Euler Gamma function $\Gamma(z)$ on $\mathbb{C}$

Question

I'm working on an exercise about the Gamma Function from Euler.

First, $\Gamma (z)= \int_0^\infty e^{-t}t^{z-1}dt$.

Now, if we consider the "similar" function $\int_{\frac{1}{n}}^\infty e^{-t}t^{z-1}dt$, why is this holomorphic in $\mathbb{C}$ for any $n$?

And then, why is $\Gamma (z)$ holomorphic in the right half plane, i.e. $Re(z)>0$?

We know that the integrand is holomorphic and its integral should be holomorphic, too, then, since we can write the integrand as a polynomial which we then integrate. But surely, this is not rigorous enough, or is it?

I'd appreciate help!

-marie

Answer 1

$\Gamma(z)$ is only defined by the integral for $\mathfrak{Re}(z)>0$ because otherwise the integral diverges at the lower bound.

The modified function, for any fixed $n > 0$, is convergent for any $z \in \mathbb{C}$: $$ g_n(z) = \int_{1/n}^\infty t^z \mathrm{e}^{-t} \frac{\mathrm{d} t}{t} $$ However, as $n$ grows, the limit would diverge for $\mathfrak{Re}(z)<0$.

$g_n(z)$ is holomorphic by Looman-Menchoff's theorem. Indeed $g_n(z)$ is uniformly convergent, and hence differentiation and integration operations can be exchanged, allowing to conclude that Cauchy-Riemann equations are verified by $g_n(z)$.

Answer 2

Recall that a uniformly convergent sequence of holomorphic functions on a compact set $X$ converges to a holomorphic function, since then we may switch the integral and limit to get $\displaystyle\int_{C}f\: dz = \lim_{n \to \infty} \int_{C}f_n\: dz = 0$ for every contour $C$ in $X$.

Now for the moment fix $n$, and for each $m\geq 1$, let $f_{n,m}(z) = \int_{1/n}^m e^{-t}t^{z-1}$. Now take any compact subset $E$ of the (open) right-hand plane. Show that the sequence $\{f_{n,m}\}_{m\geq 1}$ converges uniformly on $E$ and hence its limit in $E$ is holomorphic there. This shows that for any $n$, the function $f_n(z)=\int_{1/n}^\infty e^{-t}{t^{z-1}}$ is holomorphic everywhere in the right-hand plane. Now take again a compact subset $E$ of the right-hand plane. Show that the sequence $f_n$ converges uniformly on $E$, and hence that its limit is a function $\Gamma$ holomorphic in the right-hand plane.