Complex integral in all $ \mathbb(R) $ I have to evaluate the next integral 
$$
\int_{-\infty}^{\infty}\frac{1}{(\lambda -ip)^n}e^{ipx}dp  \quad p \in \mathbb{R}  \ \ \ n\in \mathbb{N} 
$$ I don't know what strategy to follow, because I usually work with residue integration, but when  the function inside the  intergral is real. 
 A: There are four cases.  
Case 1: $x>0$ and $\lambda<0$
Case 2: $x<0$ and $\lambda<0$
Case 3: $x>0$ and $\lambda>0$
Case 4: $x<0$ and $\lambda>0$
We shall examine Case 1 for which $x>0$ and $\lambda<0$.  

Observe the integral $I$ given by 
$$I=\oint_C\frac{e^{ixz}}{(\lambda-iz)^n}dz$$
where $C$ is comprised of the real-line from $-R$ to $R$ plus a semi-circle of radius $R$ in the upper-half $z$-plane.  
The contribution to the integral from the integration over the semi-circle tends to zero as $R \to \infty$.  Thus, 
$$\begin{align}
I&=\oint_C\frac{e^{ixz}}{(\lambda-iz)^n}dz=\int_{-\infty}^{\infty}\frac{e^{ixz}}{(\lambda-iz)^n}dz\\\\
&=2\pi i \text{Res}\left(\frac{e^{ixz}}{(\lambda-iz)^n}, z=-i\lambda\right)
\end{align}$$
For a pole of order $n$, the residue is given by 
$$\begin{align}
\text{Res}\left(\frac{e^{ixz}}{(\lambda-iz)^n}, z=-i\lambda\right)&=\frac{1}{(n-1)!}\lim_{z\to -i\lambda}\frac{d^{n-1}}{dz^{n-1}}\left((z+i\lambda)^n\frac{e^{ixz}}{(\lambda-iz)^n}\right)\\\\
&=\frac{1}{(n-1)!}\frac{(ix)^{n-1}e^{\lambda x}}{(-i)^n}
\end{align}$$
Thus, the integral of interest is 
$$I=2\pi \frac{(-1)^n\,x^{n-1}}{(n-1)!}e^{\lambda x}$$
