# Does the mean-value theorem hold for $f(x)=x^{s}$ with $s$ complex?

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.

• Shouldn't it be $\leq \frac{|s|}{n^{\operatorname{Re}(s)+1}}$?
• 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 $n\le x \le n+1$.
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 $$\mathbb{R}^2 = \mathbb{C}$$.