- This exercise gives an asymptotic formula for log n!. A more refined asymptotic formula for \(\Gamma ( s )\) as \(s \to \infty\) (Stirling’s formula) is given in Appendix A.
- Show that
\[ {\frac {d}{d x}} \int_ {x} ^ {x + 1} \log \Gamma (t) d t = \log x, \qquad {\mathrm{for}} x > 0, \]
and as a result
\[ \int_ {x} ^ {x + 1} \log \Gamma (t) d t = x \log x - x + c. \]
- Show as a consequence that log \(\Gamma ( n ) \sim n\) log n as \(n \to \infty\) . In fact, prove that \(\log \Gamma(n)=n\log n+O(n)\) as \(n\to\infty\). [Hint: Use the fact that \(\Gamma ( x )\) is monotonically increasing for all large x.]
Solution
Let \[ I(x)=\int_x^{x+1}\log\Gamma(t)\,dt. \] By Leibniz’ rule, \[ I'(x)=\log\Gamma(x+1)-\log\Gamma(x). \] Using \(\Gamma(x+1)=x\Gamma(x)\), \[ I'(x)=\log x. \] Therefore \[ I(x)=x\log x-x+c \] for a constant \(c\).
For part (b), \(\Gamma(x)\) is increasing for all sufficiently large \(x\). Hence for all sufficiently large integers \(n\), \[ \int_{n-1}^{n}\log\Gamma(t)\,dt \le \log\Gamma(n) \le \int_n^{n+1}\log\Gamma(t)\,dt. \] Applying part (a) at \(x=n-1\) and \(x=n\) gives \[ (n-1)\log(n-1)-(n-1)+c \le \log\Gamma(n) \le n\log n-n+c. \] Now \[ (n-1)\log(n-1)=n\log n+O(n), \] so both bounds imply \[ \log\Gamma(n)=n\log n+O(n). \] Dividing by \(n\log n\) yields \[ \frac{\log\Gamma(n)}{n\log n}\to1, \] that is, \(\log\Gamma(n)\sim n\log n\).