8
$\begingroup$

I would expect the $q$-Gamma function to have the property which would be the $q$-analog of the Euler reflection formula from my question title.

More concretely: $\Gamma(z)$ has simple poles at nonpositive integers, $\Gamma(1-z)$ at positive integers, and together they fold nicely to give (up to scalar multiple) $\csc(\pi z)$, the simplest periodic function with simple poles at all integers.

An analog of this I would expect would have in place of $\csc$ some form of theta function with poles at all integer powers of $q$, the $q$-analog of $1-z$ would be $q/z$, the $q$-analog of $\Gamma$ would have poles only at nonpositive powers of $q$, and then the same function of $q/z$ would have poles at positive powers of $q$, so their product would give that theta function.

Is there such an identity? I could not quite find it.

Improve this question
$\endgroup$
8
  • 6
    $\begingroup$ This article discusses $q$-analogues of the Euler reflection formula and the Euler gamma integral. $\endgroup$
    – Dietrich Burde
    Commented Aug 15, 2015 at 18:46
  • $\begingroup$ @DietrichBurde Thanks a lot, I would never find it myself! Would you make it an answer? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 15, 2015 at 21:30
  • 4
    $\begingroup$ This is a very different use of the letter "q", but the "q-aspect" version of the identity $\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin \pi z}$ in analytic number theory is the identity $\tau(\chi) \tau(\overline{\chi}) = q \chi(-1)$ for Gauss sums $\tau(\chi) = \sum_{n \in {\bf Z}/q{\bf Z}} \chi(n) e(n/q)$ of Dirichlet characters of conductor $q$. (Gauss sums show up in the functional equation for Dirichlet L-functions in much the same way that Gamma factors show up for the Riemann zeta function.) $\endgroup$
    – Terry Tao
    Commented Aug 15, 2015 at 21:51
  • $\begingroup$ @DietrichBurde After some confusion, I've checked the paper and it seems that they get the same which made me stuck: denoting $\Gamma_q(x)\Gamma_q(1-x)=f(q^x)$, one obtains$$f(qz)=-zf(z),$$whereas I would expect to obtain a function with $f(qz)=f(z)$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 16, 2015 at 21:16
  • 1
    $\begingroup$ Also, the following article discusses $q$-analogues of the Euler reflection formula. icms.kaist.ac.kr/mathnet/thesis_file/BKMS-51-4-1155-1161.pdf $\endgroup$
    – Martin Bokner
    Commented Aug 25, 2017 at 23:16

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .