Basic question regarding gamma function and contour integration Here is something basic from complex analysis, and I was wondering if someone could possibly explain me how the following worked.
Gamma function is defined for $Re(x)>0$ that
$$
\Gamma(x) = \int_0^{\infty} u^{x-1} e^{-u} du,
$$
where the integral is from $0$ to infinity on $\mathbb{R}$. Given $s \in \mathbb{C}$ with $Re(s)>0$, I want to show that by the change of variable $w = s u$ that we have the following:
$$
\int_0^{\infty} u^{x-1} e^{-su} du = s^{-x} \int w^{x-1} e^w dw = s^{-x} \Gamma(x). 
$$
The only thing I am not sure is when I do the change of variable, the integral becomes from $0$ to $\infty$ on real to $s\cdot [0,\infty]$. How does one obtain the Gamma function on the right? Thank you very much!
 A: Consider the wedge-shaped contour $\gamma$ consisting of $[\varepsilon,R]$, the arc of the circle of radius $R$ centred at $0$ joining $R$ and $sR$, the straight line joining $sR$ to $s\varepsilon$, and the arc joining $s\varepsilon$ to $\varepsilon$. Then the integrand is analytic inside $\gamma$, so Cauchy's theorem applies and
$$ \int_{\gamma} z^{x-1} e^{-z} \, dz=0. $$
The integral along the real axis is
$$ \int_{\varepsilon}^{R} u^{x-1}e^{-u} \, du \to \Gamma(x)  $$
as $\varepsilon \to 0,R \to \infty$. The integral along the other straight line is
$$ \int_R^{\varepsilon} (sy)^{x-1}e^{-sy} \, s dy = -s^{x}\int_{\varepsilon}^R y^{x-1} e^{-sy} \, dy \to -s^{x}\int_{0}^{\infty} y^{x-1} e^{-sy} \, dy $$
since the line is traversed in the negative direction. Therefore
$$ \Gamma(x)-s^{x}\int_{0}^{\infty} y^{x-1} e^{-sy} \, dy + \int_{\text{arcs}} = 0, $$
and to verify your equality, it suffices to show that the integrals along the arcs tend to zero. This is easy: since $\Re(s)>0$, we have
$$ \Re(z)>\Re(s)R $$
on the large arc, so
$$ \left\lvert \int_{R}^{sR} z^{x-1} e^{-z} \, dz \right\rvert < \tfrac{1}{2}\pi R^{\Re(x)} e^{-R\Re(s)} \to 0, $$
since the integrand is bounded above by $R^{x-1} e^{-R\Re(s)}$ for $R$ large, and the length of the contour is less than $\frac{1}{2}\pi R$. The small contour is even easier, using a similar trick to bound the integrand by $\varepsilon^{\Re(x)}$. Thus the integrals over both arcs tend to zero, and you find the result. (The same result holds if $s$ is imaginary, but you need a bit more sophistication in bounding the arcs to prove it.)
