This is a basic question from Stein and Shakarchi's book Complex Analysis.

Let $s$ be a fixed complex number with $\operatorname{Re}(s)>0$ and $f(x)=x^{-s}$ be a function from $[n,n+1]$ to $\mathbb{C}$ where $n$ is a positive integer. On page 173, it states that one can apply the mean value theorem to $f$ to obtain the inequality $$\left|\frac{1}{n^{s}}-\frac{1}{x^{s}}\right|\leq\frac{|s|}{n^{s+1}}$$ whenever $n\leq x\leq n+1$.

I am confused because $f$ is a complex valued function so the mean-value theorem would give two possibly different values $c_1$ and $c_2$ in $(n,x)$ such that $$\operatorname{Re}(f^{\prime}(c_1))+i\operatorname{Im}(f^{\prime}(c_2))=\frac{f(x)-f(n)}{x-n}$$ and so one cannot deduce the above inequality.

Many thanks for your help.

  • 2
    $\begingroup$ Shouldn't it be $\leq \frac{|s|}{n^{\operatorname{Re}(s)+1}}$? $\endgroup$
    – Gary
    Jul 26, 2022 at 22:57
  • 3
    $\begingroup$ You can argue as follows: $$ \left| {\frac{1}{{n^s }} - \frac{1}{{x^s }}} \right| = \left| {\int_n^x {\frac{{ - s}}{{t^{s + 1} }}dt} } \right| \le \int_n^x {\frac{{\left| s \right|}}{{t^{{\mathop{\rm Re}\nolimits} (s) + 1} }}dt} \le \frac{{\left| s \right|}}{{n^{{\mathop{\rm Re}\nolimits} (s) + 1} }}\int_n^x { dt } \le \frac{{\left| s \right|}}{{n^{{\mathop{\rm Re}\nolimits} (s) + 1} }}. $$ if $n\le x \le n+1$. $\endgroup$
    – Gary
    Jul 26, 2022 at 23:09
  • $\begingroup$ @Gary yes you are right- s should be Re(s). $\endgroup$
    – Joey R
    Jul 27, 2022 at 8:51

1 Answer 1


The mean value theorem you can use is the fundamental theorem of calculus: For $f \in C^1([n, x], \mathbb{R}^d)$, $$f(x) - f(n) = \int_{0}^{1}f'(n + t(x - n))(x - n)\,dt = \int_{0}^{1}f'(n + t(x - n))\,dt(x - n).$$ Here we take $d = 2$ and use $\mathbb{R}^2 = \mathbb{C}$.


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