Complex Integral with exponential I've been struggling with this:
$$\int_{0}^{\infty }\frac{e^{-px}}{x^{2}+1}\mathrm{d}x, \; \;  p\ge 0.$$  

\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*}
