Method to find inverse Fourier transform of $\frac{1}{k} \sin(k)$

Question

I wanted to find the inverse Fourier transform of $\tilde f(k)=\frac{1}{k}\sin(k)$. I found a similar question here: Calculating the Inverse Fourier Transform of $\frac{1}{\sqrt{2\pi}k}\sin k$, but I have been asked to do this using contour integration (which isn't mentioned in the answers to that question). My problem with this is if you take $\int_{-\infty}^{+\infty} \tilde f(k) e^{ikx} dk$, then this has no poles, and so nowhere to calculate a residue. Normally I'd just use a semi-circular contour, however in this case I don't think that the integral along the arc would go to $0$. Whichever contour I choose, the total integral along that contour will be $0$. This would mean that I'd get something like "the integral I want" + "something extra which isn't $0$" $= 0$. I think my main problem is finding this "something extra" - what contour should I use? Could I use the expected answer somehow to help choose that?

Answer 1

Note that we can write

$$\int_{-\infty}^\infty\frac{\sin(k)}{k}e^{ikx}\,dk=\frac{1}{2i}\int_{-\infty}^\infty\frac{e^{ik}-e^{-ik}}{k}e^{ikx}\,dk \tag1$$

Observe that for each of the principal value integrals

$$I_{\pm}(x)=\text{PV}\left(\int_{-\infty}^\infty\frac{e^{\pm ik}}{k}e^{ikx}\,dk\right)=\text{PV}\left(\int_{-\infty}^\infty\frac{e^{ ik(x\pm 1)}}{k}\,dk\right)$$

the integrand has a pole at $k=0$. To evaluate these integrals, we analyze the contour integrals

$$\begin{align} \oint_{C}\frac{e^{iz(x\pm 1)}}{z}\,dz&=\int_{-R}^{-\epsilon} \frac{e^{ik(x\pm 1)}}{k}\,dk+\int_{\epsilon}^{R} \frac{e^{ik(x\pm 1)}}{k}\,dk\\\\ &+\int_{\text{sgn}(x\pm 1)\pi}^0 \frac{e^{i\epsilon e^{i\phi}}}{\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi\\\\ &+\int_0^{\text{sgn}(x\pm 1)\pi}\frac{e^{i R e^{i\phi}}}{R e^{i\phi}}\,iR e^{i\phi}\,d\phi \tag 2 \end{align}$$

As $R\to \infty$, the last integral on the right-hand side of $(2)$ tends to $0$. As $\epsilon\to 0$, the third integral on the right-hand side of $(2)$ tends to $-\text{sgn}(x\pm 1)\pi$. Therefore, we have

$$\text{PV}\left(\int_{-\infty}^\infty\frac{e^{ ik(x+ 1)}-e^{ik(x-1)}}{k}\,dk\right)=\text{sgn}(x+1)\pi-\text{sgn}(x-1)\pi=\begin{cases}2\pi &,|x|<1\\\\0&,|x|>1\\\\\pi&,|x|=1\end{cases} \tag 3$$

Substituting $(3)$ into $(1)$ yields

$$\int_{-\infty}^\infty\frac{\sin(k)}{k}e^{ikx}\,dk=\begin{cases}\pi &,|x|<1\\\\0&,|x|>1\\\\\pi/2&,|x|=1\end{cases} $$