Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New features!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,552 answers , 19,366 comments
1,470 users with positive rep
411 active unimported users
More ...

What's the relation between the Euler $\psi$ function, the digamma function, and the hypergeometric function?

+ 6 like - 0 dislike
88 views

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.

This post imported from StackExchange Physics at 2014-03-07 13:46 (UCT), posted by SE-user user29126
asked Oct 3, 2013 in Theoretical Physics by user29126 (60 points) [ no revision ]
Could you please post a link to the paper?

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user dj_mummy
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.

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user Vibert
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 $z \to \infty$)

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user Trimok

1 Answer

+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-07 13:47 (UCT), posted by SE-user Emilio Pisanty
answered Oct 3, 2013 by Emilio Pisanty (330 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\varnothing$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...