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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?

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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. –  fair-coin tossing 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 at 10:42

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

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

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Very useful! Anyway, is my approach right? –  fair-coin tossing Jan 28 '13 at 12:27
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@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 $g$ –  fair-coin tossing Jan 28 '13 at 12:35
    
@DonAntonio can you give the function $g$? –  UnadulteratedImagination Jan 28 '13 at 13:31
    
I see it as follows: (1) For $\,t>1\,$: $$\exists M\in\Bbb N\,\,\;\text{s.t.}\;\;\,\,t>M\Longrightarrow e^{-t/2}t^{x-1}\leq 1\,\,,\,\,\text{since} e^{-t/2}t^{x-1}\xrightarrow[t\to\infty]{}0$$ and from here $\,\int\limits_M^\infty e^{-t}t^{x-1}\,dt\leq\int\limits_M^\infty e^{-t/2}dt<\infty\,$ For $\,0<t\leq 1\,$ we get $\,e^{-t}t^{x-1}\leq t^{x-1}\,$ and $\,\int\limits_0^Mt^{x-1}dt\,$ converges for $\,x-1>-1\Longleftrightarrow x>0$ –  DonAntonio Jan 28 '13 at 14:06

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