Any advice you care to offer concerning the following exercise would be greatly appreciated.

Let $\phi (a) = \Gamma (a) \Gamma (1-a) \sin(\pi a)$.
Prove that $\phi$ extends to an infinitely differentiate function on the real line.

What I know and what I've tried: I have learned that $\Gamma$ extends to a smooth function on the complement of the non positive integers. In addition if we take for granted that $\phi (a) = \phi (a+1)$ and that $\phi (0) =\pi$ (here I really mean $\phi$'s extension) then all I need to show is that $\phi$ is smooth at $0$. I have tried various obvious strategies like computing the difference quotient and trying to compute the limit of $d^k\phi/da^k$ as $a$ approaches $0$ but I haven't been successful. I would be grateful for any help you care to offer. Thanks for your time.

  • $\begingroup$ Do you know that $\Gamma$ has a simple pole at each nonpositive integer? And that the product of a meromorphic function with a pole of order $k$ and one with a zero of multiplicity $k$ at the same point has a removable singularity at that point? $\endgroup$ – Robert Israel Jul 27 '12 at 7:13
  • $\begingroup$ Actually I do! And thanks for your help. However the book im reading (lebesgue integration on euclidean space by frank jones) does not really get into complex analysis and so I feel I should be able to do this without bringing in facts from complex analysis. No mention has even been made of real analytic functions, for instance. $\endgroup$ – Phil Grantham Jul 27 '12 at 7:26

Here's a proof without using complex analysis, just the functional equation $\Gamma(x+1) = x \Gamma(x)$ and the fact that $\Gamma(x)$ is smooth for $x > 0$. From the functional equation, $$ \sin(\pi x) \Gamma(x) \Gamma(1-x) = \Gamma(x+1) \Gamma(1-x) \dfrac{\sin(\pi x)}{x}$$ Since $\sin(\pi x)/x$ extends (by giving it the value $\pi$ at $x=0$) to a smooth function (given by a convergent Maclaurin series) on $\mathbb R$, and $\Gamma(x+1)$ and $\Gamma(1-x)$ are smooth on $(-1,1)$, we get that $\phi$ is smooth on $(-1,1)$ with $\phi(0) = \pi$. From the functional equation for $\Gamma$ and $\sin(\pi (x+1)) = -\sin(\pi x)$ we get $\phi(x+1) = \phi(x)$ for non-integer $x$, so $\phi$ (redefined to be $\pi$ at the integers) is smooth on all of $\mathbb R$.
By the way, it is actually identically equal to $\pi$ by Euler's reflection formula.

  • $\begingroup$ Fantastic! Thanks so much, Robert. $\endgroup$ – Phil Grantham Jul 27 '12 at 16:54

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