Growth of Digamma function

Question

For $1\le \sigma \le 2$ and $t\ge 2$, $s=\sigma+it$ prove that $\displaystyle \frac{\Gamma'(s)}{\Gamma(s)}=O(\log t)$.

From Stirling's formula we have, $\displaystyle \Gamma(s)\approx \sqrt{2\pi}\exp\{s\log s-s-\frac 12 \log s\}$.

Then, $\displaystyle \frac{\Gamma'(s)}{\Gamma(s)}\approx\log s-\frac{1}{2s}$. From here I'm unable to estimate !! Any hint. ?

Where can I get rigorous proof ?

Edit: Wikipedia links below the question are NOT clear enough to me.

Answer 1

If you need to differentiate a holomorphic function, try to integrate instead.

Let $g(s)=\log\sqrt{2\pi}+s\log s-s-\tfrac12\log s$; so $\log\Gamma(s)=g(s)+o(1)$.

Take a circle around $s$ with radius $r=\frac12$. By Cauchy's formulas, $$ \frac{\Gamma'}{\Gamma}(s) = g'(s) + \big(\log\Gamma(s)-g(s)\big)' = \log s-\frac1{2s} + \frac1{2\pi i}\oint_{|w-s|=r} \frac{\log\Gamma(w)-g(w)}{(w-s)^2} \mathrm dw; $$ $$ \bigg|\frac{\Gamma'}{\Gamma}(s) - \log s+\frac1{2s} \bigg| \le \frac1{2\pi}\oint_{|w-s|=r} \frac{\big|\log\Gamma(w)-g(w)\big|}{r^2} |\mathrm dw| \le O(1) $$ so, $$ \frac{\Gamma'}{\Gamma}(s) = \log s + O(1). $$

Answer 2

Let's use the following formula, from Abramowitz and Stegun, valid everywhere in the strip of interest: $$ \psi(z)=-\gamma+\sum_{n=1}^{\infty}\frac{z-1}{n(n+z-1)}. $$

Letting $a=\sigma-1\in[0,1]$, $$ \psi(\sigma+it)+\gamma=\sum_{n=1}^{\infty}\frac{a+it}{n(n+a+it)}=\sum_{n=1}^{\infty}\frac{\left(a+it\right)(n+a-it)}{n\left((n+a)^2+t^2\right)}=\sum_{n=1}^{\infty}\frac{a(n+a)+int+t^2}{n\left((n+a)^2+t^2\right)}=\sum_{n=1}^{\infty}\frac{a(n+a)}{n\left((n+a)^2+t^2\right)}+i\sum_{n=1}^{\infty}\frac{t}{(n+a)^2+t^2}+\sum_{n=1}^{\infty}\frac{t^2}{n\left((n+a)^2 + t^2\right)}. $$ The first term is no more than $2a\sum_{n=1}^{\infty}1/n^2=\pi^2/3$, independent of $t$. The second and third terms can be bounded by integrals; the third integral must be split into two sums first. Specifically, $$ \sum_{n=1}^{\infty}\frac{t}{(n+a)^2+t^2}\le \sum_{n=1}^{\infty}\frac{1/t}{(n/t)^2+1}\le\int_{0}^{\infty}\frac{dx}{x^2+1}=\frac{\pi}{2}, $$ and $$ \sum_{n=t+1}^{\infty}\frac{t^2}{n\left((n+a)^2+t^2\right)}\le \sum_{n=t+1}^{\infty}\frac{1/t}{(n/t)\left((n/t)^2+1\right)}\le\int_{1}^{\infty}\frac{dx}{x(x^2+1)}=\log\left(\frac{x}{\sqrt{x^2+1}}\right)\Bigg\vert_{1}^{\infty}=\frac{1}{2}\log 2. $$ Note the restricted range of summation. Finally, $$ \sum_{n=1}^{t}\frac{t^2}{n\left((n+a)^2+t^2\right)}\le\sum_{n=1}^{t}\frac{1}{n}=O(\log t), $$ so we conclude that $\psi(\sigma+it)$ is $O(\log t)$, with bounded imaginary part.