A short study of the gamma function

In the last exercise of the course Topology and Multivariable Calculus (MAA202) last semester, we have a kinda nice (but skipped) exercise. So I'll try to tackle it today.

Exercise

For $x > 0$, we define Euler’s Gamma function by

$$\Gamma(x) = \int_0^{+\infty} t^{x-1} e^{-t} \mathrm{d}t$$

1. Prove that $\Gamma \in \mathcal{C}^0(\mathbb{R}_+^*)$.

2. (a) Show that $$\Gamma(x+1) = x\Gamma(x)$$ for all $x > 0$.

2. (b) Deduce that $\Gamma(n) = (n-1)!$ for all $n \in \mathbb{N}^*$.

So, let me try...

My attempt

1. Ahh... What to do... It's obvious :( but I can't prove it! Hmm, maybe DCT? Let's try unfolding the term $e^{-t}$. Since $e^x = \exp(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$, we have

$$\Gamma(x) = \int_0^{+\infty} t^{x-1} \sum_{k=0}^{\infty} \frac{(-1)^k t^k}{k!} \mathrm{d}t = \int_0^{+\infty} \sum_{k=0}^{\infty} \frac{(-1)^k t^{k+x-1}}{k!} \mathrm{d}t$$

No... This doesn't look so good. (I planned to swap the sum with the integral, using some theorems...) If they're swapped (even if it works), then one must (infinitely) sum the integrals $\int_0^{+\infty} \frac{(-1)^k t^{k+x-1}}{k!} \mathrm{d}t$ for all $k$, which, very clearly, the integral diverges (since the antiderivative of $t \mapsto t^c$ is $t \mapsto \frac{t^{c+1}}{c}$, and $\lim_{t \to +\infty} \varepsilon t^{c+1} = \pm \infty$ (diverges, as a simple evaluation of polynomials at infinity) for any $\varepsilon > 0$.

Let us observe $\Gamma(x)$ for $x \in ]0, 1[$. We have $\Gamma(1) = \int_0^{+\infty} t^0 e^{-t} \mathrm{d}t = [-e^{-t}]_0^{+\infty} = 0-(-1) = 1$. Ahh it doesn't tell me anything.

Can we prove that $\Gamma$ is differentiable everywhere in $]0, +\infty[$ first? Then it would follow that it is continuous... No. It's too complicated swapping the derivative with the improper integral.

Uh... How about let $\Gamma_n(x) = \int_n^{n+1} t^{x-1} e^{-t} \mathrm{d}t$? Then $\Gamma(x) = \sum_{n=0}^\infty \Gamma_n(x)$. Note that for us to be sure, we'd need to prove first that the integral from the definition of $\Gamma$ converges as it tends to infinity (because $\lim_{x \to +\infty} \phi(x) = \ell$ implies $\lim_{n \to +\infty} \phi(n) = \ell$).

Hmm... No. We don't really need to. The function $\phi_x(y) = \int_0^y t^{x-1} e^{-t} \mathrm{d}t = \sum_{k=0}^{\lfloor y \rfloor -1} \Gamma_k(x) + \int_{\lfloor y \rfloor}^y t^{x-1} e^{-t} \mathrm{d}t$ is increasing because $t^{x-1} e^{-t} > 0$ for all $t > 0$, and bounded by $\phi_x(\lceil y \rceil)$.

No, no, stop, stop. This is too complicated than what it should be. Let's recall Proposition 4.1.3.

The sad thing is that, in this chapter, the lecture notes didn't fully prove things, so there are a lot of gaps and it's a bit hard to follow. bruhhh sad. Anyway, let us go back to the 1.

Let $x_0 \in ]0, +\infty[$ be any number.

Let $f(x, t) = t^{x-1} e^{-t}$ be defined on $I \times J$ where $I := ]0, +\infty[$ and $J := [0, +\infty[$. Then,

• for all $x \in I, t \mapsto f(x, t) = t^{x-1} e^{-t}$ is continuous on $J$, since it is a product of continuous functions
• for all $t \in J, x \mapsto f(x, t) = t^{x-1} e^{-t}$ is continuous at $x_0$, since it is a product of continuous functions
• let $\varphi(t) = t^{x_0} e^{-t}$, then $\varphi(t)$ is integrable on any compact segment of $J$, for noncompact interval $[a, +\infty[$, we see that since $t \mapsto t^{x_0}$ is in $\mathcal{O}(e^{\frac{t}{2}})$ as $t \to +\infty$, then for all $\varepsilon > 0$ there exists $M > 0$ such that $\frac{t^{x_0}}{e^{\frac{t}{2}}} < \varepsilon$ for all $t \geq M$. So $t^{x_0} e^{-t} < \varepsilon e^{-\frac{t}{2}}$ for all $t \geq M$. Choose $\varepsilon = 1$ and let $M_1$ be the obtained value of $M$. Hence, $$\int_a^{+\infty} \varphi \\\leq \int_0^{+\infty} \varphi \\= \int_0^{M_1} \varphi + \int_{M_1}^{+\infty} \varphi \\< \int_0^{M_1} \varphi + \int_{M_1}^{+\infty} e^{-\frac{t}{2}} \mathrm{d}t \\= \int_0^{M_1} \varphi + [-2e^{-\frac{t}{2}}]_{M_1}^{+\infty}\\=\int_0^{M_1} \varphi + 2e^{-\frac{M_1}{2}} \\< +\infty$$ Now, there exists a neighborhood $V$ of $x_0$ such that for all $x \in V, t \in J, |f(x, t)| \leq \varphi(t)$. Let us prove this explicitly. Hmm... maybe take $V = [x_0 - \frac{x_0}{C}, x_0 + \frac{x_0}{C}]$, then, for $t \geq 1$, $|f(x, t)| = t^{x-1}e^{-t} \leq t^{x_0 + \frac{x_0}{C}-1} e^{-t} = f(x_0, t) t^{\frac{x_0}{C}}$... No, no. If $x_0$ is very big then we can't bound that... Let us consider two cases of $x_0$.
  
  Case 1. $0 < x_0 < 1$, then $V = [x_0 - \frac{x_0}{2}, x_0 + \frac{x_0}{2}]$ should work? For $t \geq 1$, $|f(x, t)| = t^{x - 1} e^{-t} \leq t^{x_0 + \frac{x_0}{2}-1} e^{-t} = \varphi(t) t^{\frac{x_0}{2}-1} \leq \varphi(t)$. For $0 \leq t < 1, |f(x, t)| = t^{x-1}e^{-t} \leq t^{x_0 - \frac{x_0}{2} - 1} e^{-t} = \varphi(t) t^{-\frac{x_0}{2}-1} \color{red}{\leq \varphi(t)}$ (this is wrong).
  
  Case 2. $x_0 \geq 1$, then let $V = [x_0 - \frac{1}{2}, x_0 + \frac{1}{2}]$, so if $t \geq 1$, $|f(x, t)| \leq t^{x_0 + \frac{1}{2} - 1} e^{-t} = \varphi(x_0) t^{-\frac{1}{2}} \leq \varphi(x_0)$. Finally, if $0 \leq t \leq 1$, then $|f(x, t)| \leq t^{x_0-\frac{1}{2}-1} e^{-t} = \varphi(x_0) t^{-\frac{3}{2}} \color{red}{\leq \varphi(x_0)}$ (also wrong).

Ok. Let me restart this from square zero, all over again, for $\varphi$. Divide into three cases. Case 1. $0 < x_0 < 1$. Let $\varphi(t) = t^{x_0} e^{-t}$. Ah shit this is fucking insane. Perturbation doesn't work.

Let's do it the ugly way. No cases on $x_0$ now. Define

$$\varphi(t) = \begin{cases} 1 & \text{ if } t \leq 1 \\ t^{x_0} e^{-t} & \text { otherwise} \end{cases}$$

Clearly, $|f(x, t)| = t^{x-1} e^{-t} \leq 1$ for all $0 \leq t \leq 1$ (no matter the neighborhood of $x$; we don't even need to care). Now, for $t > 1$, $|f(x, t)| = t^{x-1} e^{-t} \leq t^{x_0} e^{-t} = \varphi(t)$ for all $x-1 \leq x_0$, so we can easily assign $V = [\frac{x_0}{2}, x_0+1]$ as a neighborhood which gives $|f(x, t)| \leq \varphi(t)$. And we've proved that the old $\varphi$ is integrable on any interval (not necessary compact) of $[0, +\infty[$, now the new $\varphi$ only differs from the old one in $[0, 1]$, so the new one's integral is bounded by the old one plus one, hence integrable.

By 4.1.3., we have $\Gamma(x) = g(x) = \int_0^{+\infty} t^{x-1} e^{-t} \mathrm{d}t$ well-defined, and continuous at $x_0$. Since we assumed $x_0$ to be any positive real, $g$ is continuous everywhere in $]0, +\infty[$. Therefore, $\Gamma \in \mathcal{C}^0(\mathbb{R}_+^*)$. $\square$

2. (a) Hmm... I think we can do integration by parts, right? Let $f \colon t \mapsto t^x$ and let $g$ be such that $g' \colon t \mapsto e^{-t}$, then $\int_a^b f g' = [fg]_a^b - \int_a^b gf'$. (Note that $f$ and $g$ are defined on $]0, +\infty[$). Clearly, $g$ is $t \mapsto -e^{-t}$. So $$\int_0^{+\infty} t^x e^{-t} \mathrm{d}t = [-t^x e^{-t}]_0^{+\infty} - \int_0^{+\infty} -e^{-t} x t^{x-1} \mathrm{d}t$$ for $x > 0$.

Now, let us calculate $[-t^x e^{-t}]_0^y$ for $y \in ]0,+\infty[$.

$$[-t^xe^{-t}]_0^y = -y^x e^{-y} - (-0^xe^{-0}) = -y^x e^{-y}$$

Clearly, as $y \to +\infty$, this quantity tends to zero (because $(y \mapsto y^x) \in \mathcal{O}(y \mapsto e^{y})$. So, $\lim_{y \to +\infty} [-t^x e^{-t}]_0^y = 0$.

Coming back to the integration by parts, we have

$$\Gamma(x+1) = \int_0^{+\infty} t^x e^{-t} \mathrm{d}t = -\int_0^{+\infty} -e^{-t}xt^{x-1}\mathrm{d}t = x\int_0^{+\infty} e^{-t}t^{x-1}\mathrm{d}t = x \Gamma(x)$$ for any $x > 0$. This completes the proof. $\square$

2. (b) This is very easy. Compute $\Gamma(1) = 1$, then let us do an induction that $\Gamma(n) = (n-1)!$. Suppose $\Gamma(n) = (n-1)!$, then let us prove that $\Gamma(n+1) = n!$. By 2. (a), we have $\Gamma(n+1) = n \Gamma(n) = n (n-1)! = n!$. This completes the induction. $\square$

Afterthought

It took me way too long to finish the 1. and I don't see how one can do it faster (with approximately the same level of rigor).

Ok, so now I'm ready to watch this. Hmmm too lazy to watch everything lol (skipped the logarithm part), but it's a nice video! What if I try to reconstruct a function $\widetilde{\Gamma} \in \mathcal{C}^0(\mathbb{R}_+^*)$ by myself, such that it satisfies the property of the gamma function... Ah no, well, the function is unique anyway, so it's left to prove that it's unique.

Here I'll give myself two bonus exercises:

Bonus 1 (Bohr–Mollerup theorem)

Prove that the function $f \colon \mathbb{R_+^*} \to \mathbb{R}$ defined by the following properties:

• $f(0) = 1$
• $f(x) = x f(x-1)$
• $f$ is logarithmically convex
• $f(x) > 0$ for all $x > 0$

is unique. Note that I thought it would be just a simple exercise, but a quick Wikipedia search shows that this is a full textbook theorem! Maybe too hard for me lol.

Bonus 2

The map $$x \mapsto \lim_{N \to \infty} N^x \prod_{k=1}^N \frac{k}{x+k}$$ defined on $]0, +\infty[$ is exactly the same as the map $$x \mapsto \int_0^{+\infty} t^{x-1} e^{-t} \mathrm{d}t$$ on $]0, +\infty[$.

Comment. Well, I think it's enough to prove that both satisfy the characterization of the gamma function, and by applying Bohr-Mollerup theorem, since the function is unique, they must be the same.

Well, for now I'm too lazy to do this, but hopefully I'll come back to this one later.

Bonus 3

(Really bonus in the sense that I don't really think I can make it but well, why not state it here) Find the analytic continuation of the gamma function defined on $]0, +\infty[$ (to the whole domain $\mathbb{C} \setminus \mathbb{Z}_-$).