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
$$\int_{-\infty}^\infty\frac{\sin(k)}{k}e^{ikx}\,dk=\begin{cases}\pi &,|x|<1\\\\0&,|x|>1\\\\\pi/2&,|x|=1\end{cases} $$
Best Answer
First, let's compute the FT of $\text{sech}{(\pi x)}$, which may be derived using the residue theorem. We simply set up the Fourier integral as usual and comvert it into a sum as follows:
$$\begin{align}\int_{-\infty}^{\infty} dx \, \text{sech}{(\pi x)} \, e^{i k x} &= 2 \int_{-\infty}^{\infty} dx \frac{e^{i k x}}{e^{\pi x}+e^{-\pi x}}\\ &= 2 \int_{-\infty}^0 dx \frac{e^{i k x}}{e^{\pi x}+e^{-\pi x}} + 2 \int_0^{\infty}dx \frac{e^{i k x}}{e^{\pi x}+e^{-\pi x}}\\ &= 2 \sum_{m=0}^{\infty} (-1)^m \left [\int_0^{\infty}dx \, e^{-[(2 m+1) \pi+i k] x} +\int_0^{\infty}dx \, e^{-[(2 m+1) \pi-i k] x} \right ] \\ &= 2 \sum_{m=0}^{\infty} (-1)^m \left [\frac{1}{(2 m+1) \pi-i k} + \frac{1}{(2 m+1) \pi+i k} \right ]\\ &= 4\pi \sum_{m=0}^{\infty} \frac{(-1)^m (2 m+1)}{(2 m+1)^2 \pi^2+k^2}\\ &= \frac{1}{2 \pi}\sum_{m=-\infty}^{\infty} \frac{(-1)^m (2 m+1)}{\left (m+\frac12\right)^2+\left(\frac{k}{2 \pi}\right)^2} \end{align}$$
By the residue theorem, the sum is equal to the negative sum of the residues at the non-integer poles of
$$\pi \csc{(\pi z)} \frac{1}{2 \pi}\frac{2 z+1}{\left ( z+\frac12\right)^2+\left (\frac{k}{2 \pi}\right)^2}$$
which are at $z_{\pm}=-\frac12 \pm i \frac{k}{2 \pi}$. The sum is therefore
$$-\frac12\csc{(\pi z_+)} - \frac12 \csc{(\pi z_-)} = -\Re{\left [\frac{1}{\sin{\pi \left (-\frac12+i \frac{k}{2 \pi}\right )}}\right ]} = \text{sech}{\left ( \frac{k}{2}\right)}$$
By this reasoning, the FT of $\operatorname{sech}{x}$ is $\pi\, \text{sech}{\left ( \frac{\pi k}{2}\right)}$.