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?
 A: 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.
A: 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 $x=\pm 1$.
