(\mathbb{R})$ to the original function, and will converge pointwise to the mean of the left- and right-hand limits of the original function. Because $\sin(k)/k$ is even, the above may be written as \begin{align} &\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\frac{1}{\sqrt{2\pi}}\frac{\sin k}{k}\cos(kx)dk \\ & =\lim_{R\rightarrow\infty}\frac{1}{\pi}\int_{0}^{R}\frac{\sin(k)}{k}\cos(kx)dk \\ & =\lim_{R\rightarrow\infty}\frac{1}{\pi}\int_{0}^{R}\frac{\sin(k+kx)+\sin(k-kx)}{2k}dk \\ & =\lim_{R\rightarrow\infty}\frac{1}{2\pi}\left[\int_{0}^{R(1+x)}\frac{\sin(u)}{u}du+\int_{0}^{R(1-x)}\frac{\sin(u)}{u}du\right] \end{align} If $x > 1$, the above gives $$\frac{1}{2\pi}\left[\int_{0}^{\infty}\frac{\sin(u)}{u}du+\int_{0}^{-\infty}\frac{\sin(u)}{u}du\right]=0.$$ The same is true if $x < -1$. If $-1 < x < 1$, the value is $$\frac{1}{2\pi}\left[2\int_{0}^{\infty}\frac{\sin(u)}{u}du\right] = \frac{1}{2\pi}\left[2\frac{\pi}{2}\right] = \frac{1}{2}.$$ For $x=-1$ or $x=1$, the value is $\frac{1}{4}$, which is the mean of the left- and right-hand limits of $\frac{1}{2}\chi_{[-1,1]}$ at $x=\pm 1$.