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This was from part of a proof of making an analytic continuation of $\zeta(s)$, we have:

$$ \zeta(s) = s \int_1^\infty \frac{[x]-x+\frac{1}{2}}{x^{s+1}} \, dx + \frac{1}{s-1} + \frac{1}{2} $$ The integral is convergent for $\sigma >0$ and uniformly convergent in any finite region to the right of $\sigma = 0$. It therefore defines an analytic function of $s = \sigma + it$, regular for $\sigma >0$ and a simple pole at $s=1$. (p14, Titchmarsh)

My question is:

Why is the integral $$g(s):=\int_1^\infty \frac{[x] - x + \frac{1}{2} }{x^{s+1} } \, dx $$ is analytic for $\sigma > 0$. Is the integrand even analytic? I don't understand what "uniformly convergent integral" in this case means - and how this implies the analyticity of the function.


EDIT: My proof for analyticity of $\int_{a}^{a+1} F(x,s) \, dx $. (Is this right?)

We let $F(x,s) = \frac{[x]-x+\frac{1}{2}}{x^{s+1}}$. We first show $\int_{a}^{a+1} F(x,s) \, dx $ is analytic, for some fixed $a \in \mathbb{N}$. For each $n \in \mathbb{N}$, on $[a, a+1-\frac{1}{n}]$ the function $$A_n(s) := \int_a^{a+1-\frac{1}{n}} F(x,s) \, dx $$ is analytic on $B_r(p) \subseteq \mathbb{C}$ a ball of radius $r$ around $p$. We see that $|F(x,s)| \le M$ for all $s \in B_r(p)$ and $x \in [a,a+1]$. and $ \lim_{n \rightarrow \infty} F(x,s) \chi_{[a,a+1-\frac{1}{n}] } = F(x,s) \chi_{[a,a+1)} $. that is, a.e. convergence to $F(x,s) \chi_{[a,a+1]}$. By Dominated Convergence Theorem, $$ \lim_{n \rightarrow \infty} A_n(s) = \int_a^{a+1} F(x,s) \, dx := A(s) $$ Further $A_n(s)$ converges compactly, in fact it is uniform, as for all $n,m \ge N$, for all $ s \in B_{r}(p) $, $$ \Big| A_n(s) - A_m(s) \Big| \le M \frac{2}{N} \rightarrow 0 $$ So the convergence is compact, and $A(s)$ is analytic on $B_r(p)$.

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  • $\begingroup$ Here $F(s) = \frac{\zeta(s)}{s}-\frac{1}{s-1}+\frac{1}{2s} = \int_1^\infty (\lfloor x \rfloor-x+\frac12)x^{-s-1}dx$ is holomorphic for $\Re(s) > 0$ since $F'(s) = \int_1^\infty (\lfloor x \rfloor-x+\frac12)x^{-s-1}(-\log x)dx$. Also $F(s) = \sum_{k=0}^\infty \frac{(s-s_0)^k}{k!}F^{(k)}(s_0) =\sum_{k=0}^\infty \frac{(s-s_0)^k}{k!}\int_1^\infty (\lfloor x \rfloor-x+\frac12)x^{-s_0-1}(-\log x)^kdx$ for $|s-s_0| < |\Re(s_0)|$ so it is analytic there $\endgroup$ – reuns Jul 4 '17 at 18:12
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Theorem 1: If $F: \Omega\times[a,b] \to \mathbb{C}$ is continuous and such that $F(s,x)$ is holomorphic in $s$ for each fixed $x$, then $G(s) := \int_a^b f(s,x)dx$ is holomorphic on $\Omega$.

Theorem 2: If $(f_n)_n$ is a sequence of holomorphic functions on a domain $\Omega$ such that $f_n$ converges uniformly to $f$ on every compact subset of $\Omega$, then $f$ is holomorphic on $\Omega$.

Definition: An integral of the form $\int_a^\infty f(s,x)dx$ is "uniformly convergent" if $\int_a^N f(s,x) \to \int_a^\infty f(s,x)$ uniformly in $s$ as $N \to \infty$.

To show the integral is analytic at any point $s=\sigma+it$ for $\sigma > 0$, it suffices to show it is open in any compact subset $K$ contained in $\{s=\sigma+it : \sigma> 0\}$. Combining Theorems 1 and 2, it suffices to show that the integral $\int_1^\infty \frac{[x]-x+\frac{1}{2}}{x^{s+1}}dx$ is uniformly convergent for $s \in K$. Note $|\int_N^\infty \frac{[x]-x+\frac{1}{2}}{x^{s+1}}dx| \le \int_N^\infty |\frac{[x]-x+\frac{1}{2}}{x^{s+1}}|dx$ and since $|[x]-x+\frac{1}{2}| \le 1$, we bound the integral by $\int_N^\infty \frac{1}{|x^{s+1}|}dx = \int_N^\infty \frac{1}{x^{\sigma+1}}dx \le \int_N^\infty \frac{1}{x^{\sigma_0+1}}dx$ where $\sigma_0$ is the smallest value that $Re s$ takes on for all $s$ in $K$. The point is, this last quantity is independent of $s\in K$ (thus uniform in $s$) and is finite, since $\sigma_0 > 0$ since $K$ is compact.

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  • $\begingroup$ Just to clarify, to show analyticity of the function $f_n = \int_1^n f(s,x) \, dx$, and to apply Theorem 1, we split the integral up? Since $f(s,x)$ is piecewise continuous in [1,n] wrt $x$ in our case. $\endgroup$ – CL. Jul 4 '17 at 9:05
  • $\begingroup$ Ah yes. I should have mentioned that. Yes, each $f_n$ is a finite sum of the integrals that Theorem 1 applies to. $\endgroup$ – mathworker21 Jul 4 '17 at 9:50
  • $\begingroup$ Is everything satisfactory now? $\endgroup$ – mathworker21 Jul 4 '17 at 22:21
  • $\begingroup$ Yes, thanks a lot mathworker21, do you mind also looking at this question: math.stackexchange.com/questions/2347475/… ? Thank you so much. $\endgroup$ – CL. Jul 6 '17 at 6:51
  • $\begingroup$ Actually I noticed something I am unsure of. Can we actually apply theorem 1? We need continuity of integrand $f$ on $[a,a+1] \times \Omega$. But $f$ is not continuous at $a+1$. I recall the proof of theorem 1 requires continuity on the closed interval. $\endgroup$ – CL. Jul 12 '17 at 11:22

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