How to find the inverse Fourier transfmation of $\exp(-sk)/k$. I've tried this with the help of hint given by one of my friend.He told me to first find the Inverse fourier transformation of $\exp(-sk)$ which is 
$$ \frac{\sqrt2}{\sqrt \pi}\frac{x}{x^2+ s^2}$$
After this I proceed by applying integration by parts, but I'm not getting the desired result
which is $$\frac{\sqrt2}{\sqrt \pi}  \arctan \left( \frac{x}{s} \right)$$ 
Please help me .I've invested lots of time in this.
Any kind of Hints are welcome.
 A: Define the Fourier Transform of $f(x)$ as
$$F(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx$$
Then, the inverse Fourier Transform of $F(k)$ is given by
$$f(x)=\frac1{2\pi}\int_{-\infty}^\infty F(k) e^{ikx}\,dk$$
We seek to find the inverse Fourier Transform for $F(k)=\frac{e^{-s|k|}}{k}$ where $s\in \mathbb{R}$ and $s>0$.
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METHODOLOGY $1$:
We interpret the inverse Fourier Transform as the Cauchy Principal Value.  Then, we have
$$\begin{align}
f(x)&=\frac1{2\pi}\text{PV}\int_{-\infty}^\infty F(k) e^{ikx}\,dk\\\\
&=\frac1{2\pi}\text{PV}\int_{-\infty}^\infty \frac{e^{-s|k|}}{k} e^{ikx}\,dk \tag 1\\\\
&=\frac{i}{\pi}\int_{0}^\infty \frac{e^{-sk}}{k} \sin(kx)\,dk \tag 2\\\\
&=\frac{i}{\pi}\int_{0}^\infty \left(\int_s^\infty e^{-s'k}\,ds'\right) \sin(kx)\,dk\\\\
&=\frac{i}{\pi}\int_s^\infty \int_{0}^\infty  e^{-s'k} \sin(kx)\,dk \,ds'\\\\
&=\frac{i}{\pi}\int_s^\infty \frac{x}{s'^2+x^2}\,ds'\\\\
&=\frac{i}{2}\text{sgn}(x)-\frac{i}{\pi}\arctan(s/x)\\\\
&=\frac{i}{\pi}\arctan(x/s) \tag 3
\end{align}$$
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METHODOLOGY $2$:
The inverse Fourier Transform of $G(k)=e^{-s|k|}$ for $s\in \mathbb{R}$ with $s>0$ is 
$$\begin{align}
g(x)&=\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-s|k|}e^{ikx}\,dk\\\\
&=\frac{1}{2\pi}\int_{-\infty}^{0} e^{(s+ix)k}\,dk+\frac{1}{2\pi}\int_{0}^{\infty} e^{(-s+ix)k}\,dk\\\\
&=\frac{1}{2\pi}\left(\frac{1}{s+ix}+\frac{1}{s-ix}\right)\\\\
&=\frac{1}{2\pi}\frac{2s}{s^2+x^2}
\end{align}$$
Note that $g(x)=-i\frac{df(x)}{dx}$ where $f(x)$ is given by $(1)$.  Noting from $(2)$ that $f(0)=0$, we find that 
$$f(x)=i\int_0^x g(x')\,dx'=\frac{i}{\pi}\arctan(x/s)$$
which agrees with the result in $(3)$.
