# When calculating integrals, why replacing factorials with $\Gamma$ so often works?

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):

1. 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.

2. 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!

• What are your thoughts on splitting out your questions into multiple posts? In particular, the counterexample question would be a good one for a separate post... May 17, 2020 at 18:22
• @ColmBhandal: I can certainly do that. I'm not very sure about putting them both into one question myself. May 17, 2020 at 18:26
• Your question, your choice :) I think it may get more traction as two separate ones though May 17, 2020 at 18:27
• @ColmBhandal: I'll probably leave them in one. I find that I can't come up with a good and short title for the counterexample question. May 17, 2020 at 18:47

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 $$\Re z >-\frac{1}{2}$$.

• That's a very nice theorem. However, I'm still not sure what singles $\Gamma$ out. Is it true that $\Gamma$ is the only factorial extension that is holomorphic (at least with $\operatorname{Re} z > -1$)? (To be honest, I haven't ever seen any other such extension, I only know there are infinitely many.) That would make it clear. May 19, 2020 at 16:15
• Look up Hadamard's gamma function.
– Gary
May 19, 2020 at 16:21

It boils down to why factorials arise in the context, the gamma function uses a very clever trick to make a recurrence relation work outside of it’s normal habitat. $$\Gamma(z+1)=z\Gamma(z)$$ $$\Gamma(1)=1$$ From those two alone we just get that $$\Gamma(n)=(n-1)!$$ the full gamma function uses a clever trick to make this much less “recursive”, if we take repeated derivatives of $$x^n$$ we get a pretty clear pattern. $$\begin{matrix} 1 & 0\\ x & 1& 0\\ x^2& 2x& 2 & 0\\ x^3&3x^2&6x&6&0\\ x^4&4x^3&12x^2&24x&24&0\\ \end{matrix}$$ Before every 0 comes (n-1)! But how to use this for an integral?
Simple, use integration by parts to lock it until all x’s got derivativated to continue.
$$\Gamma(z)=\int^\infty_0x^{z-1}e^{-x}dx$$ The bounds are to make the UV term of integration by parts 0 and the -x to make the integral part positive and not negative.
Back to the question, it works because most problems who get a factorial get it because of a reduction formula, recurrence relation, which $$\Gamma$$ extends very well.
Like that every time $$\pi$$ shows up there is a hidden circle, every time $$\Gamma$$ or ! Show up there is a hidden recurrence relation.
In your case: Recurrence relation for the integral, $I_n=\int\frac{dx}{(1+x^2)^n}$.
(The indefinite case). most recurrence relations are of the form: $$I_{n+1}=I_nP(n)R(x),I_0=k$$ where P and R are rational functions.
or similar that can be transformed or build out of functions in the form.
by scaling all terms by k we can get it simplified.
we can split it into two parts:$$I_n=I^1_{n-1}I^2_{n-1}$$ 1.$$I^1_n=R(x)I^1_{n-1}$$ which is simply $$I^1_n=R(x)^n$$ and extended to $$I^1_n=R(x)^a$$.
2.$$I^1_n=P(n)I^1_{n-1}$$ this one is a little more complicated. $$P(n)=a\frac{\overbrace{(n+b_1)(n+b_2)(n+b_3)...}^{P_1(n)}}{\underbrace{(n+c_1)(n+c_2)(n+c_3)...}_{P_2(n)}}$$ $$I^1_n=a^n \frac{I^{1_1}_n}{I^{1_2}_n}$$ simplifying the problem to solving: $$k_n=k_{n-1}T(n)$$ for some polynomial T.
all polynomials are factorable on $$C$$ so $$T(n)=(n+t_1)(n+t_2)...$$
splitting $$k_n$$ to $$k^1_nk^2_n...$$ one for every factor we narrow the problem to extending: $$m_n=m_{n-1}(n+r)$$. when finding what integral would give such a recurrence relation we can use integration by parts: $$\int^b_au'vdx=uv^b_a-\int^b_auv'dx$$ we want the right to be the integral for n-1 times (n+r) so we need $$uv^b_a=0$$ it's logical to choose the bounds to be 0 and $$\infty$$ to match gamma and choosing $$v=e^{-x}$$ to make the sign positive. $$\int^{\infty}_0ue^{-x}dx=u(0) + \int^{\infty}_0u'e^{-x}dx$$ the u satisfying the requirements is $$u=x^{r+a}$$ thus, our integral for it is $$\int^{\infty}_0x^{a+r}e^{-x}dx = \Gamma(a+r)$$ To sum up, most recurrence relations involving factorial can be broken down into a form that can be broken down into a form that can be broken down into a form that can be broken down into Gamma functions.
even yours, if you do a little transforming.
2. Probably not in a non-trivial way, it seems to me like factorial will never arise out of an integral not already containing it unless it’s a recurrence relation, in which case, $$\Gamma$$ will most likely extend it.