Compute $\int_0^{+\infty} u^3 e^{-u(1+i)}du $ Let $i^2=-1$. I want to compute the integral:
$$I=\int_0^{+\infty} u^3 e^{-u(1+i)}du .$$
My idea is to substitute variables with $v=(1+i)u$ and link to the Gamma function. But then the region goes from $\mathbb{R}^+$ to something in the complex plane. 
Any advice ?
 A: Note
$$I(a)=\int_0^\infty e^{-au}du= \frac 1a$$
and,
$$\int_0^{\infty} u^3 e^{-au}du = -\frac{d^3I(a)}{da^3}=\frac{6}{a^4}$$
Thus, 
$$\int_0^{\infty} u^3 e^{-u(1+i)}du = \frac{6}{(1+i)^4}=-\frac32$$
A: Using your substitution we have,
$$\begin{align*}
I &= \int_0^\infty u^3 e^{-(1+i)u}\,\mathrm du\\ 
 &= \frac{1}{(1+i)^4}\int_0^{(1+i)\infty}v^3 e^{-v}\,\mathrm dv,\qquad v=(1+i)u.\\ 
\end{align*}$$
 
Now we'd really like to evaluate the integral
$$J=\int_0^{(1+i)\infty}v^3 e^{-v}\,\mathrm dv.$$
We can go about doing this by considering the contour integral
$$\oint_{C_R} z^3 e^{-z}\,\mathrm dz$$
where $C_R$ is the following pizza-slice contour of radius $R$ in the 1st quadrant of the complex plane:
         
             
  
 
Bounding the absolute value of the integral along the semicircular arc shows that it (the integral along the arc) goes to $0$ as $R\rightarrow\infty$, then we find that
$$\lim_{R\rightarrow\infty}\oint_{C_R} z^3 e^{-z}\,\mathrm dz=J+\int_\infty^0 x^3 e^{-x}\,\mathrm dx.$$
But since $z^3 e^{-z}$ is entire on the region enclosed by $C_R$ we have that, by Cauchy's integral theorem,
$$\begin{align*}
 0 &= J+\int_\infty^0 x^3 e^{-x}\,\mathrm dx \\ 
 &\Rightarrow J=\int_0^\infty x^3 e^{-x}\,\mathrm dx \\ 
 &= \Gamma(4) 
\end{align*}$$
so
$$\begin{align*}
I &= \frac{\Gamma(4)}{(1+i)^4}\\ 
 &= -\frac{3}{2}.
\end{align*}$$
