Can somebody help me out with the intermediate details of eqn. (2.5) in this paper?

Generalized gravitational entropy. Aitor Lewkowycz and Juan Maldacena. arXiv:1304.4926.

Is the Euler $\psi$ function appearing in the above equation the same as the digamma function?

I can't seem to figure out the relation between this Euler $\psi$ function and the hypergeometric function ${}_2F_1(a,b;c;z)$. It must be related to the derivative of the hypergeometric function but I can't figure out exactly how.

  • 1
    $\begingroup$ It's definitely the digamma function. How to prove it, not sure. It's better to put state the problem explicitly in your post and ask if it can be migrated to mathematics stackexchange. $\endgroup$ – Vibert Oct 3 '13 at 16:22
  • $\begingroup$ Some formulae may be useful to you : Abramowitz and Stegun 1 formula $15.2.1$, and Abramowitz and Stegun 2 formulae $15.3.11, 15.3.12, 15.3.14$ (looking at $z \to \infty$) $\endgroup$ – Trimok Oct 3 '13 at 17:06

The Euler $\psi$ function is exactly the same as the digamma function, $$\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}.$$ (For its use as "Euler $\psi$ function" in the literature, see e.g. this paper.) While the Gamma function, the Pochhammer symbols, and the like, are very useful in constructing the hypergeometric functions, it is not possible to express either $\psi$ or $\Gamma$ as special cases of the hypergeometric family.

The digamma function is mildly useful in finding the derivatives of ${}_2F_1(a,b;c;z)$ with respect to the parameters $a,b$ and $c$, though: since $${}_2F_1(a,b;c;z)=\sum_{n=0}^\infty \frac{\Gamma(a+n)}{\Gamma(a)}\frac{(b)_n}{(c)_n}\frac{z^n}{n!},$$ differentiating by $a$ you can reduce this to $$\frac{\partial}{\partial a}{}_2F_1(a,b;c;z) =-\psi(a){}_2F_1(a,b;c;z) +\sum_{n=0}^\infty \psi(a+n) \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}.$$ I don't know what good it'll do you, though.

In normal circumstances I would refer you to the Digital Library of Mathematical Functions, which supersedes Abramowitz and Stegun, and particularly chapter 5. For the moment, though, you'll have to make do with the print version.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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