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I found the following curious identity for the gamma function on Wikipedia for which I'd like to know some references (proof, history, etc).

The identity is as follows: $$\Gamma(t) = x^t \sum_{n=0}^\infty \frac{L_n^{(t)}(x)}{t+n}$$ Here $L_n^{(t)}(x)$ are the generalized Laguerre polynomials, and the expression seems to be valid for $\mathrm{Re}(t) < \frac12$. (That's all which is specified on Wiki).

EDIT: There are now two very helpful answer, but before accepting one of them I'd like to broaden the question a little bit further in asking for a representation theoretic interpretation of the gamma function identity.

Thanks a lot in advance!

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    $\begingroup$ Your edit sufficiently changes your question that it is no longer the same question. I recommend that you create a new question. $\endgroup$ – Eric Towers Jul 7 '14 at 15:13
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Generalized Laguerre polynomials have the generating function $$\sum_{n=0}^{\infty}\xi^n L_n^{(t)}(x)=(1-\xi)^{-t-1}e^{-\frac{\xi x}{1-\xi}}.$$ Multiplying this identity by $\xi^{t-1}$ and integrating w.r.t. $\xi$ from $0$ to $1$, one finds \begin{align}\sum_{n=0}^{\infty}\frac{L_n^{(t)}(x)}{t+n}&=\int_0^1\left(\frac{\xi}{1-\xi}\right)^{t-1} e^{-\frac{\xi x}{1-\xi}}\frac{d\xi}{(1-\xi)^2} =x^{-t}\int_0^{\infty}s^{t-1}e^{-s}ds=x^{-t}\Gamma(t), \end{align} where the second step is achieved by the change of variables $s=\frac{\xi x}{1-\xi}$. $\blacksquare$

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  • $\begingroup$ thank you very much for your answer! do you know of a representation theoretic interpretation of this description of the gamma function? $\endgroup$ – user5831 Jul 7 '14 at 9:28
  • $\begingroup$ You are very welcome! and sorry, I cannot suggest any representation-theoretic interpretation. $\endgroup$ – Start wearing purple Jul 7 '14 at 21:27
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A reference: Chaudhry, M.A. et al. Asymptotics and closed form of a generalized incomplete gamma function.

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  • $\begingroup$ I don't really support this rule , but someone on stackexchange once told me to not use links as an answer, but use link as a side to the answer, not as the main subject of the answer. $\endgroup$ – Joao Jul 7 '14 at 7:19
  • $\begingroup$ thank you very much for your answer! $\endgroup$ – user5831 Jul 7 '14 at 9:28
  • $\begingroup$ @Joao: At the time of this answer, there was an adequate demonstration, but zero references, as requested in the OP. $\endgroup$ – Eric Towers Jul 7 '14 at 14:52
  • $\begingroup$ @EricTowers Ok thanks for clearing everthing up. $\endgroup$ – Joao Jul 9 '14 at 2:54

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