# Limit of the gamma function

I have to prove that $$\lim_{x\to 0^{+}}\Gamma(x)=\lim_{x\to+\infty}\Gamma(x)=+\infty$$ where $\Gamma$ is the gamma function.

As for the $x\to\infty$ I would like to show that $\Gamma$ is increasing on interval $(a,\infty)$ for some $a>0$. Increasing function has a limit and $$\lim_{x\to+\infty}\Gamma(x)=\lim_{n\to+\infty}\Gamma(n)=\lim_{n\to+\infty}(n-1)!=\infty$$

But I don't know how to prove the function is increasing.

As for the $x\to 0$ my idea is

$$\lim_{x\to 0^{+}}\Gamma(x)=\lim_{x\to 0^{+}}\int_{0}^{\infty}e^{-t}t^{x-1}dt=\int_{0}^{\infty}\lim_{x\to 0^{+}}e^{-t}t^{x-1}dt=$$$$=\int_0^{\infty}e^{-t}t^{-1}dt\ge\int_0^1e^{-t}t^{-1}dt\ge e^{-1}\int_0^1t^{-1}dt=\infty$$

But I don't know whether I can change the order of limit and integral (second equality). Maybe if I showed that $\Gamma$ is decreasing on interval $(0,a)$ for some $a>0$ I could replace the limit $x\to 0$ with $n\to\infty$ and use some Lebesgue theorem.

• As for $x\to 0^+$, you can interchange limit and integral, but I don't know if you already have the theorems to do that (monotone convergence theorem for example can be used). It's perhaps easier to first cut down the domain of the integral to $(0,1]$, then use $e^{-t} \geqslant e^{-1}$, and explicitly evaluate $\int_0^1 t^{x-1}\,dt$. Commented Feb 15, 2016 at 11:49
• For $x\to \infty$, you can do with less than monotonicity. For $x\in [1,2]$, we have $\Gamma(x) \geqslant c:= \int_0^1 te^{-t}\,dt + \int_1^\infty e^{-t}\,dt > 0$. Then for $x\in [n,n+1)$ by the functional equation we have $\Gamma(x) \geqslant c\cdot (n-1)!$. Commented Feb 15, 2016 at 13:01

The $\Gamma$ function is log-convex, hence convex, by the Bohr-Mollerup theorem (we may take that theorem as the definition of the $\Gamma$ function, too, since it gives the Euler product as a by-product).

The $\Gamma$ function is positive on $(0,1)$ as the integral of a positive function, hence the functional relation $\Gamma(x+1)=x\cdot\Gamma(x)$ gives that $\Gamma(x)>0$ for any $x>0$. $\Gamma(x)$ is increasing over $(2,+\infty)$ because:

$$\frac{d}{dx}\log\Gamma(x) = \psi(x) = -\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{x-1+n}\right)$$ and the RHS, given $x>2$, is positive: $$-\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{x-1+n}\right)\geq -\frac{2}{3}+\sum_{n\geq 1}\frac{1}{n(n+1)} = \frac{1}{3}.$$ Moreover, since $\Gamma(1)=1$ and we have continuity in a neighbourhood of $1$ (as a consequence of convexity),

$$\lim_{x\to 0^+}\Gamma(x) = \lim_{x\to 0^+}\frac{\Gamma(x+1)}{x} = \lim_{x\to 0^+}\frac{1}{x} = +\infty.$$

You can prove using integration by parts that for all $$x>0$$ $$\Gamma(x+1)=x\Gamma(x)$$ since $$\Gamma$$ is continuous, then $$\lim_{x\to 0^+}x\Gamma(x)=\lim_{x\to 0^+}\Gamma(x+1)=\Gamma(1)=1.$$ Thus $${\displaystyle \Gamma(x)\underset{0}{\sim}\frac{1}{x}}$$ which means $$\Gamma(x)$$ and $$\displaystyle\frac{1}{x}$$ have the same limit at $$0$$.

You can also interchange limit and integral using monotone convergence theorem as suggested by
Daniel Fischer, or use Fatou's lemma.

• I'm 3 years late, but I'm interested in this since I have a similar exercise as homework. How would he exchange the limit and integral using monotone convergence when it's a theorem for sequences? Or can it be generalized over to functions as well? Commented Apr 21 at 17:10