How to Show $\int_n^{n+1}(\frac{1}{n^s}-\frac{1}{x^s})dx$ is Analytic

Question

I found in a note that $F_n(s)=\int_n^{n+1}(\frac{1}{n^s}-\frac{1}{x^s})dx$, then $F_n(s)$ is analytic in $\mathbb{C}$ for every $n \in \mathbb{N}$.

Now, from the definition of analytic function found in Wikipedia (click here), I am having trouble to see how $F_n(s)$ is analytic, specially, the integral is defined by the variable $x$, but we are dealing with complex plan and complex number $s= \sigma +it.$

I request to explain and show elaborately why $F_n(s)$ is analytic.

Thanks.

Answer 1

Let $N \in \Bbb{N}$,then:

$$\int_N^{N+1}\frac{1}{N^s}dx=(\frac{1}{N})^s=e^{-(\ln{N})s}$$ which is analytic on $\Bbb{C}$

Now let $h_n \to 0$ and $s \in \Bbb{C}$

We assume that $|h_n| \leq 1,\forall n \in \Bbb{N}$

Then $$g_n(x):=\frac{\frac{1}{x^{s+h_n}}-\frac{1}{x^s}}{h_n} \to -\ln{x}e^{-s\ln{x}}=-\ln{x} \frac{1}{x^s}$$

We have $|g_n(x)|=\frac{|e^{-h_n\ln{x}}-1|}{|x^{\Re{s}}||h_n|}$

and using the inequality: $|e^z-1| \leq|z|e^{|z|}$ we have that $$|g_n(x)| \leq \frac{|\ln{x}|e^{|h_n||\ln{x}|}}{|x^{\Re{s}}|}\leq \frac{|\ln{x}|e^{|\ln{x}|}}{N^{\Re{s}}}\leq |\ln{x}|e^{|\ln{x}|} \in L^1([N,N+1])$$

So by dominated convergence you have the differentiability as $s$.

Since $s$ is arbitrary we finally have that $F_N(s)$ is differentiable on the whole plane as difference of differentiable functions on the whole plane.

So $F_N(s)$ is entire as a difference of entire functions.