Uniform convergence about Dirichlet integral $f(s):=\int_1^\infty\frac{a(x)}{x^s}\,dx =\lim\limits_{T\to\infty}\int_1^T\frac{a(x)}{x^s}\,dx$

Question

On page 87 of Ingham's book: The Distribution Of Prime Numbers, the author asserts the following results, but does not give proof.

Let $a(x)$ be a bounded and integrable function over any finite interval $1\leq x\leq X$, and $s=\sigma+it$ is a complex variable. If $f(s)=\int_1^\infty\frac{a(x)}{x^s}\,dx$ converges for $s=s_0=\sigma_0+it_0$, it converges uniformly in any fixed angle of the type $$|\arg(s-s_0)|\le\alpha<\frac{\pi}{2}.$$

But I can only prove that for any fixed $\delta>0$ and any fixed angle $0<\alpha<\frac{\pi}{2}$, $f(s)$ is convergent uniformly in $$\{s\in\mathbb{C}, \Re(s)\geq \sigma+\delta\}\cap\{s\in\mathbb{C}, |\arg(s-s_0)|\leq\alpha<\frac{\pi}{2}\}.$$

This is my proof:

Let $A(T)=\int_1^T\frac{a(x)}{x^{s_0}}\,dx$, since $\int_1^\infty\frac{a(x)}{x^{s_0}}\,dx$ is convergent, $A(T)$ is bounded uniformly for $T\geq1$ and let $A>0$ such that $|A(T)|\leq A$ for all $T\geq1$.

Let $f_N(s)=\int_1^N\frac{a(x)}{x^s}dx$. Then for $\Re(s)>\Re(s_0)$, we have that \begin{eqnarray*} % \nonumber to remove numbering (before each equation) f(s)-f_N(s)&=&\int_N^\infty\frac{a(x)}{x^s}dx=\int_N^\infty\frac{dA(x)}{x^{s-s_0}} =\frac{A(x)}{x^{s-s_0}}\Big|_N^{+\infty}+(s-s_0)\int_N^\infty\frac{A(x)}{x^{s-s_0+1}}dx \\ &=&-\frac{A(N)}{N^{s-s_0}}+(s-s_0)\int_N^\infty\frac{A(x)}{x^{s-s_0+1}}dx. \end{eqnarray*} Then $$|f(s)-f_N(s)|\leq \frac{A}{N^{\sigma-\sigma_0}}+\frac{|s-s_0|}{\sigma-\sigma_0}\cdot\frac{A}{N^{\sigma-\sigma_0}} =\frac{A}{N^{\sigma-\sigma_0}}\left(\frac{|s-s_0|}{\sigma-\sigma_0}+1\right).$$ Hence for any $\delta>0$ and any fixed angle $0<\alpha<\frac{\pi}{2}$, when $s$ belongs to $$\{s\in\mathbb{C}, \Re(s)\geq \sigma+\delta\}\cap\{s\in\mathbb{C}, |\arg(s-s_0)|\leq\alpha<\frac{\pi}{2}\},$$ $$|f(s)-f_N(s)|\leq \frac{A}{N^{\delta}}\left(\sec(\alpha)+1\right).$$ Hence $f_N(s)$ converges to $f(s)$ uniformly in $$\{s\in\mathbb{C}, \Re(s)\geq \sigma+\delta\}\cap\{s\in\mathbb{C}, |\arg(s-s_0)|\leq\alpha<\frac{\pi}{2}\}.$$ Hence $f(s)$ converges and is analytic for $\Re(s)>\Re(s_0)$, with $$f'(s)=-\int_1^\infty\frac{a(x)\log x}{x^s}\,dx.$$

(For Dirichlet integrals, you can also refer to Keith Conrad's lecture notes about Dirchlet series, page 10, exercise 14)

My Question is: How to prove the assertion of Ingham's book?

Answer 1

Let $b(y)=\int_y^\infty\frac{a(x)}{x^{s_0}}\,dx$. Then, similarly to your derivation, $$f(s)-f_N(s)=\frac{b(N)}{N^{s-s_0}}-(s-s_0)\int_N^\infty\frac{b(x)}{x^{s-s_0+1}}\,dx.$$ Now let $r(x)=\sup\limits_{y>x}|b(y)|$, then $|b(x)|\leqslant r(x)\leqslant r(N)$ for $x\geqslant N$, and $$|f(s)-f_N(s)|\leqslant\frac{r(N)}{\color{gray}{N^{\sigma-\sigma_0}}}\left(\frac{|s-s_0|}{\sigma-\sigma_0}+1\right),$$ again much like the way you did. Now use the fact that $r(N)\to 0$ as $N\to\infty$.