State and prove an asymptotic formula for \(\Gamma'(x)\) as \(x\to+\infty\) along the real axis. Equivalently, find an elementary function \(A(x)\) such that \[ \frac{\Gamma'(x)}{A(x)}\longrightarrow1. \]
Solution
1. The function\[ \phi(x)=\log\Gamma(x),\qquad x>0, \]is convex.
Using the integral representation \[ \Gamma(x)=\int_0^\infty e^{-t}t^{x-1}\,dt, \] let \(x,y>0\) and \(0<\theta<1\). Then \[ \Gamma(\theta x+(1-\theta)y) =\int_0^\infty \bigl(e^{-t}t^{x-1}\bigr)^\theta \bigl(e^{-t}t^{y-1}\bigr)^{1-\theta}\,dt. \] Hölder’s inequality gives \[ \Gamma(\theta x+(1-\theta)y) \le \Gamma(x)^\theta\Gamma(y)^{1-\theta}. \] Taking logarithms proves the convexity of \(\phi\).
2. For every \(x>1\),\[ \log(x-1) \le \frac{\Gamma'(x)}{\Gamma(x)} \le \log x. \]
For a differentiable convex function, the derivative at \(x\) lies between the backward and forward secant slopes over intervals of length \(1\). Hence \[ \phi(x)-\phi(x-1) \le \phi'(x) \le \phi(x+1)-\phi(x). \] Using \(\Gamma(x)=(x-1)\Gamma(x-1)\) and \(\Gamma(x+1)=x\Gamma(x)\) gives \[ \phi(x)-\phi(x-1)=\log(x-1), \qquad \phi(x+1)-\phi(x)=\log x. \] Since \(\phi'(x)=\Gamma'(x)/\Gamma(x)\), the stated bounds follow.
3. The logarithmic derivative satisfies\[ \frac{\Gamma'(x)}{\Gamma(x)}\sim\log x \qquad(x\to+\infty). \]
For \(x>2\), divide the inequalities in 2 by \(\log x>0\): \[ \frac{\log(x-1)}{\log x} \le \frac{\Gamma'(x)}{\Gamma(x)\log x} \le1. \] The left-hand side tends to \(1\), so the squeeze theorem gives \[ \frac{\Gamma'(x)}{\Gamma(x)\log x}\longrightarrow1. \]
4. Therefore one may take\[ \boxed{ A(x)=\sqrt{\frac{2\pi}{x}}\left(\frac{x}{e}\right)^x\log x }. \]Equivalently,\[ \Gamma'(x) \sim \sqrt{\frac{2\pi}{x}}\left(\frac{x}{e}\right)^x\log x. \]
Stirling’s formula gives \[ \Gamma(x) \sim \sqrt{\frac{2\pi}{x}}\left(\frac{x}{e}\right)^x. \] Multiplying this by the asymptotic relation in 3 yields \[ \frac{\Gamma'(x)}{ \sqrt{2\pi/x}(x/e)^x\log x} = \frac{\Gamma(x)}{\sqrt{2\pi/x}(x/e)^x} \cdot \frac{\Gamma'(x)}{\Gamma(x)\log x} \longrightarrow1. \]