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

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?

• The Gamma function of x is not continuous (but is defined) for negative x – Elements in Space Jan 28 '13 at 11:23
• The domain of $\Gamma$ (as written above) is $\mathbb R_+$. I'm asking if it is continuous in its domain. – Dubious Jan 28 '13 at 11:26
• Yes. I think so. – Bombyx mori Jan 28 '13 at 11:27
• If there exists a function $f$ such that $\Gamma(x)\leq h$ forall $x>0$ such that $\int_\Omega h \; d\mu < \infty$ then it is. – UnadulteratedImagination Jan 28 '13 at 11:34
• Can you find the measurable $g$ function that $t^{x-1}e^{-t}\leq g$? – Felipe Feb 7 '14 at 10:42

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 $\,[0,b]\,$ : $$\left|\Gamma(x)-\Gamma(x_0)\right|=\left|\lim_{b\to\infty}\int\limits_0^b \left(t^{x-1}-t^{x_0-1}\right)e^{-t}\,dt\right|\leq$$

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

• Very useful! Anyway, is my approach right? – Dubious Jan 28 '13 at 12:27
• @Galoisfan, yes it is...but if you're using the DCT then I think it'd be better if you specifically show the integrable function $\,g(x)\,$ s.t. $\,\left|t^{x_0-1}e^{-t}\right|\leq |g(x)|\,$ . Not that this is hard to do in this case. – DonAntonio Jan 28 '13 at 12:31
• I have some problems to show the integrable function $g$ – Dubious Jan 28 '13 at 12:35
• @DonAntonio can you give the function $g$? – UnadulteratedImagination Jan 28 '13 at 13:31
• I don't think that the "basic argument proof" is correct. Here $\delta$ depends on $\epsilon$ and $b$. We later allow $b$ to go to infinity. – user60898 Feb 4 '13 at 6:10

Since we want to consider continuity at a single fixed point $x_0$, we may assume that $x_0\in(a,b)$ for certain $0<a<1<b$.

Assume $x\in(a,b)$. Using the Mean Value Theorem, there exists $\xi$ between $x$ and $x_0$ such that $$t^{x-1}-t^{x_0-1}=(\log t)\,t^{\xi-1}\,(x-x_0).$$ So $$|t^{x-1}-t^{x_0-1}|\leq|(\log t)\,(t^{1-a}+t^{b-1})|\,|x-x_0|.$$ The estimate for $t^{\xi-1}$ is obtained by considering $t<1$ and $t>1$ respectively.

Now \begin{align} \left|\Gamma(x)-\Gamma(x_0)\right| &\leq\int_0^\infty |t^{x-1}-t^{x_0-1}|\,e^{-t}\,dt\\ \ \\ &\leq|x-x_0|\,\int_0^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt\tag{1}\\ \ \\ &\leq\,|x-x_0|\,\left(\frac1{(2-a)^2}+\frac1{b^2}+\frac{2c}{\sqrt e}\right) \end{align}

(see below), and so $\Gamma$ is continuous at $x_0$. Note that $(1)$ is a Lipschitz condition, but it is local: in the sense that the constant depends on $a$ and $b$.

Note that $$\tag{2} \left|\int_0^1 t^r\,\log t\,dt\right|<\infty \iff r>-1.$$

Since $0<a<1$ and $b>0$, we have $1-a>0$ and $b-1>-1$. By $(2)$, then $$\tag{3} \int_0^1|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt\leq-\int_0^1\log t\,(t^{1-a}+t^{b-1})\,dt=\frac1{(2-a)^2}+\frac1{b^2}$$ Let $$c=\sup\{|\log t\,(t^{1-a}+t^{b-1})e^{-t/2}|:\ t\in(0,\infty)\}$$ (the sup exists because the function is continuous and bounded, since both the limits at $0$ and $\infty$ exist). Then \begin{align}\tag{4} \int_1^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt &=\int_1^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t/2}\,e^{-t/2}\,dt\\ \ \\ &\leq c\,\int_1^\infty e^{-t/2}\,dt=\frac{2c}{\sqrt e}. \end{align} Using $(3)$ and $(4)$, we get $$\int_0^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt\leq\frac1{(2-a)^2}+\frac1{b^2}+\frac{2c}{\sqrt e}.$$