Value of a Sum linked with Beta Dirichlet Function and Zeta Function

Question

Recently, I tried to calculate this double sum:

$$ F(s) = \sum_{(a,b) \in \mathbb{Z}^2 \backslash (0,0) } \frac{1}{(a^2 + b^2)^s}$$

For $ s \in \mathbb{C}, Re(s) > 1$

I think i found the value of this function with the following reasoning. Indeed, $F(s)$ is :

$$ F(s) = \sum_{n = 1}^{\infty} \frac{r_2(n)}{n^s} $$ where $r_2$ is the "two-square function" : the number of possibilities to write $n$ as sum of two squares.

The Jacobi two-square theorem give us the well-known formula : $r_2(n) = 4(d_1(n) - d_3(n))$. Another way to write this equality is : $$r_2(n) = 4 \sum_{d | n} \sin \left( \frac{1}{2} \pi d \right)$$

Therefore the Mobius inversion formula give us :

$$ 4 \sin \left( \frac{1}{2} \pi n \right) = \left( r_2 \star \mu \right) (n) $$

Using the compatibility properties of dirichlet series with the convolution product, we have :

$$ 4 \sum_{n = 1}^{\infty} \frac{\sin \left( \frac{1}{2} \pi n \right)}{n^s} = F(s) \left( \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} \right)$$

And we can see that : $$ \sum_{n = 1}^{\infty} \frac{\sin \left( \frac{1}{2} \pi n \right)}{n^s} = \beta(s) $$ It's the Dirichlet Beta function. And we know that : $$ \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = \frac{1}{\zeta (s) }$$

Finally the value i found is :

$$ F(s) = \sum_{(a,b) \in \mathbb{Z}^2 \backslash (0,0) } \frac{1}{(a^2 + b^2)^s} = 4 \beta(s) \zeta(s) $$

Is this reasoning correct ? And can we generalize this result for others sums, like this one :

$$ \sum_{(a,b) \in \mathbb{Z}^2 \backslash (0,0) } \frac{1}{(a^3 + b^3)^s} $$

?

Answer 1

Recall the definition of the Dedekind $\zeta$-function, and let us specialize to the case: $$ K=\Bbb Q(i)\ ,\qquad\text{ where }i=\sqrt{-1} $$ is an abstract root of the polynomial $X^2+1$. Then $K$ is the field of gaussian numbers, the ring of integers $\mathcal O_K=\Bbb Z[i]$ is the ring of all $a+bi\in K$ with $a,b\in\Bbb Z$. This ring is a unique factorization domain, each ideal $I$ is principal, i.e. $I=(a+ib)$ for some suitable $a+ib\in \mathcal O_K$, and two ideals $(a+ib)$ and $(a'+ib')$ are equal, iff the $\mathcal O_KK$-numbers $a+ib$ and $a'+ib'$ are associated, i.e. are obtained from each other by multiplication with a unit. There are exactly four units, $\pm 1$, $\pm i$. There are four units. So the Dedekind $\zeta$-function $\zeta_K$ is given by: $$ \zeta_K(a)= \sum_{(0)\ne I\le \mathcal O_K}N(I)^{-s} =\sum_{\substack{a+ib\ne 0\\\text{modulo}\\\text{association}}}N(a+ib)^{-s} =\frac 14\sum_{a+ib\ne 0}(a^2+b^2)^{-s}\ . $$ Here, $N=N_{K:\Bbb Q}$ is the norm of $K$ over $\Bbb Q$. On the other side, we can also write the Euler product for $\zeta_K$. Then group the terms in this product as the come from lifting/splittin $\Bbb Z$-primes $p$ to $K$.

• A $\Bbb Z$-prime $p$ which is $1$ modulo four, written as $p=1[4]$, splits uniquely as $p+i0=(a+ib)(a-ib)$ (modulo units) in two prime( ideal)s of $\mathcal O_K$, so $N(a+ib)=N(a-ib)=p$.
• A $\Bbb Z$-prime $p$ which is $3$ modulo four, written as $p=3[4]$, is inert, it is still prime when seen in $\mathcal O_K$, and $N(p)=N(a-ib)=p$.
• The remained $\Bbb Z$-prime $2$ ramifies as $2=(1+i)(1-i)\sim(1\pm i)^2$, so $N(1\pm i)=2$, and the prime $1\pm i$ contributes once with norm $2$,

So using Euler products, $$ \begin{aligned} \zeta_K(s) &= \prod_{\mathfrak p\text{ $\mathcal O_K$-prime}} \frac 1{1-N(\mathfrak p)^{-s}} \\ &= \frac 1{1-2^{-s}} \left( \prod_{\substack{p\text{ $\Bbb Z$-prime}\\p=1[4]}} \frac 1{(1-p^{-s})^2} \right) \left( \prod_{\substack{p\text{ $\Bbb Z$-prime}\\p=3[4]}} \frac 1{1-(p^2)^{-s}} \right) \\ &=\zeta_{\Bbb Q}(s) \left( \prod_{\substack{p\text{ $\Bbb Z$-prime}\\p=1[4]}} \frac 1{1-p^{-s}} \right) \left( \prod_{\substack{p\text{ $\Bbb Z$-prime}\\p=3[4]}} \frac 1{1+p^{-s}} \right) \\ &=\zeta_{\Bbb Q}(s)\cdot L(s,\chi) \ , \end{aligned} $$ where $\chi$ is the corresponding Dirichlet character modulo four.

So far, we are recovering the OP-result using relations between $L$-functions of number fields. The result is correct.

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Regarding the transposition of the one or the other argument to the denominator base $(a^3+b^3)$ instead of $(a^2+b^2)$ (taken with same power under the same summation), i would say that there is no transposition.

• The $L$-series argument is broken, for instance because $(a,b)\to a^3+b^3$ is no norm for some number field (realized as a two dimensional vector space over $\Bbb Q$), not even multiplicative for some subring of some number field.
• The successful count of representations of an integer as sum of two squares can be traced back to the fact that a suitable sum over $q^{n^2}$ is a modular form. However, a similar sum over $q^{n^3}$ has no longer any modular properties. See also the short discussions under the MO question 57717.

In order to say something in the new case "$a^3+b^3$" a new method, a new insight are needed.