From the book "Asymptotic Analysis and Perturbation Theory" by William Paulsen,
Problem 23 p.67.
Show that $\Gamma'(1)=-\gamma$.
Hint: First use the recursion property of $\Gamma(x)$ to show that, for $n$ an integer $>1$,
$$\frac{\Gamma'(n)}{\Gamma(n)}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}+\Gamma'(1)$$Then use Stirling's formula to show that,
$$\frac{\Gamma'(x)}{\Gamma(x)}\sim \ln(x)-\frac{1}{2x}+\cdots\ \text{as}\ x\to\infty$$
I did some searching and found many posts using the Leibniz integral rule to show $\Gamma'(1)=-\gamma$ which I understand, but I cannot seem to finish the problem using the hints they give. Here is my attempt.
We know $\Gamma(n)=(n-1)!$. Taking the logarithmic derivative,
$$\begin{align*}
\frac{\Gamma'(n)}{\Gamma(n)}&=\frac{\mathrm{d}}{\mathrm{d}n}\left(\ln(n-1)!\right) \\
&=\frac{\mathrm{d}}{\mathrm{d}n}\left(\ln\left(\prod_{k=1}^{n-1}(n-k)\right)\right) \\
&=\frac{\mathrm{d}}{\mathrm{d}n}\left(\sum_{k=1}^{n-1}\ln\left(n-k\right)\right) \\
&=\sum_{k=1}^{n-1}\left(\frac{1}{n-k}\right)
\end{align*}$$
Writing out the terms,
$$\frac{\Gamma'(n)}{\Gamma(n)}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}$$
Though there is some resembelnce, this does not agree with the hint as I am missing the $\Gamma'(1)$ term. I think the problem comes from when I took the derivative but I don't know the exact issue. Can someone help me? Thanks in advance.
Best Answer
Using the recursion property $\Gamma(x + 1) = x \Gamma(x)$, $$\frac{\Gamma'(x)}{\Gamma(x)} = \frac{d}{dx} \log(\Gamma(x)) = \frac{d}{dx} \log\bigl( (x-1) \Gamma(x-1) \bigr) = \frac{1}{x - 1} + \frac{d}{dx} \log(\Gamma(x - 1)).$$ Now repeat this many times until we have $$ \begin{align*} \frac{\Gamma'(x)}{\Gamma(x)} &= \frac{1}{x - 1} + \frac{1}{x - 2} + \frac{d}{dx} \log(\Gamma(x - 2)) \\ &= \dots = \frac{1}{x - 1} + \frac{1}{x - 2} + \dots + \frac{1}{2} + 1 + \frac{d}{dx} \log(\Gamma(x - (n - 1))). \end{align*}$$ This last expression is also (just by chain rule/viewing the logarithmic derivative as what it is) $$ \frac{d}{dx} \log(\Gamma(x - (n - 1))) = \frac{\Gamma'(x - (n - 1))}{\Gamma(x - (n - 1))}. $$ Now evaluate everything at $x = n$ and remember $\Gamma(1) = 1$.