I am currently reading some material that makes heavy usage of Hypergeometric functions, and there is one particular point about applying Stirling's approximation to various terms consisting of Gamma-Functions that is not very clear to me.

We have the classical Stirling's approximation formula for the Gamma-Function in the form: $$ \Gamma(z)=\sqrt{2\pi}e^{-z}z^{z-1/2}\left(1+O\left(\frac{1}{|z|}\right)\right) $$

for $|\arg(z)|<\pi$ as $|z|\to\infty$. In absolute value: $$ |\Gamma(z)|=\sqrt{2\pi}e^{-\Re(z)}|z|^{\Re(z)-1/2}e^{-\Im(z)\arg(z)}\left(1+O\left(\frac{1}{|z|}\right)\right) $$

Now, there is also the shifted Stirling's approximation for the Gamma-Function due to C. Rowe, if I am not mistaken, that says: $$ \Gamma(z+a)=\sqrt{2\pi}e^{-z}z^{z+a-1/2}\left(1+O\left(\frac{1}{|z|}\right)\right) $$

uniformly for $|\arg(z)|\leq\pi-\varepsilon$, $a$ in a compact subset of $\mathbb{C}$ and some suitable fixed $\varepsilon>0$, as $|z|\to\infty$. In absolute value: $$ |\Gamma(z+a)|=\sqrt{2\pi}e^{-\Re(z)}|z|^{\Re(z+a)-1/2}e^{-\Im(z+a)\arg(z)}\left(1+O\left(\frac{1}{|z|}\right)\right) $$

My first question refers to terms of the type $$ \Gamma(az+b)\Gamma(cz+d) $$ for some complex numbers $a,b,c$ and $d$.

(Q1) Using the classical Stirling's approximation formula (i.e. not the shifted one), how can one obtain a meaningful aggregated asymptotics for the above expression?

I am asking this because there are several places that apply the non-shifted version of Stirling's formula to shifted Gamma factors without mentioning any details, which leaves the impression that this is a fairly standard or even trivial argument. Unfortunately, at this point I am unable to see its triviality. What bothers me here is the term $$ (az+b)^{az+b-1/2}(cz+d)^{cz+d-1/2} $$ as well as $\arg(az+b)$ and $\arg(cz+d)$, since the shifts by $c$ and $d$ break any easy manipulations.

I am naturally assuming that I am missing something very obvious here (as usual).

I have intentionally not specified anything about the parameters $a,b,c$ and $d$ because my question rather aims at the principle of applying the non-shifted Stirling's approximation to Gamma terms like the above one.

(Q2) Are there any other versions or forms of the Stirling's approximation for $\Gamma(z)$ that could be particularly useful for computing such kinds of asymptotics?

I will be extremely thankful if someone could give some insight in (principle of) the application of Stirling's approximatioin formula(s) to terms composed of Gamma factors!

  • $\begingroup$ Check out the book of Paris and Kaminsky $\endgroup$ – Sasha Sep 9 '11 at 15:29
  • $\begingroup$ The Digital Library of Mathematical Functions entry on asymptotics of the gamma function may also be useful. $\endgroup$ – Mike Spivey Sep 9 '11 at 17:39
  • $\begingroup$ @Mike Spivey: thank you for pointing me to this link. It seems to be a very useful reference for various things related to special functions and asymptotics in complex analysis. $\endgroup$ – M.G. Sep 12 '11 at 21:59
  • $\begingroup$ @ex-falso-quodlibet: You're welcome. I use the DLMF all the time and am happy to introduce it to others! $\endgroup$ – Mike Spivey Sep 12 '11 at 22:00

Let me take a stab at (Q1). I find it better working for $\log\Gamma(z)$ rather than $\Gamma(z)$, since $\Gamma(z) = \exp( \log\Gamma(z))$.

Stirling formula reads as follows: $$ \log\Gamma(z) \sim (z-\frac{1}{2}) \log(z) - z + \frac{1}{2} \log(2 \pi) + o(1) $$ for $\vert z \vert \to \infty$ and $ \vert \arg(z) \vert < \pi - \epsilon$.

Notice that shifted formula is a simple consequence of the above:

$$ \begin{eqnarray} \log\Gamma(z+a) &\sim& ( z+a -\frac{1}{2}) \log(z+a) - z - a + \frac{1}{2} \log(2 \pi) + o(1) \\ &\sim& ( z+a -\frac{1}{2}) \log(z) + ( z+a -\frac{1}{2}) \log(1+\frac{a}{z}) - z - a + \frac{1}{2} \log(2 \pi) + o(1) \\ &\sim& ( z+a -\frac{1}{2}) \log(z) + z \log(1+\frac{a}{z}) - z - a + \frac{1}{2} \log(2 \pi) + o(1) \\ &\sim& ( z+a -\frac{1}{2}) \log(z) - z + \frac{1}{2} \log(2 \pi) + o(1) \end{eqnarray} $$

Now $$ \begin{eqnarray} \log\Gamma(a z + b) + \log\Gamma(c z + d) & \sim & ( a z +b -\frac{1}{2}) \log(a z) - a z + \frac{1}{2} \log(2 \pi) + \\ & & ( c z +d -\frac{1}{2}) \log(c z) - c z + \frac{1}{2} \log(2 \pi) + o(1) \end{eqnarray} $$

Then it is a matter of recombining terms as $ ( \mathcal{A} + \mathcal{B} z) + (\kappa z + \rho -\frac{1}{2}) \log (\kappa z) - \kappa z + \frac{1}{2} \log(2 \pi) + o(1) $.

See the book of Paris and Kaminski to fill in the details.

  • $\begingroup$ Thank you for this clarification, it is much appreciated! The $\log$-notation certainly has its advantages, especially when having to keep track of various coefficients. Also, you appear to prefer equivalence notation instead of $=$-notation, my guess is because you'd like to avoid abuse of notation and because this way you also don't have to write the bounded $arg$-term anyway. Your comment about recombining terms suggests about an useful application of Hadamard's Factorization Theorem (or more generally, Weierstrass Factorization Theorem for entire functions of order $>1$) :-) $\endgroup$ – M.G. Sep 12 '11 at 21:49

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