# Show that $\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}}$

I wonder why this is true

$$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}}$$

Where the sum omits the case $n = m = 0$ ofcourse.

• For sure, $h(-232)=2$ (see mathworld.wolfram.com/ClassNumber.html) hence there are just two binary reduced quadratic forms of discriminant $-232$, namely $m^2+58n^2$ and $2m^2+29n^2$, so the problem boils down to computing the value of a Dirichlet L-function. – Jack D'Aurizio Oct 12 '17 at 23:53
• The discriminant of $a^2+58b^2$ is $-4\cdot 58=-232$. Similar results do not hold if $58$ is replaced by a random integer, such series encode the arithmetic structure of certain quadratic fields in a very strict way. – Jack D'Aurizio Oct 13 '17 at 0:46
• But a homogeneous quadratic polynomial is not that different. The discriminant of $Ax^2+Bxy+Cy^2$ is defined as $B^2-4AC$ as expected. Would you mind mentioning why you are digging into the theory of integer quadratic forms (and beyond) and your actual attempts? – Jack D'Aurizio Oct 13 '17 at 0:56
• I was Just amazed and inspired. – mick Oct 13 '17 at 1:02
• Using Kronecker limit formula (see previous comment) we get the general formula $$\sum_{m, n\in \mathbb {Z}}'\frac{(-1)^{m}}{m^{2}+pn^{2}}=-\frac{\pi\log(2g_{p} ^{4})}{\sqrt{p}}$$ where $p>0$ and $g_{p}$ denotes one of Ramanujan's class invariants. If $p$ is rational then $g_{p}$ is algebraic. – Paramanand Singh Oct 13 '17 at 12:14

• Let $F=\mathbb{Q}(\sqrt{-58}),\mathcal{O}_F=\mathbb{Z}[\sqrt{-58}]$. Its ideal class group is $C_F= \{ (1),(2,\sqrt{-58})\}$

thus the ideals with their norm are $N((n+\sqrt{-58}m))= n^2+58m^2$, $N(\frac{2n+\sqrt{-58}m}{2}(2,\sqrt{-58}))= \frac12(4n^2+58m^2)$

then your series is $\ \ 2 \ L(1,\psi)$

where $\psi((n+\sqrt{-58}m))=1, \psi(\frac{2n+\sqrt{-58}m}{2}(2,\sqrt{-58}))=-1$ is the Hecke character induced by the non-trivial character of $C_F$.
$$L(s,\psi) = \sum_I \psi(I) N^{-s} =\frac{1}{|\mathcal{O}_F^\times|} \sum_{n,m \in \mathbb{Z}^2}' N((n+\sqrt{-58}m))^{-s}- 2^{s} N((2n+\sqrt{-58}m))^{-s}$$

• Using $|C_F| = 2$ and some class field theory, we find that $H = F(\sqrt{-2})$ is the Hilbert class field of $F$ and $$\zeta_H(s) = \zeta_F(s) L(s,\psi)$$

Now by chance it happens that $H/\mathbb{Q}$ is itself an abelian extension. Thus we can write $\zeta_H$ as a product of Dirichlet L-functions

$$\zeta_H(s) = \zeta(s)\ L(s,{\left(\frac{-58}{.}\right)})\ L(s,{\left(\frac{-2}{.}\right)})\ L(s,{\left(\frac{29}{.}\right)})$$ Together with $\displaystyle\zeta_F(s) = \zeta(s)\ L(s,{\left(\frac{-58}{.}\right)})$ it means $$L(1,\psi) = L(1,{\left(\frac{-2}{.}\right)})\ L(1,{\left(\frac{29}{.}\right)})$$ and we use quadratic reciprocity to write $(\frac{-d}{.}) = (\frac{.}{\Delta})$ and conclude.

• Funny to see a rather innocuous looking sum solved using Hecke Characters and Class Field Theory. +1 – user335907 Oct 13 '17 at 4:45

(Too long for a comment.)

We have,

$$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 10n^2} = - \frac{2\pi \ln( \sqrt2\; U_{5})}{\sqrt {10}}$$

$$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{2\pi \ln( \sqrt2\; U_{29})}{\sqrt {58}}$$

with fundamental units $U_5 = \frac{1+\sqrt5}2$ and $U_{29} = \frac{5+\sqrt{29}}2$.

P.S. Presumably there might be a family for $d = 5,\,13,\,37$.

Courtesy of a comment by Paramanand Singh, we have the closed-form in terms of the Dedekind eta function $\eta(\tau)$ as,

$$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + pn^2} = - \frac{2\pi \ln(\sqrt2\,g_p^2)}{\sqrt {p}} =- \frac{\pi \ln(2\,g_p^4)}{\sqrt {p}}$$ where,

$$g_p = 2^{-1/4}\frac{\eta(\tfrac12\sqrt{-p})}{\eta(\sqrt{-p})}$$

In particular, $$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 6n^2} = - \frac{2\pi \ln\big(\sqrt2\,(1+\sqrt2)^{1/3}\big)}{\sqrt{6}}$$ $$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 22n^2} = - \frac{2\pi \ln\big(\sqrt2\,(1+\sqrt2)\big)}{\sqrt{22}}$$ and more complicated ones for $d=5,\,13,\,37$.

• Glad to see your update. When there are too my comments people may not read all of them. Putting that comment as a part of answer (with additional details) will make it more accessible. I had already delivered my +1 so just thanks for now. – Paramanand Singh Oct 13 '17 at 19:19
• @ParamanandSingh: I upvoted your comment with the formula so it goes into the top 5. – Tito Piezas III Oct 14 '17 at 12:51
• @ParamanandSingh: I believe there should be a similar formula for $$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{am^2 +amn+p n^2}$$ especially for $a=1$ and $a=2$. – Tito Piezas III Oct 14 '17 at 15:21
• I am not sure, perhaps some algebraic number theory expert can help. You may ask it as a separate question. It might be possible to transform $am^{2}+amn+pn^{2}$ into $rM^{2}+sN^{2}$ where $M, N$ are linear combinations of $m, n$ in such a manner that if $m, n$ take all integer values then $M, N$ also take all integer values. This will reduce the problem to current one. – Paramanand Singh Oct 14 '17 at 15:33
• @ParamanandSingh: I've asked the question in this post. – Tito Piezas III Oct 18 '17 at 16:26

Finally I managed to sum this series using Ramanujan's class invariants. We have the definition $$g(q) = 2^{-1/4}q^{-1/24}\prod_{n = 1}^{\infty}(1 - q^{2n - 1}), \, g_{p} = g(e^{-\pi\sqrt{p}})\tag{1}$$ Ramanujan established that if $p$ is a positive rational number then $g_{p}$ is an algebraic real number. Moreover we have the non-trivial identity $$g_{4/p} = 1/g_{p}\tag{2}$$ Another ingredient we need is the formula $$\sin \pi z = \pi z\prod_{n = 1}^{\infty}\left(1 - \frac{z^{2}}{n^{2}}\right)\tag{3}$$ Taking logs and differentiating with respect to $z$ we get $$\pi\cot\pi z = \frac{1}{z} -2z \sum_{n = 1}^{\infty}\frac{1}{n^{2} - z^{2}}$$ Replacing $z$ by $iz$ we get $$\pi\coth\pi z = \frac{1}{z} + 2z\sum_{n = 1}^{\infty}\frac{1}{n^{2} + z^{2}}$$ The above sum can be written as $$\sum_{n = 1}^{\infty}\frac{1}{n^{2} + z^{2}} = \frac{\pi}{2z} - \frac{1}{2z^{2}} + \frac{\pi e^{-2\pi z}}{z(1 - e^{-2\pi z})}\tag{4}$$ Consider the sum in question \begin{align} S(p) &= \sum_{m, n\in\mathbb{Z}}'\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\ &= 2\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2}} + 2\sum_{n = 1}^{\infty}\sum_{m\in\mathbb{Z}}\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\ &= 2\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2}} + \frac{2}{p}\sum_{n = 1}^{\infty}\frac{1}{n^{2}}+ 4\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\ &=\frac{(2- p)\pi^{2}}{6p} + 4\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\ &=\frac{(2- p)\pi^{2}}{6p} + 4\sum_{m = 1}^{\infty}(-1)^{m}\sum_{n = 1}^{\infty}\frac{1}{m^{2} + pn^{2}}\notag\\ &= \frac{(2- p)\pi^{2}}{6p} + \frac{4}{p}\sum_{m = 1}^{\infty}(-1)^{m}\sum_{n = 1}^{\infty}\frac{1}{m^{2}(1/p) + n^{2}}\notag\\ &= \frac{(2- p)\pi^{2}}{6p} + \frac{4}{p}\sum_{m = 1}^{\infty}(-1)^{m}\left(\frac{\pi\sqrt{p}}{2m} - \frac{p}{2m^{2}} + \frac{\pi\sqrt{p} e^{-2\pi m/\sqrt{p}}}{m(1 - e^{-2\pi m/\sqrt{p}})}\right)\text{ (using equation }(4)) \notag\\ &=\frac{\pi^{2}}{3p}-\frac{2\pi\log 2}{\sqrt{p}}+\frac{4\pi}{\sqrt{p}}\sum_{m=1}^{\infty} \frac{(-1)^{m}q^{m}}{m(1-q^{m})},\,q=e^{-2\pi/\sqrt{p}}\notag\\ &=\frac{\pi^{2}}{3p}-\frac{\pi\log 4}{\sqrt{p}}+\frac{4\pi}{\sqrt{p}}(a(q^{2})-a(q))\notag\\ &\, \, \, \, \,\,\,\,\text{where }a(q) =\sum_{n=1}^{\infty} \frac{q^{n}} {n(1-q^{n})}=-\log\prod_{n=1}^{\infty}(1-q^{n})\notag\\ &=\frac{\pi^{2}}{3p}-\frac{\pi\log 4} {\sqrt{p}}+\frac{4\pi}{\sqrt{p}}\log\prod_{n=1}^{\infty}(1-q^{2n-1}) \notag\\ &=\frac{\pi^{2}}{3p}-\frac{\pi}{\sqrt {p}} \log\frac{2}{q^{1/6}g^{4}(q)}\text{ (via equation }(1)) \notag\\ &= -\frac{\pi\log(2/g_{4/p}^{4})}{\sqrt{p}}\notag\\ &= -\frac{\pi\log (2g_{p}^{4})}{\sqrt{p}}\text{ (using equation }(2))\notag \end{align} It is well known that $g_{58} = \sqrt{(5 + \sqrt{29})/2}$ and this gives the desired closed form for $S(58)$. The above technique can also be used (with some more effort) to prove the Kronecker's second limit formula.

The function $a(q)$ used above is related to work of Simon Plouffe. See this answer for details.

• nice answer (+1)...i was fiddeling around with dedekind etas but gave up. they seem to be quiet closely related to the class invariants you are using.. – tired Oct 21 '17 at 9:40
• @tired: oh yes the class invariants can be expressed in terms of eta functions. For the current case just note that $\prod(1-q^{2n-1})=\prod(1-q^{n})/(1-q^{2n})$ so that it is essentially $\eta(q) /\eta(q^{2})$. – Paramanand Singh Oct 21 '17 at 10:30