My Math Forum Improper Gamma Integral

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 August 5th, 2012, 10:49 PM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Improper Gamma Integral Okay, I know that $\int_{1}^{\infty} \Gamma(x) dx$ and $\sum_{n=1}^{\infty} \Gamma(n)$ both diverges. But what about this limit: $\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \Gamma(k) - \int_{1}^{n} \Gamma(x) dx$ ? Is this also divergent? Or is it convergent? Even if I had a graph of $F(x)= \sum_{k=1}^{x} \Gamma(k) - \int_{1}^{x} \Gamma(t) dt$, It would mean a lot to me. Can anybody give me a graph?(Oh, I know that's not possible ) Any help will be greatly appreciated. Balarka .
 August 6th, 2012, 12:34 AM #2 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: Improper Gamma Integral Hi ! I think that F(n) is not converging because, for n tending to infinity, the Sum is equivalent to Gamma(n) and the Integral is equivalent to Gamma(n)/ln(n))
August 6th, 2012, 12:39 AM   #3
Math Team

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Re: Improper Gamma Integral

Quote:
 Originally Posted by JJacquelin for n tending to infinity, the Sum is equivalent to Gamma(n) and the Integral is equivalent to Gamma(n)/ln(n))
Wow! Can you prove that (I mean, can you show me the proof?)?

August 6th, 2012, 06:51 AM   #4
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Re: Improper Gamma Integral

Quote:
 Originally Posted by mathbalarka Wow! Can you prove that (I mean, can you show me the proof?)?
The proof for the integral requires somme background about polygamma functions (especially asymptotic expansion)
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August 6th, 2012, 06:53 AM   #5
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Re: Improper Gamma Integral

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 August 6th, 2012, 10:03 AM #6 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Improper Gamma Integral Ok. Lots of Thanks, Balarka .
 August 6th, 2012, 10:57 AM #7 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Improper Gamma Integral I found another interesting thing about $F(x)= \int \Gamma(x) dx$ : It is well known that: $\int \frac{\pi}{\sin(\pi x)} dx= \ln$$\tan\(\frac{\pi x}{2}$$\) \,\,\,\,\,\,\, (1)$ Now, the euler's reflection formula for the gamma function states that: $\frac{\pi}{\sin$$\pi x$$}= \Gamma(x)\Gamma(1-x)$ Substituting this into (1) gives $\ln$$\tan\(\frac{\pi x}{2}$$\)= \int \Gamma(x)\Gamma(1-x) dx = \Gamma(x) F(1-x) - \int F(1-x) \Gamma#39;(x) dx$ From this, $F(1-x)= \frac{1}{\Gamma(x)} $\ln$$\tan\(\frac{\pi x}{2}$$\) + \int F(1-x) \Gamma(x) \psi_0(x) dx$$ So, F(1-x) diverges for all $x \in \mathbb{Z} \cup [0]$ I am currently very sleepy now so please forgive me if I am wrong Balarka .

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