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Here is a Dirichlet series. What is the domain of convergence? Let $f(a,b) = a^4 + b^4$.

$$ L(f, s) = \sum_{(a,b) \in \mathbb{Z}^2} \frac{a^4 + b^4}{(a^2 + b^2)^s} $$

Here $(a,b) \neq (0,0)$. For $s > 10$ this series certainly converges. So I am looking for smaller values of $s$. This motivate careful look at my number theory textbooks:

Theorem Suppose that $\alpha(s) = \sum a_n n^{-s}$ converges at a point $s = s_0$. Then the series converges uniformly in the range: $$s = \sigma + i t \in \big\{\sigma > \sigma_0 ,\; |t - t_0| \leq H (\sigma - \sigma_0)\big\} $$ Here $H$ is any arbitrary constant.

This is an infinite triangle-shaped region in $\mathbb{C}$ with a vertex at $s = s_0$. I am saying that $s_0 = 10$ is feasible. I beleive $s > 6$ is also possible. I think this is called the $\color{#388A3E}{\text{abscissa of convergence}}$ (or ``line of convergence").

Then there's the question of how this uniform convergence changes with $H$. To the right of $s = s_0$, in that half-plane we get convergence, letting $H \to \infty$. So this Dirichlet series is locally uniformly convergent.

Once we have convergence to the right of a certain, line, we could try to analytically continued. I am trying to find a reasonable value for the badly divergent series:

$$ L(f, \tfrac{1}{2}) \; "=" \sum_{(a,b)\in \mathbb{Z}^2} \frac{a^4 + b^4}{\sqrt{a^2 + b^2}}$$

This is just the analytic continuation of the zeta function to $s = \frac{1}{2}$. Could be done using the Euler Maclaurin formula or Ramanujan summation.

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    $\begingroup$ You want to sum over $(a,b)\in\mathbb{Z}^2$ I think? $\endgroup$
    – Shashi
    Dec 17 '17 at 18:06
  • $\begingroup$ The series is undefined if $a,b=0.$ Also, pay attention to @Shashi's comment. Finally, I don't see why you introduce $f.$ $\endgroup$
    – zhw.
    Dec 17 '17 at 18:46
  • $\begingroup$ @cactus314, why are you trying to analytically continue this to $s = 1/2$? As I have mentioned to you before, Dirichlet $L$-series are only arithmetically interesting when their coefficients are multiplicative. $\endgroup$ Dec 18 '17 at 11:30
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Sketch: $(a,b)$ will denote an element of $\mathbb Z^2.$ I'll consider

$$\sum_{(a,b) \ne \{(0,0)\}} \frac{a^4 + b^4}{(a^2+ b^2)^s}, \text { for } s>0.$$

We can simplify a bit by noting that $a^4+ b^4$ is bounded above and below by constant multiples of $(a^2+b^2)^2.$ Thus we need only consider

$$\sum_{(a,b)\ne (0,0)} \frac{(a^2+b^2)^2}{(a^2+ b^2)^s} =\sum_{(a,b)\ne (0,0)} (a^2+b^2)^{2-s}.$$

Now the last expression looks like an integral, namely

$$\int_{x^2+y^2 \ge 1}(x^2+y^2)^{2-s}\, dx\,dy.$$

That's perfect for polar coordinates. We get

$$2\pi \int_1^\infty (r^2)^{2-s}\, r\, dr.$$

The last integral converges iff $s>3.$ I've left out some details, to be sure, but $s>3$ seems right to me for the domain of convergence.

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  • $\begingroup$ Can we compute $\sum_{a,b\in\mathbb{Z}}'\frac{a^2 b^2}{(a^2+b^2)^s}$ in explicit terms? $\endgroup$ Dec 17 '17 at 19:44
  • $\begingroup$ @JackD'Aurizio You sir would know better than I, as your laboratory of special functions is much vaster than mine. $\endgroup$
    – zhw.
    Dec 17 '17 at 19:47
  • $\begingroup$ I almost managed to compute it for $s=4$: it involves some non-trivial series, namely $$\sum_{b\geq 1}\frac{1}{\sinh^6(\pi b)}\quad\text{and}\quad \sum_{b\geq 1}\frac{1}{b^2\sinh^2(b\pi)}.$$ $\endgroup$ Dec 17 '17 at 20:04
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At the $8n$ points where $\max(|a|,|b|)=n$, we know that $$ n^2\le a^2+b^2\le2n^2 $$ and $$ n^4\le a^4+b^4\le2n^4 $$ Therefore, $$ \sum_{n=1}^\infty8n\frac{n^4}{2^sn^{2s}}\le\sum_{\substack{(a,b)\in\mathbb{Z}^2\\(a,b)\ne(0,0)}}^\infty\frac{a^4+b^4}{\left(a^2+b^2\right)^s}\le\sum_{n=1}^\infty8n\frac{2n^4}{n^{2s}} $$ Thus, we have convergence precisely when $2s-5\gt1$; that is, $s\gt3$.

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  • $\begingroup$ How did you turn $a^2 + b^2 \asymp n$ into the entire series? $\endgroup$
    – cactus314
    Dec 18 '17 at 0:14
  • $\begingroup$ When $n\gt0$, there are precisely $8n$ terms where $\max(|a|,|b|)=n$. Thus, I just use the max and min of the values in the numerator and denominator for each $n$. $\endgroup$
    – robjohn
    Dec 18 '17 at 0:16
  • $\begingroup$ I guess you're right. Also $2\pi < 8$. $\endgroup$
    – cactus314
    Dec 18 '17 at 0:22
  • $\begingroup$ I'm summing over boundaries of squares, so I am not using $\pi$. $\endgroup$
    – robjohn
    Dec 18 '17 at 0:24
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Well, it is pretty simple to compute the abscissa of convergence of the following integral:

$$ \iint_{x^2+y^2\geq 1}\frac{x^4+y^4}{(x^2+y^2)^s}\,dx \,dy = \int_{0}^{2\pi}\int_{1}^{+\infty}\rho^{5-2s}\left(\cos^4\theta+\sin^4\theta\right)\,d\rho\,d\theta$$ which is $\sigma=3$. By series-integral comparison, $\sum_{a,b\in\mathbb{Z}}'\frac{a^4+b^4}{(a^2+b^2)^s}$ is divergent if $\text{Re}(s)\leq 3$.
On the other hand, assuming $\text{Re}(s)>3$ we have

$$\left|\sum_{a,b\in\mathbb{Z}}'\frac{a^4+b^4}{(a^2+b^2)^s}\right|\leq \sum_{a,b\in\mathbb{Z}}'\frac{1}{(a^2+b^2)^{\sigma-2}}=\sum_{n\geq 1}\frac{r_2(n)}{n^{\sigma-2}}=\sum_{n\geq 1}\frac{4(\chi_4*1)(n)}{n^{\sigma-2}} $$ and the RHS equals $4\,\zeta(\sigma-2)\,L(\chi_4,\sigma-2)<+\infty.$ It follows that the abscissa of convergence of your series is exactly $\color{red}{3}$.

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