# Proving analyticity of gamma function

1. Jun 23, 2011

### nonequilibrium

Hello,

In our course of complex analysis we proved that the gamma function,
$$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \mathrm d t$$
for 0 < Re(z), is analytical.

We did this by defining
$$f_{\epsilon,R}(z) = \int_\epsilon^R t^{z-1} e^{-t} \mathrm d t$$
about which we can prove that it is analytical by rewriting t^(z-1) as a power series (by using the definition of the complex power in function of the exponential and ln) and then bringing the sum outside of the integral by using the uniform convergence of the power series (on compact subsets).

Then we proved that $$f_{\epsilon,R} \to \Gamma$$ uniformly on compact subsets (of the right half-plane).

And apparently analyticity of the gamma function follows, but I don't see why. Surely not because gamma is the uniform limit of analytical functions, because a uniform limit doesn't preserve analicity. So I'm guessing there is an implicit last step I'm overlooking.

Can anybody help?
Thank you.

EDIT: personally, I think I could prove it the following way: first take the derivative of $$f_{\epsilon,R}$$, what I got is the same function but with an extra ln(t) in the integrand. However, I think one can still prove that (also) converges uniformly, but now to $$\int_0^\infty t^{z-1} e^{-t} \ln t \mathrm d t$$, and then, by the uniform limit of derivatives, the limit of the original function is differentiable and this also gives an explicit form of the derivative of the gamma function. However, this doesn't seem like the thought process that the above proof was using, and I'm interested in understanding the above proof.

Last edited: Jun 23, 2011
2. Jun 23, 2011

### micromass

Staff Emeritus
Hi mr. vodka!

If a sequence of analytic functions converges (locally) uniformly then the limit function is analytic. In particular, if $f_n\rightarrow f$, then we have that

$$f_n^\prime\rightarrow f^\prime$$

this is not true in the real case, but it is true in complex analysis!

3. Jun 23, 2011

### nonequilibrium

Interesting!

Why is this so? Do you know where I can find a proof?

4. Jun 23, 2011

### micromass

Staff Emeritus
The analycity is a consequence of Morera's theorem. Recall that Morera's theorem says that if a continuous function has the property that

$$\int_\gamma f(z)dz=0$$

for every closed C1-curve gamma, then f is analytic.

The clue is that uniform convergence allows you to exchange limit and integral, thus

$$\int_\gamma \lim_{n\rightarrow +\infty}{f_n(z)}dz=\lim_{n\rightarrow +\infty}{\int_\gamma f_n(z)}=0$$

thus the limit is analytic. The statement about the derivatives follows from the Cauchy integral formula.

As a reference, I can recommend the third chapter of Freitag and Busam's "Complex analysis"...

5. Jun 23, 2011

### nonequilibrium

Brilliant :)

<3 Complex Analysis

thank you