In the English translation of The Gamma Function by Emil Artin (1964 - Holt, Rinehart and Winston) there appears to be a mistake in the formula given for the gamma function on page 24:

$$\Gamma(x) = \sqrt{2\pi}x^{x-1/2}e^{-x+\mu(x)}$$ $$\mu(x)=\sum_{n=0}^\infty(x+n+\frac{1}{2})\text{log}(1+\frac{1}{x+n})-1=\frac{\theta}{12x},\ \ \ \ \ 0 < \theta < 1$$

and on page 22 where this is derived, it is noted that '$\theta$ is a number independent of $x$ between 0 and 1'.

This sounds incorrect, as $\theta$ does depend on $x$, but since the wording is a little ambiguous it may just be an unclear translation. The original German might have meant that $0< \theta(x) < 1$ for any $x$. That the variable $x$ is suppressed from $\theta$ could be just confusing notation, or someone's misunderstanding (possibly mine.)

The preface does mention that a (different) formula had to be corrected for the English reprint.

I would like to know if there are mistakes in this book, and if so, whether they exist in the German edition. Is there an available list of errata?


1 Answer 1


It seems clear that $\theta$ can indeed be chosen to be a number independent of $x$ as stated, to get Stirling's formulas for the gamma function when $x$ is large. The wording, at least in English, is not too helpful in this section. But I'm less clear about where in the formula on page 24 there is supposed to be a mistake. Here as in any mathematics book (especially a translation) one has to be wary about misprints or errors. Probably there is no publicly available list of errata for this small monograph published originally in 1931 in German and later republished in 1964 in an English translation by Michael Butler. This English version is included in the 2007 AMS softcover book Exposition by Emil Artin: A Selection edited by Michael Rosen. (There is an older 1965 book The Collected Papers of Emil Artin published by Addison-Wesley and edited by Lang & Tate. This contains Artin's research papers, in the original German or English.) As Zavosh observes, the 1964 preface by Edwin Hewitt reprinted here does indicate one formula corrected in the translation: " ... a small error following formula (59) (this edition) was corrected..." However, the formula seems to be the one actually numbered (5.9). Caveat lector.

  • $\begingroup$ I think $\theta$ actually converges to 1 quickly as $x$ grows large. For just an asymptotic formula for $\Gamma(x)$ (as in the common version of Stirling's formula), there would be no reason to mention $\mu(x)$ or $\theta$ at all, since the $e^{-\mu(x)}$ term converges to 1. It sounds like the author is stating on page 24 an exact formula involving a constant $\theta$, when it's actually not a constant. I suspect it's a mistake of the translator originating from a misunderstanding on page 22. $\endgroup$
    – Zavosh
    Commented Apr 4, 2010 at 16:35
  • $\begingroup$ I looked at the AMS 2007 version and it's exactly the same. I will mark your answer as accepted soon, unless someone else comes up with a miraculously clarifying answer (which is unlikely.) Thanks very much. $\endgroup$
    – Zavosh
    Commented Apr 4, 2010 at 16:38
  • 1
    $\begingroup$ Since this section of Artin's book is concerned with approximating the gamma function, there may be some tendency to use the equals sign loosely. I guess the point is to find a convenient elementary function giving a good approximation for large $x$; the choice might be fine-tuned in various ways. But in the era before computers the shape of an approximating function would have been the most interesting question for many people. $\endgroup$ Commented Apr 4, 2010 at 17:29
  • $\begingroup$ P.S. At the moment the AMS book mentioned earlier is on sale online: Exposition by Emil Artin: A Selection - Michael Rosen, Brown University, Editor - AMS | LMS, 2006, 346 pp., Softcover, ISBN-10: 0-8218-4172-6, ISBN-13: 978-0-8218-4172-3, List: US$59, All AMS Members: US$47, Sale Price: US$38, HMATH/30 $\endgroup$ Commented May 9, 2010 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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