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

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?

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 $$N\to\infty$$.

• $A(y)$ is defined by $A(y)=\int_1^y\frac{a(x)}{x^{s_0}}dx$ and $\lim\limits_{y\to\infty}A(y)=\int_1^\infty\frac{a(x)}{x^{s_0}}dx$. Why $R(N)\to0$ as $N\to\infty$, Could you explain it?
– HGF
Commented Mar 24, 2022 at 12:40
• @HGF: Yes, I missed $\int_1^y\mapsto\int_\infty^y$ in rush. Fixed. Commented Mar 24, 2022 at 13:12
• I am sure your proof is correct, but what is the essential difference between the two proofs? Why is the final result of the two proofs very different and the result obtained by your method is much stronger.
– HGF
Commented Mar 24, 2022 at 13:19
• Maybe I understand, this is essentially the same method as proving that the dirichlet series converges uniformly. Thank you.
– HGF
Commented Mar 24, 2022 at 13:44
• @HGF: Basically, this is the freedom of choice of the antiderivative of $a(x)/x^{s_0}$ (followed by IBP). The uniform bound works for any choice, but gives a weaker estimate. Choosing the one that tends to $0$ as $x\to\infty$ allows one to exploit this property. Commented Mar 24, 2022 at 15:41