Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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? – Robert Israel Jul 27 '12 at 7:13
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. – Phil Grantham Jul 27 '12 at 7:26
up vote 1 down vote accepted

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.

share|cite|improve this answer
Fantastic! Thanks so much, Robert. – Phil Grantham Jul 27 '12 at 16:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.