I have to compute the followings integrals $\forall\; b\in \mathbb{C},\; \text{Re} \;b \gt0,p\gt 0$
$$ \int_{-\infty}^\infty \frac{e^{ipx}}{x-ib}$$ $$ \int_{-\infty}^\infty \frac{e^{ipx}}{x+ib}$$
using residue theroem, and then deduce Laplace Transforms
$$ \int_{-\infty}^\infty \frac{b\cos(px)}{x^2+b^2} = \int_{-\infty}^\infty \frac{x\sin(px)}{x^2+b^2} = \pi e^{-pb} $$
I find
$$ \int_{-\infty}^\infty \frac{e^{ipx}}{x-ib} = 2\pi i e^{-pb}$$ $$ \int_{-\infty}^\infty \frac{e^{ipx}}{x+ib} = 0$$
But then I don't know how to handle these equations : for example, I could try this :
$$ \int_{-\infty}^\infty \frac{b\cos(px)}{x^2+b^2} = \int_{-\infty}^\infty \frac{b(e^{ipx}+e^{-ipx})}{2(x+ib)(x-ib)}$$ but in each term of the RHS I can't get rid of the $\frac{1}{x+ib}$ or $\frac{1}{x-ib}$ that differs from the equation I computed.
If $p > 0$ and $\Im z > 0$, then $|e^{ipz}|=e^{-p\Im z}$ allows you to close the contour in the upper half-plane.
Let $C_{R}$ be the positively oriented contour composed of the linear segement from $-R+i0$ to $R+i0$ combined with the semicircular contour $Re^{i\theta}$ for $0 \le \theta \le \pi$. Then, for $R$ large enough so that $ib$ is inside $C_{R}$, $$ \frac{1}{2\pi i}\int_{C_{R}}\frac{1}{z-ib}e^{ipz}\,dz = e^{-pb},\\ \frac{1}{2\pi i}\int_{C_{R}}\frac{1}{z+ib}e^{ipz}\,dz = 0. $$ The integral over the semicircular part of the arc vanishes as $R\rightarrow \infty$. So you end up with two Cauchy principle value integrals that you can subtract to get an absolutely convergent integral, $$ \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{b}{t^{2}+b^{2}}e^{ipt}\,dt = e^{-pb}, $$ or you can add to get another Cauchy principle value integral $$ c.p.v \frac{1}{\pi i}\int_{-\infty}^{\infty}\frac{t}{t^{2}+b^{2}}e^{ipt}\,dt = e^{-pb} $$ If you take $b$ to be real and positive, then the imaginary parts of both integrals are $0$ and the real parts are both $e^{-pb}$, which gives $$ \int_{-\infty}^{\infty}\frac{b\cos(p t)}{t^{2}+b^{2}}\,dt=\pi e^{-pb} = \int_{-\infty}^{\infty}\frac{t\sin(pt)}{t^{2}+b^{2}}\,dt, \;\;\;\mbox{$b >0$ real, $p > 0$ real}. $$ The first integral is absolutely convergent. The second is an improper Riemann integral that converges as the limits tend to $\pm \infty$. The first integral can be extended from real $b$ to any complex $b$ for which $b^{2}$ does not lie on the negative real axis; and the extension is holomorphic in $b$. So the first identity extends to all such $b$ by the Identity Theorem. The second identity is trickier to extend because the convergence is not absolute.