TITLE: When calculating integrals, why replacing factorials with $\Gamma$ so often works?
QUESTION [10 upvotes]: There are lots of definite integrals that depend on a parameter $n \in \mathbb N$ and whose result contains factorials of $n$ or some simple functions of it. For instance,
$$ \int\limits_{-\infty}^\infty \frac{\mathrm{d} x}{(1+x^2)^n} = \pi \frac{(2n-2)!}{2^{2n-2} [(n-1)!]^2} \qquad (n \in \mathbb N),$$
but there are loads and loads of similar ones.
Now let's say we allow $n$ to be a real number instead of a natural number. It looks like that in most of those cases, it's enough to replace the factorials with gamma functions, giving generalizations like
$$ \int\limits_{-\infty}^\infty \frac{\mathrm{d} x}{(1+x^2)^\alpha} = \pi \frac{\Gamma(2\alpha-1)}{2^{2\alpha-2} \Gamma(\alpha)^2} \qquad (\alpha \in \mathbb R),$$
and a quick numeric integration for a couple of $\alpha$ shows that this could be correct. And if it works, then it will work for complex $\alpha$ as well.
The question(s):

Why this works so often? Obviously the integral should depend on $\alpha$ continuously, so it makes sense to replace the factorial with some function that extends it continuously. However there are many of those. What's so special about $\Gamma$ that those integrals seem to generalize only to expressions with $\Gamma$ and not with any other possible factorial extension? I know that $\Gamma$ is the only factorial extension that is log-convex, but I can't see how it connects to this.
Would anybody be able to show a counterexample where this simple replacement with $\Gamma$'s fails? Is there any theorem that would give conditions for this to work?

Thanks!

REPLY [4 votes]: One possible way to prove such extensions is by using the following theorem of Carlson: If $f(z)$ is holomorphic in the sector $|\arg z|\leq \alpha$ with $\alpha\geq \frac{\pi}{2}$, $|f(z)|=\mathcal{O}(e^{c|z|})$ with some $c < \pi$ in this sector, and if $f(n) = 0$ when $n = 1, 2, 3,\ldots$, then $f(z) \equiv  0$.
For more on this theorem, see G. H. Hardy, On two theorems of F. Carlson and S. Wigert, Acta Math. 42 (1920), pp. 327–339.
In your example, let
$$
f(z)=\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{(1 + x^2 )^{z + 1} }}}  - \pi \frac{{\Gamma (2z + 1)}}{{2^{2z} \Gamma (z + 1)^2 }},\quad \Re z \ge 0.
$$
Now
$$
\left| {\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{(1 + x^2 )^{z + 1} }}} } \right| \le \int_{ - \infty }^{ + \infty } {\frac{{dx}}{{(1 + x^2 )^{\Re z + 1} }}}  \le \int_{ - \infty }^{ + \infty } {\frac{{dx}}{{1 + x^2 }}}  = \pi ,
$$
and, by Stirling's formula,
$$
\pi \frac{{\Gamma (2z + 1)}}{{2^{2z} \Gamma (z + 1)^2 }} = \sqrt {\frac{\pi }{z}} \left( {1 + \mathcal{O}\!\left( {\frac{1}{z}} \right)} \right).
$$
Thus, $|f(z)|=\mathcal{O}(1)$ for $\Re z\geq 0$. As you noted, $f(n)=0$ for $n = 1, 2, 3,\ldots$. Thus, by Carlson's theorem, $f(z)$ is identically zero. Accordingly,
$$
\int_{ - \infty }^{ + \infty } {\frac{{dx}}{{(1 + x^2 )^{z + 1} }}} = \pi \frac{{\Gamma (2z + 1)}}{{2^{2z} \Gamma (z + 1)^2 }},
$$
for $\Re z \ge 0$. Now, you may use analytic continuation to extend this identity to the larger region $\Re z >-\frac{1}{2}$.