I am stuck with something that may be simple, but I just cannot find the answer.
We define, for $z$ a complex number of real part strictly positive, $\Gamma(z) = \int_{0}^{\infty} t^{z - 1}exp(-t) dt$. How do we show that $\Gamma$ is an analytic function on $\{ z \in \mathbb{C}, \Re(z) > 0 \}$ ?
I started by rewritting the expression of $\Gamma$ this way:
$\Gamma(z) = \int_{- \infty}^{\infty} exp(zu)exp(-e^{u}) du$
Then, if $a$ is a complex number with $\Re(a) > 0$, we have:
$\Gamma(z) = \int_{- \infty}^{\infty} exp((z - a)u)exp(au)exp(-e^{u}) du = \int_{- \infty}^{\infty} \sum \limits_{n = 0}^{+\infty} \frac{(z-a)^{n}}{n!}u^{n}exp(-e^{u}) du$.
However after that, I do not see how I could exchange $\sum$ and $\int$ ...
Thank you for your help.
It is analytical, because it is differentiable as a function of a complex variable $z$. The integral corresponding to the derivative converges uniformly, so you can differentiate with respect to the parameter.