Continuity of the (real) $\Gamma$ function.

Question

Consider the real valued function $$\Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t}dt$$ where the above integral means the Lebesgue integral with the Lebesgue measure in $\mathbb R$. The domain of the function is $\{x\in\mathbb R\,:\, x>0\}$, and now I'm trying to study the continuity. The function $$t^{x-1}e^{-t}$$ is positive and bounded if $x\in[a,b]$, for $0<a<b$, so using the dominated convergence theorem in $[a,b]$, I have:

$$\lim_{x\to x_0}\Gamma(x)=\lim_{x\to x_0}\int_0^{\infty}t^{x-1}e^{-t}dt=\int_0^{\infty}\lim_{x\to x_0}t^{x-1}e^{-t}dt=\Gamma(x_0)$$

Reassuming $\Gamma$ is continuous in every interval $[a,b]$; so can I conclude that $\Gamma$ is continuous on all its domain?

Answer 1

You could also try the basic approach by definition.

For any $\,b>0\,\,\,,\,\,\epsilon>0\,$ choose $\,\delta>0\,$ so that $\,|x-x_0|<\delta\Longrightarrow \left|t^{x-1}-t^{x_0-1}\right|<\epsilon\,$ in $\,[0,b]\,$ : $$\left|\Gamma(x)-\Gamma(x_0)\right|=\left|\lim_{b\to\infty}\int\limits_0^b \left(t^{x-1}-t^{x_0-1}\right)e^{-t}\,dt\right|\leq$$

$$\leq\lim_{b\to\infty}\int\limits_0^b\left|t^{x-1}-t^{x_0-1}\right|e^{-t}\,dt<\epsilon\lim_{b\to\infty}\int\limits_0^b e^{-t}\,dt=\epsilon$$

Answer 2

Note that it suffices to show the continuity of $\Gamma(x)$ for all $x>1$ because we can use the known property $\Gamma(x+1)=x\Gamma(x)$ to conclude continuity of those $x$ with $0<x\leq1$. I will break down the proof into two steps:

$1.)$ I show that $\Gamma_k(x):=\int\limits_0^k t^{x-1}e^{-t}dt$ is continuous at point $a>1$.

$2.)$ I show that $\Gamma_k(x)\to\Gamma(x)$ uniformly on a neighbourhood of $a$ so that the limit function inherits the continuity on this neighbourhood.

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Proof:

$1.)$ The integrand $K:(1,\infty)\times(0,k)$, where $K(x,t)=t^{x-1}e^{-t}$ is continuous in each point. For a given $\epsilon>0$ and a point $(a,t')$, there exists a $\delta_{t'}$ such that $$\Big\Vert{x\choose t}-{a\choose t'}\Big\Vert<\delta_{t'}\implies | K(x,t)-K(a,t'|<\frac{\epsilon}{2k}.$$ We define a neigbourhood of $t'$ by $$U_{\delta_t'}(t'):=\{t\in(0,k)\mid|t-t'|<\delta_{t'}\}.$$

The union $\bigcup\limits_{t\in(0,k)}U_{\delta_t}(t)$ covers the interval $(0,k)$ and due to compactness there exists a finite cover $\bigcup\limits_{i=1}^nU_{\delta_{t_i}}(t_i)$ of $(0,k)$. Now we set $\delta:=\min\{\delta_1\cdots, \delta_n\}$. Hence, for all $x>1$ with $|x-a|<\delta$ it follows: $$ |\Gamma_k(x)-\Gamma_k(a)|=\left|\int\limits_0^k K(x,t)dt-\int\limits_0^k K(a,t)dt\right|\leq\int\limits_0^k \left|K(x,t)-K(a,t)\right|dt\\ \leq\int\limits_0^k \left|K(x,t)-K(a,t_i)|+|K(a,t_i)-K(a,t)\right|dt<\int\limits_0^k\frac{\epsilon}{k} dt<\epsilon. $$ Here we used that $K$ is continuous in each $(a,t_i)$ and that $t$ lies witihin a $U_{\delta_{t_i}}(t_i)$ so that $\Vert{x\choose t}-{a\choose t_i}\Vert<\delta_i$ and that $\Vert{a\choose t_i}-{a\choose t_i}\Vert<\delta_i$. Hence, $\Gamma_k(x)$ is continuous.

$2.)$ Now let's consider only those $x$ with $1<x<a+1$. Then, we see that $$ \left|\Gamma_k(x)-\Gamma(x)\right|\leq \int\limits_k^{\infty}\left| t^{x-1}e^{-t}dt\right|dt\leq\int\limits_k^{\infty}\left| t^{a}e^{-t}dt\right|dt=\Gamma_k(a+1)-\Gamma(a+1)<\epsilon $$ as $\lim\limits_{k\to\infty}\Gamma_k(a)=\Gamma(a)$. Hence, we have uniform convergence on the neighbourhood of $a$ and $\Gamma(x)$ inherits the continuity at point $a$.