Help with integration of $\mathrm{exp}(jxt)/(x^2 + a^2)$ I am facing issue in solving this integral:
$$ \int_{-\infty}^{\infty} \frac{e^{(jxt)}}{x^2 + a^2} dx $$
I have tried to apply euler formula to seperate $e^{jxt} = cos(xt) + jsin(xt)$ then wolfram gives me result but I don't know how ?
 A: 
I thought that it would be instructive to present a solution that relies on real analysis only.  In the following development, we assume that $a\in \mathbb{R}$ and $t\in \mathbb{R}$.

Let $f(t,a)$ be represented by the integral
$$\begin{align}
f(t,a)&=\int_{-\infty}^\infty \frac{e^{ixt}}{x^2+a^2}\,dx \\\\
&=\int_{-\infty}^\infty \frac{\cos(xt)}{x^2+a^2}\,dx+i\int_{-\infty}^\infty \frac{\sin(xt)}{x^2+a^2}\,dx \tag 1
\end{align}$$
Exploiting the even symmetry and odd symmetry of the integrands in the integrals on the right-hand side of $(1)$, we can write
$$f(t,a)=2\int_0^\infty \frac{\cos(xt)}{x^2+a^2}\,dx \tag 2$$
Enforcing the substitution $x\to |a|x$ in $(2)$ reveals

$$\bbox[5px,border:2px solid #C0A000]{f(t,a)=\frac{2}{|a|}\int_0^\infty \frac{\cos(|a|tx)}{x^2+1}\,dx} \tag 3$$


Next, we use Feynman's Trick for differentiating under the integral and differentiate $f(t,a)$, as given in $(3)$, with respect to $t$.  Proceeding, we find that
$$\begin{align}
\frac{\partial f(t,a)}{\partial t}&=-2\int_0^\infty \frac{x\sin(|a|tx)}{x^2+1}\,dx\\\\
&=-2\int_0^\infty \frac{(x^2+1)\sin(|a|tx)-\sin(|a|tx)}{x(x^2+1)}\,dx\\\\
&=-2\int_0^\infty \frac{\sin(|a|tx)}{x}\,dx+2\int_0^\infty \frac{\sin(|a|tx)}{x(x^2+1)}\,dx\\\\
&=-\pi \text{sgn}(t)+2\int_0^\infty \frac{\sin(|a|tx)}{x(x^2+1)}\,dx
\end{align}$$
Differentiating once more with respect to $t$, we have for $t\ne 0$, we obtain
$$\begin{align}
\frac{\partial^2f(t,a)}{\partial t^2}&=2|a|\int_0^\infty \frac{\cos(|a|tx)}{x^2+1}\,dx\\\\
&=a^2f(t,a)
\end{align}$$
Therefore, $f(t,a)$ satisfies the differential equation 
$$\frac{\partial^2f(t,a)}{\partial t^2}-a^2f(t,a)=0 \tag 4$$
along with the initial conditions $\displaystyle f(0,a)=\pi/|a|$ and $\displaystyle \left.\frac{\partial f(t,a)}{\partial t}\right|_{t=0^{\pm}}=\mp \pi$.  
Solving the differential equation in $(4)$ and enforcing the initial conditions yields

$$\bbox[5px,border:2px solid #C0A000]{f(t,a)=\frac{\pi e^{-|at|}}{|a|}}$$

And we are done!
A: 
Herein, we use complex analysis to evaluate the integral of interest.  We assume that $t\in \mathbb{R}$ and $a\in \mathbb{R}$.  The more general cases for which $t$ and $a$ are complex can be analyzed analogously with special care to account for the signs on the real parts of $t$ and $a$.

Let $f(t,a)$ be represented by the integral 

$$\bbox[5px,border:2px solid #C0A000]{f(t,a)=\int_{-\infty}^\infty \frac{e^{ixt}}{x^2+a^2}\,dx }\tag 1$$


Next we analyze the integral $g(t,a)$ as given by
$$g(t,a)=\oint_C \frac{e^{izt}}{z^2+a^2}\,dz \tag 2$$
where $C$ is the closed contour comprised of $(i)$ the real line segment from $-R$ to $R$ and (ii) the semi-circle centered at the origin with radius $R$ in the upper (lower) half plane for $t>0$ ($t<0$).   

Applying the residue theorem to $(2)$, we find that for $R>|a|$
$$\begin{align}
g(t,a)&=2\pi i\,\text{sgn}(t)\, \text{Res}\left(\frac{e^{izt}}{z^2+a^2}, z
=i|a|\text{sgn}(t) \right)\\\\
&=\frac{\pi e^{-|at|}}{|a|} \tag3
\end{align}$$
We can write $(2)$ 
$$g(t,a)=\int_{-R}^R \frac{e^{ixt}}{x^2+a^2}\,dx+\text{sgn}(t)\,\int_0^{\pi\text{sgn}(t)} \frac{e^{itRe^{i\phi}}}{R^2e^{i2\phi}+a^2}\,iRe^{i\phi}\,d\phi \tag 4$$ 

As $R\to \infty$, the second integral on the right-hand side of $(4)$ approaches zero while the first integral approaches $f(t,a)$ as given by $(1)$.  

Putting together $(3)$ and $(4)$, we arrive at the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{e^{ixt}}{x^2+a^2}\,dx =\frac{\pi e^{-|at|}}{|a|}} $$

which agrees with the result reported in This Answer, which used only real analysis.
