Complex Integral with exponential I've been struggling with this:
$$\int_{0}^{\infty }\frac{e^{-px}}{x^{2}+1}\mathrm{d}x, \; \;  p\ge 0.$$  
 A: \begin{align*}
\int_0^\infty \frac{e^{-px}}{x^2 + 1}dx &\overset{(1)}{=} \int_0^\infty \int_0^\infty e^{-px} e^{-sx} \sin(s)ds dx \\ 
 &\overset{(2)}{=} \int_0^\infty \int_0^\infty e^{-(p+s)x} \sin (s)dx ds\\ 
 &\overset{(3)}{=} \int_0^\infty \frac{\sin(s)}{(p+s)}  ds  \\ 
 &\overset{(4)}{=} \text{Ci}(p) \sin (p)+\frac{1}{2} (\pi -2 \text{Si}(p)) \cos (p)\\ 
\end{align*}
$\displaystyle(1): \int_0^{\infty} e^{-sx} \sin(s)dx = \frac{1}{1+x^2}$   
$(2):$ change of order of integration.  
$\displaystyle(3): \int_{0}^\infty e^{-(p+s)x} dx = \frac{1}{(p+s)}$
$(4):$
 \begin{align*}
\int_0^\infty \frac{\sin(s)}{(p+s)}  ds &=  \int_p^\infty \frac{\sin(y - p)}{y }dy \\ 
 &= \int_p^\infty \frac{\cos(p)\sin(y) - \cos(y) \sin(p)}{y }dy \\ 
 &= - \sin(p) \int_p^\infty \frac{\cos(y)}{y}dy  + \cos(p)\int_p^\infty \frac{\sin(y)}{y}dy\\ 
 &=  \sin(p)  \text{Ci}(p) + \cos(p) \left( \int_0^{\infty } \frac{\sin(y)}{y}dy -  \int_0^{p } \frac{\sin(y)}{y}dy\right )\\ 
 &= \text{Si}(p)\cos(p) + \frac \pi 2 \cos(p)  - \sin(p)  \text{Ci}(p) 
\end{align*}
A: This definite integral can be 'solven' by using integral Fourier transform or Laplace transform. Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by
$$
\begin{align}
F(\omega)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\\
&=\int_{-\infty}^{\infty}e^{-a|t|}e^{-i\omega t}\,dt\\
&=\int_{-\infty}^{0}e^{at}e^{-i\omega t}\,dt+\int_{0}^{\infty}e^{-at}e^{-i\omega t}\,dt\\
 &=\lim_{u\to-\infty}\left. \frac{e^{(a-i\omega)t}}{a-i\omega} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+i\omega)t}}{a+i\omega} \right|_{0}^{t=v}\\
&=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}\\
&=\frac{2a}{\omega^2+a^2}.
\end{align}
$$
Next, the inverse Fourier transform of $F(\omega)$ is
$$
\begin{align}
f(t)=\mathcal{F}^{-1}[F(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega\\
e^{-a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{\omega^2+a^2}e^{i\omega t}\,d\omega\\
\frac{\pi e^{-a|t|}}{a}&=\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{\omega^2+a^2}\,d\omega.
\end{align}
$$
Comparing the last integral to the problem, we have $a=1$ and $it=-p\;\rightarrow\;t=ip$. Therefore
$$
\int_{-\infty}^{\infty}\frac{e^{-px}}{x^2+1}\,dx=\pi e^{-ip}.
$$
I cannot assure that
$$
\int_{-\infty}^{\infty}\frac{e^{-px}}{x^2+1}\,dx=2\int_0^{\infty}\frac{e^{-px}}{x^2+1}\,dx,
$$
except for $p=0$ that yields
$$
\int_0^{\infty}\frac{1}{x^2+1}\,dx=\frac{\pi}{2}.
$$

$$
\text{# }\mathbb{Q.E.D.}\text{ #}
$$
A: The answer, done with Maple, is not simple:
$$ -1/2\, \left( -\sin \left( 2\,p \right) {\it Ci} \left( p \right) +
\cos \left( 2\,p \right)  \left( {\it Si} \left( p \right) -1/2\,\pi 
 \right) +{\it Si} \left( p \right) -1/2\,\pi  \right) \cos \left( p
 \right) +1/2\, \left( -\cos \left( 2\,p \right) {\it Ci} \left( p
 \right) -\sin \left( 2\,p \right)  \left( {\it Si} \left( p \right) -
1/2\,\pi  \right) +{\it Ci} \left( p \right)  \right) \sin \left( p
 \right) 
$$
See MapleHelp concerning the notation.
