Moments of elliptic integrals and critical $L$-values

Abstract

We compute the critical $L$-values of some weight 3, 4, or 5 modular forms, by transforming them into integrals of the complete elliptic integral $K$. In doing so, we prove closed-form formulas for some moments of $K'^3$. Many of our $L$-values can be expressed in terms of Gamma functions, and we also obtain new lattice sum evaluations.

1 Introduction

Let $K$ and $K'$ denote the complete elliptic integrals of the first kind, defined as follows:

$$\begin{aligned} K(k):&= \int _{0}^{1}\frac{{\mathrm d}u}{\sqrt{(1-u^2)(1-k^2u^2)}} =\frac{\pi }{2}\, {_2F_1}\left( {\begin{array}{c} \frac{1}{2},\frac{1}{2}\\ 1 \end{array};k^{2}}\right) , \\ K'(k):&= K(k'), \quad k' := \sqrt{1-k^2}. \end{aligned}$$

One of the present authors (Wan) studied integrals of elliptic integrals, or “moments of elliptic integrals” in [17]. That paper includes several conjectures which have since been settled. The last remaining conjecture is

$$\begin{aligned} \int _{0}^{1}K'(k)^3 {\mathrm d}k=\frac{\Gamma ^8\left( \frac{1}{4}\right) }{128\pi ^2}. \end{aligned}$$

(1)

There are several versions of (1). For instance, we can reformulate the integral using transformations for $K'$. We can also use integration by parts, because derivatives of elliptic integrals equal linear combinations of elliptic integrals. Some examples found in [17] include

$$\begin{aligned} \int _0^1 K'(k)^3 {\mathrm d}k = \frac{10}{3} \int _0^1 K(k)^3 {\mathrm d}k = 5 \int _0^1 k K'(k)^3 {\mathrm d}k. \end{aligned}$$

(2)

In the first part of this paper, we prove formula (1), and provide some intuition about how to discover related integrals. We also settle some additional integrals involving $K'^3$, and give closed-form evaluations of integrals containing higher powers of $K'$ (see Eqs. (20) and (21)). In the second part of the paper, we study a more general phenomenon, where critical $L$-values of odd weight modular forms can be expressed in terms of Gamma values. Some new lattice sums are produced from our investigation.

We note that Y. Zhou, in a 2013 paper [19], used methods based on spherical harmonics to prove both Eqs. (1) and (12). Zhou also applied his ideas to many other integrals.

2 Critical $L$-values

One of the main goals of this paper is to connect integrals like (1) to critical $L$-values of modular forms. We say that a function $f(\tau )$ is a modular form of weight $k$ and level $N$, if it satisfies

$$\begin{aligned} f\left( \frac{a\tau +b}{c\tau +d}\right) =(c\tau +d)^k f(\tau ), \end{aligned}$$

whenever $\left( {\begin{array}{c} a \\ c \end{array}\quad \begin{array}{c} b \\ d \end{array}}\right) \in SL_2(\mathbb {Z})$, and $c\equiv 0\mod N$. We also require that $f(\tau )$ be holomorphic in the upper half plane, and that it vanish at the cusps (that is, it is a cusp form). The most interesting modular forms are the Hecke eigenforms, whose Fourier series

$$\begin{aligned} f(\tau )=\sum _{n=1}^{\infty }a_n e^{2\pi i n \tau } \end{aligned}$$

have multiplicative coefficients $a_n$. If we attach an $L$-series to $f(\tau )$,

$$\begin{aligned} L(f,s)=\sum _{n=1}^{\infty }\frac{a_n}{n^s}, \end{aligned}$$

(3)

then $L(f,s)$ has a meromorphic continuation to the complex plane. We say that $L(f,j)$ is a critical $L$-value if $j\in \{1,2,\dots ,k-1\}$. These numbers typically hold arithmetic significance; some properties of critical $L$-values are summarized in [9]. For instance, the Birch and Swinnerton-Dyer conjecture predicts the value of $L(f,1)$, whenever $f(\tau )$ is a weight 2 cusp form attached to an elliptic curve. In Theorem 1, we prove that Eq. (1) is equivalent to an explicit formula for a critical $L$-value of a weight $5$ cusp form. In particular, we prove

$$\begin{aligned} 30L(g,4)=\int _{0}^{1}K'(k)^3 {\mathrm d}k, \end{aligned}$$

(4)

where

$$\begin{aligned} g(\tau ):=\eta ^4(\tau ) \eta ^2(2\tau )\eta ^4(4\tau ), \end{aligned}$$

and $\eta (\tau )$ is the usual Dedekind eta function. We then use properties of $g(\tau )$ to prove Eq. (1) in Theorem 2.

Formula (4) naturally suggests looking at critical $L$-values of different modular forms. Martin has classified all of the possible multiplicative eta quotients [10]. While it is possible to consider a much larger class of modular forms than just multiplicative eta quotients, these are typically the easiest examples to work with. Martin’s list contains precisely two examples of weight $5$. The first is $g(\tau )$ above, and the second example is

$$\begin{aligned} h(\tau ):=\frac{\eta ^{38}(8\tau )}{\eta ^{14}(4\tau )\eta ^{14}(16\tau )}. \end{aligned}$$

It is standard to prove, using formulas such as (9) below, that

$$\begin{aligned} 192L(h,4)=\int _{0}^{1}\frac{K'(k)^3}{\sqrt{k}(1-k^2)^{3/4}}{\mathrm d}k, \end{aligned}$$

and based on the previous example, we guess that $L(h,4)$ should also be related to Gamma values. We discovered the following identity after a brief numerical search:

$$\begin{aligned} \int _{0}^{1}\frac{K'(k)^3}{\sqrt{k}(1-k^2)^{3/4}}{\mathrm d}k = \frac{3\Gamma ^8(\frac{1}{4})}{32\sqrt{2}\pi ^2}. \end{aligned}$$

(5)

We started by calculating the integral on the left to high numerical precision (denoted $I$), and then we used the PSLQ algorithm to search for a linear dependencies in the set

$$\begin{aligned} \left\{ \log |I|,~\log \pi ,~\log \Gamma (1/3),~\log \Gamma (1/4),\dots ,~\log 2,~\log 3,~\log 5,\dots \right\} . \end{aligned}$$

Bailey and Borwein [1] used PSLQ to discover many identities among moments of elliptic integrals – far more than what is currently proven.

The crucial property which allows us to prove (1) and (5) is that the attached modular forms are also binary theta functions. It is often possible to rewrite the $L$-functions of the modular forms as Hecke $L$-functions involving Grossencharacters, but this connection is not usually essential. In Section 4, we describe some additional integrals which arise from weight $3$ cusp forms. We think that it is noteworthy that there appears to be fewer interesting formulas for $L$-values attached to even weight cusp forms. We discuss this in the conclusion.

3 Proof of the conjectures

We relate conjecture (1) to critical $L$-values in the following theorem. The proof is typical of the approach we use for subsequent integrals, so we spell out the details.

Theorem 1

Let $g(\tau )=\eta ^4(\tau )\eta ^2(2\tau )\eta ^4(4\tau )$. Formula (1) is equivalent to

$$\begin{aligned} L(g,4)=\frac{\Gamma ^8\left( \frac{1}{4}\right) }{3840 \pi ^2}. \end{aligned}$$

(6)

Proof

The proof follows from Ramanujan-style manipulations. Set $k=\sqrt{\alpha }$, and notice that

$$\begin{aligned} \int _{0}^{1}K'(k)^3 {\mathrm d}k=\frac{\pi ^3}{16}\int _{0}^{1}{_2F_1}\biggl (\begin{array}{c} \frac{1}{2},\frac{1}{2} \\ 1 \end{array};1-\alpha \biggr )^3 \frac{{\mathrm d}\alpha }{\sqrt{\alpha }}. \end{aligned}$$

Now we make a change of variables. Set

$$\begin{aligned} q=\exp \biggl (-\pi \frac{{_2F_1}\bigl (\begin{array}{c} \frac{1}{2},\frac{1}{2}\\ 1 \end{array};1-\alpha \bigr )}{{_2F_1}\bigl (\begin{array}{c} \frac{1}{2},\frac{1}{2}\\ 1 \end{array};\alpha \bigr )}\biggr ), \end{aligned}$$

(7)

and notice that [3, Eqs. (2.3.10)]

$$\begin{aligned} {\mathrm d}\alpha =\alpha (1-\alpha ){_2F_1}\biggl (\begin{array}{c} \frac{1}{2},\frac{1}{2} \\ 1 \end{array};\alpha \biggr )^2\frac{{\mathrm d}q}{q}. \end{aligned}$$

(8)

It is standard to show that $q\in (0,1)$ when $\alpha \in (0,1)$, and $q$ is monotone with respect to $\alpha$. The integral becomes

$$\begin{aligned} \int _{0}^{1}K'(k)^3 {\mathrm d}k=-\frac{1}{16}\int _{0}^{1}\sqrt{\alpha }(1-\alpha ){_2F_1}\biggl (\begin{array}{c} \frac{1}{2},\frac{1}{2} \\ 1 \end{array};\alpha \biggr )^5\log ^3 q\frac{{\mathrm d}q}{q}. \end{aligned}$$

Consider the Dedekind eta function with respect to $q$, where $q=e^{2\pi i \tau }$:

$$\begin{aligned} \eta (q):=q^{1/24}\prod _{n=1}^{\infty }(1-q^n). \end{aligned}$$

By [2][p. 124, Entry 12], we have

$$\begin{aligned} \sqrt{\alpha }(1-\alpha ){_2F_1}\biggl (\begin{array}{c} \frac{1}{2},\frac{1}{2} \\ 1 \end{array};\alpha \biggr )^5=4\frac{\eta ^{14}(q^2)}{\eta ^4(q^4)}, \end{aligned}$$

(9)

and the integral reduces to

$$\begin{aligned} \int _{0}^{1}K'(k)^3 {\mathrm d}k=-\frac{1}{4}\int _{0}^{1}\frac{\eta ^{14}(q^2)}{\eta ^4(q^4)}\log ^3 q\frac{{\mathrm d}q}{q}=-4\int _{0}^{1}\frac{\eta ^{14}(q^4)}{\eta ^4(q^8)}\log ^3 q\frac{{\mathrm d}q}{q}. \end{aligned}$$

From [16], Entry $t_{8,12,48}$ and Entry $t_{8,18,60a}$], we can show that

$$\begin{aligned} \frac{\eta ^{14}(q^4)}{\eta ^4(q^8)}=4\eta ^4(q^2)\eta ^2(q^4)\eta ^4(q^8) + \eta ^4(q)\eta ^2(q^2)\eta ^4(q^4). \end{aligned}$$

Therefore, the integral becomes

$$\begin{aligned} \int _{0}^{1}K'(k)^3 {\mathrm d}k=&-16\int _{0}^{1}\eta ^4(q^2)\eta ^2(q^4)\eta ^4(q^8)\log ^3 q\frac{{\mathrm d}q}{q}-4\int _{0}^{1} \eta ^4(q)\eta ^2(q^2)\eta ^4(q^4)\\ \log ^3 q\frac{{\mathrm d}q}{q}=&-5\int _{0}^{1}\eta ^4(q)\eta ^2(q^2)\eta ^4(q^4)\log ^3 q\frac{{\mathrm d}q}{q}. \end{aligned}$$

Switching to the more traditional notation for $\eta$ in terms of $\tau$ , we have

$$\begin{aligned} \int _{0}^{1}K'(k)^3 {\mathrm d}k=30L(g,4), \end{aligned}$$

where $g(\tau )=\eta ^4(\tau )\eta ^2(2\tau )\eta ^4(4\tau )$. It follows that (1) and (6) are equivalent. $\square$

In order to prove (6), we require the fact that $g$ is a binary theta function. Glaisher [7] showed that

$$\begin{aligned} g(\tau )=\frac{1}{4}\sum _{(n,m)\in \mathbb {Z}^2}(n-i m)^4 q^{n^2+m^2}, \end{aligned}$$

(10)

where as usual $q=e^{2\pi i \tau }$.

Theorem 2

Formula (6) is true.

Proof

Using the definition (3) and Eq. (10), we have

$$\begin{aligned} L(g,4)=\frac{1}{4}\sum _{(n,m)\ne (0,0)}\frac{(n-i m)^4}{(n^2+m^2)^4}=\frac{1}{4}\sum _{(n,m)\ne (0,0)}\frac{1}{(n+i m)^4}. \end{aligned}$$

The Weierstrass invariant $g_2(\tau )$ can be defined by

$$\begin{aligned} g_2(\tau ):=60 \sum _{(n,m)\ne (0,0)}\frac{1}{(n+\tau m)^4}, \end{aligned}$$

and can be calculated using

$$\begin{aligned} g_2(\tau )=\frac{64}{3}(1-k^2+k^4)K^4(k), \end{aligned}$$

(11)

where $k, K$, and $\tau$ obey the classical relations $k=\theta _2^2(e^{\pi i \tau })/\theta _3^2(e^{\pi i \tau })$ and $K = \frac{\pi }{2}\theta _3^2(e^{\pi i \tau })$. In the language of Eisenstein series, $g_2(\tau )=120 \zeta (4) E_4(\tau )$.

Standard evaluations show that $k=1/\sqrt{2}$ when $\tau =i$, and thus

$$\begin{aligned} L(g,4)=\frac{1}{240}g_2(i)=\frac{1}{15}K^4\biggl (\frac{1}{\sqrt{2}}\biggr ). \end{aligned}$$

Since $K(1/\sqrt{2})=\Gamma ^2(\frac{1}{4})/(4\sqrt{\pi })$, we obtain

$$\begin{aligned} L(g,4)=\frac{\Gamma ^8(\frac{1}{4})}{3840\pi ^2}, \end{aligned}$$

completing the proof. $\square$

Remark 1

Wan used numerical experiments to observe that the moments of $K'^3$ and $K^2 K'$ are related by a rational factor [17]:

$$\begin{aligned} \int _0^1 K'(k)^3\mathrm dk = 3 \int _0^1 K(k)^2 K'(k)\mathrm dk. \end{aligned}$$

(12)

We sketch a proof of Eq. (12) here. We can use Tricomi’s Fourier series [17, Sect. 6]

$$\begin{aligned} K(\sin t) = \sum _{n=0}^\infty \frac{\Gamma (n+\frac{1}{2})^2}{\Gamma (n+1)^2} \sin ((4n+1)t), \end{aligned}$$

to deduce that

$$\begin{aligned} \sum _{n=0}^\infty \frac{\Gamma ^2(n+\frac{1}{2})}{\Gamma ^2(n+1)}\int _0^{\pi /2} \cos ((4n+2)t)\cos (t)F(\sin t)\mathrm dt = \int _0^1 \bigl (k' K'(k)-k K(k)\bigr )F(k)\mathrm dk, \end{aligned}$$

(13)

whenever $F$ is selected so that summation and integration are interchangeable. To derive (13), perform the change of variables $k \mapsto \cos t$ on the right-hand side, and then use the trigonometric identity

$$\begin{aligned} \cos (t)\cos ((4n+1)t)-\sin (t)\sin ((4n+1)t) = \cos ((4n+2)t). \end{aligned}$$

We then set $F(k) = K^2(k)/k'$ in (13). The right-hand side simplifies under the transformation $k \mapsto k'$, and cancellations occur on the left-hand side due to orthogonality. After simplifying, we have

$$\begin{aligned} \frac{\pi ^2}{8}\sum _{n=0}^\infty \frac{\Gamma ^4(n+\frac{1}{2})}{\Gamma ^4(n+1)} \, _4F_3\biggl ( \begin{array}{c} \frac{1}{2},\frac{1}{2},-n,-n\\ 1,\frac{1}{2}-n,\frac{1}{2}-n \end{array};1\biggr )= \int _0^1 K'(k)^3-K(k)^2 K'(k)\mathrm dk. \end{aligned}$$

The sum is precisely $2\int _0^1 K(k)^2 K'(k)\mathrm dk$, as established near the end of [17]. This proves (12).

Closed forms such as (1) and (12), together with the use of Legendre’s relation in [17], produce additional evaluations involving the complete elliptic integral of the second kind $E$, for example

$$\begin{aligned} \int _0^1 E(k)K'(k)^2 \mathrm dk = \frac{\pi ^3}{12}+\frac{\Gamma ^8(\frac{1}{4})}{384\pi ^2}. \end{aligned}$$

$\Diamond$

Before proving (5), we note that there are various reformulations of the integral. For example, a quadratic transformation gives

$$\begin{aligned} \int _{0}^{1}\frac{K'(k)^3}{\sqrt{k}(1-k^2)^{3/4}}{\mathrm d}k =\frac{1}{8\sqrt{2}} \int _0^1 \frac{(1+k)^3K'(k)^3}{k^{3/4}\sqrt{1-k}}{\mathrm d}k. \end{aligned}$$

Experimentally, we discover the binary theta expansion

$$\begin{aligned} h(\tau )&= \prod _{n=1}^\infty \frac{(1-(-1)^n q^{4n})^{14}}{(1-q^{8n})^4} \end{aligned}$$

(14)

$$\begin{aligned}&= \frac{1}{2} \sum _{m, n = -\infty }^\infty (-1)^m (2n+1-2i m)^4 \, q^{(2m)^2+(2n+1)^2}. \end{aligned}$$

(15)

The proof of (15) is slightly tedious. We may show that both sides are modular forms on $\Gamma _0(16)$, that their Fourier coefficients agree for sufficiently many terms, and then appeal to the valence formula. Alternatively, the double sum can be built up from formulas of derivatives of $\theta _2(q^4)$ and $\theta _4(q^4)$ (Ramanujan’s $\varphi (-q^4)$ and $\psi (q^8)$) in [2, pp. 122–123]. The result can be written in terms of $k$ and $K$, and then reduced to the product (14).

Theorem 3

Formula (5) is true.

Proof

From (15), we have the (lattice) sum

$$\begin{aligned} 2L(h,4)= \sum _{m, n = -\infty }^\infty \frac{(-1)^m(2n+1-2im)^4}{\bigl ((2n+1)^2+(2m)^2\bigr )^4} = \sum _{m, n = -\infty }^\infty \frac{(-1)^m}{(2n+1+2im)^4}, \end{aligned}$$

so it remains to show

$$\begin{aligned} \sum _{m, n = -\infty }^\infty \frac{(-1)^m}{(2n+1+2im)^4} = \frac{\Gamma ^8(\frac{1}{4})}{1024\sqrt{2} \pi ^2}. \end{aligned}$$

(16)

This can be achieved using routine manipulations on $g_2(\tau )$. By (11), we have

$$\begin{aligned} \sum _{(m,n) \ne (0,0)} \frac{(-1)^m}{(n+m\tau )^4} = \frac{1}{60}\bigl (2g_2(2\tau )-g_2(\tau )\bigr ). \end{aligned}$$

This leads to

$$\begin{aligned} \sum _{m,n} \frac{(-1)^m}{(2n+1+m\tau )^4}&= \frac{1}{60}\bigl (2g_2(2\tau )-g_2(\tau )\bigr ) - \frac{1}{2^4}\frac{1}{60}\bigl (2g_2(\tau )-g_2(\tau /2)\bigr ) \nonumber \\&= \frac{1}{960}\bigl (g_2(\tau /2)-18g_2(\tau )+32g_2(2\tau )\bigr ). \end{aligned}$$

(17)

Now (16) follows by setting $\tau =2i$ in (17) and simplifying. The required values of $k$ (which are the 1st, 4th, and 16th singular values) and $K$ at these values can be found in [2, 3], and [21]. $\square$

Example 1

We may look at other critical $L$-values of $g$ and $h$. Simple calculations give

$$\begin{aligned} L(g,3)&= \frac{1}{2\pi } \int _0^1 k K'(k)^2 K(k){\mathrm d}k, \\ L(g,2)&= \frac{1}{\pi ^2} \int _0^1 k K(k)^2 K'(k){\mathrm d}k, \\ L(g,1)&= \frac{1}{\pi ^3} \int _0^1 k K(k)^3{\mathrm d}k. \end{aligned}$$

One may notice the similarity between $L(g,3)$ and $L(g,2)$; indeed they are related by the substitution $k \mapsto k'$. This similarity can be explained, since $g$ is a weight 5 modular form, so there is a functional equation connecting $L(g, s)$ and $L(g,5-s)$ (with some Gamma factors). In view of (12), (2), and [17, Thm. 5], the values $\pi ^{5-s} L(g,s)$ where $s \in \{1,2,3,4\}$ are all related to each other by rational constants.

Similarly, we have $L(h,1) = 384/\pi ^3 \, L(h,4)$, which can be explained by either using the functional equation or a substitution in the integral. Experimentally, we also observe that $L(h,4) = \frac{\pi }{4} L(h,3)$; an integral formulation for this is

$$\begin{aligned} \int _{0}^{1}\frac{K'(k)^2K(k)}{\sqrt{k}(1-k^2)^{3/4}}{\mathrm d}k = \frac{\Gamma ^8(\frac{1}{4})}{32\sqrt{2}\pi ^2}, \end{aligned}$$

(18)

and an equivalent formulation as a sum is

$$\begin{aligned} \sum _{m, n = -\infty }^\infty \frac{(-1)^m(2n+1-2mi)}{(2n+1+2im)^3} = \frac{\Gamma ^8(\frac{1}{4})}{256\sqrt{2} \pi ^2}. \end{aligned}$$

We are very grateful to Y. Zhou, who has kindly supplied us with a proof of (18) (plus some generalizations) using techniques from [19]; since such techniques differ significantly from the ones in the current paper, we do not include the proof here. Our approach to (18) is modular (using some ideas from [13]), but is more complicated, and we will present it in future work. In general, we expect the critical $L$-values of odd weight modular forms to be related by algebraic constants and powers of $\pi$, though the computation of these constants is not trivial—see the conclusion for discussion. $\Diamond$

Remark 2

Generalizations of (1) are possible. We start with the level 4, weight 9 modular form [18, Part 1C]

$$\begin{aligned} f(q) = \frac{1}{4} \sum _{m,n} (m-in)^8 q^{m^2+n^2}. \end{aligned}$$

Expanding the brackets in $f(q)$ binomially, we see that each sum in the expansion is a product of $\theta _3(q) = \sum _n q^{n^2}$ and its derivatives. Using $k=\theta _2^2(q)/\theta _3^2(q)$, we conclude that

$$\begin{aligned} f(q) = \frac{8(4k^2k'^2+k^4k'^4)}{\pi ^9}K^9(k). \end{aligned}$$

(19)

We also have the following $L$-value:

$$\begin{aligned} L(f,8) = \frac{1}{4} \sum _{(n,m)\ne (0,0)} \frac{1}{(m+in)^8} = \frac{\Gamma ^{16}(\frac{1}{4})}{2^{10}\,525\,\pi ^4}; \end{aligned}$$

the last equality holds, since as an Eisenstein series the sum equals $\frac{1}{2} \zeta (8) E_8(i)$ and $E_8=E_4^2$. Writing $L(f,8)$ as an integral using (19), we obtain

$$\begin{aligned} \int _0^1 k(4+k^2-k^4)K'(k)^7\mathrm {d}k = \frac{3\, \Gamma ^{16}(\frac{1}{4})}{2^{12}\,5\,\pi ^4}. \end{aligned}$$

(20)

Experimentally, $L(f,8-i)/\pi ^i$ are all related by rational constants. Since

$$\begin{aligned} \sum _{(n,m)\ne (0,0)} \frac{1}{(m+in)^{4k}} \end{aligned}$$

can always be expressed as a rational number times a power of $K(1/\sqrt{2})$ [8], we have found generalizations of the above result which involve higher powers of $K'$, for instance

$$\begin{aligned} \int _0^1 k(16-92k^2+93k^4-2k^6+k^8)K'(k)^{11}\mathrm {d}k = \frac{189\, \Gamma ^{24}(\frac{1}{4})}{2^{15}\,65\,\pi ^6}. \end{aligned}$$

(21)

Moments of higher powers of $K'$, the observation about rational constants above, as well as many other integrals involving the complete elliptic integrals will be elaborated in a future paper.¹ $\Diamond$

4 Weight $3$ cases and lattice sums

In this section, we note some additional formulas for critical $L$-values of weight $3$ cusp forms. The ideas for the proof below are borrowed from [11].

Theorem 4

Suppose that $f(\tau )=\eta ^3(r\tau )\eta ^3(s \tau )$, where $r+s\equiv 0 \ (\mathrm {mod} \ 8)$. Then,

$$\begin{aligned} L(f,2)= \frac{8\pi ^2}{\sqrt{rs^3}}\biggl (\sum _{n=0}^{\infty }(-1)^{\frac{n(n+1)}{2}}q^{\frac{(2n+1)^2}{8}}\biggr )^4 = \frac{8}{\sqrt{rs^3}} \, k_{r/s} \, k'_{r/s} \, K^2(k_{r/s}), \end{aligned}$$

(22)

where $q=e^{-\pi \sqrt{r/s}}$, and $k_p$ denotes the $p$th singular value of $K$.

Proof

The Jacobi triple product gives

$$\begin{aligned} \eta ^3(\tau )=\sum _{n=1}^{\infty }n\chi _{-4}(n)e^{\frac{2\pi i n^2 \tau }{8}}. \end{aligned}$$

If we write $L(f,2)$ as a real-valued integral, then

$$\begin{aligned} L(f,2)=4\pi ^2\int _{0}^{\infty }\sum _{n,k\ge 1}n k \chi _{-4}(n)\chi _{-4}(k)e^{-\frac{2\pi r n^2 u}{8}-\frac{2\pi s k^2 u}{8}}u {\mathrm d}u. \end{aligned}$$

Applying the involution for the weight $3/2$, theta function leads to

$$\begin{aligned} L(f,2)=\frac{4\pi ^2}{\sqrt{s^3}}\int _{0}^{\infty }\sum _{n,k\ge 1}n k \chi _{-4}(n)\chi _{-4}(k)e^{-\frac{2\pi r n^2 u}{8}-\frac{2\pi k^2}{8s u}}\frac{{\mathrm d}u}{\sqrt{u}}. \end{aligned}$$

By absolute convergence, we evaluate the integral first using standard results – which produces a Bessel $K$ function with order $1/2$. Simplifying, we get

$$\begin{aligned} L(f,2)=&\frac{8\pi ^2}{\sqrt{r s^3}}\sum _{n,k\ge 1}k \chi _{-4}(k)\chi _{-4}(n)q^{\frac{n k}{2}} =\frac{8\pi ^2}{\sqrt{r s^3}}\sum _{k\ge 1}\frac{k \chi _{-4}(k)q^{\frac{k}{2}}}{1+q^k}\\ =&\frac{8\pi ^2}{\sqrt{r s^3}}\biggl (\sum _{k=0}^{\infty }(-1)^{\frac{k(k+1)}{2}}q^{\frac{(2k+1)^2}{8}}\biggr )^4. \end{aligned}$$

The final Lambert series identity follows from [20, Table 1]. The connection with singular values follows from standard theta function manipulations, leading to

$$\begin{aligned} L(f,2) = \frac{2\pi ^2}{\sqrt{rs^3}} \theta _2^2(q)\theta _4^2(q), \end{aligned}$$

and from the fact that $k_p = \theta _2^2(e^{-\pi \sqrt{p}})/\theta _3^2(e^{-\pi \sqrt{p}})$. $\square$

Theorem 5

The following evaluations are true:

$\varvec{f(\tau )}$	$\varvec{L(f,2)}$
$\eta ^6(4\tau )$	$\displaystyle \frac{\Gamma ^4(\frac{1}{4})}{64\pi }$
$\eta ^3(2\tau )\eta ^3(6\tau )$	$\displaystyle \frac{\Gamma ^6(\frac{1}{3})}{2^{17/3} \pi ^2}$
$\eta ^3(\tau )\eta ^3(7\tau )$	$\displaystyle \frac{ \Gamma ^2(\frac{1}{7})\Gamma ^2(\frac{2}{7})\Gamma ^2(\frac{4}{7})}{224\pi ^2}$
$\eta ^3(3\tau )\eta ^3(5\tau )\pm \eta ^3(\tau )\eta ^3(15\tau )$	$\displaystyle \frac{\Gamma (\frac{1}{15})\Gamma (\frac{2}{15})\Gamma (\frac{4}{15})\Gamma (\frac{8}{15})}{30\sqrt{54\mp 6}\,\pi }$
$\displaystyle \frac{\eta ^5(4\tau )\eta ^5(8\tau )}{\eta ^2(2\tau )\eta ^2(16\tau )}$	$\displaystyle \frac{\Gamma ^2(\frac{1}{8})\Gamma ^2(\frac{3}{8})}{64\sqrt{2}\pi }$
$\displaystyle \frac{\eta ^{18}(8\tau )}{\eta ^6(4\tau )\eta ^6(16\tau )}$	$\displaystyle \frac{\Gamma ^4(\frac{1}{4})}{32 \sqrt{2} \pi }$
$\eta ^2(\tau )\eta (2\tau )\eta (4\tau )\eta ^2(8\tau )$	$\displaystyle \frac{\Gamma ^2(\frac{1}{8}) \Gamma ^2(\frac{3}{8})}{192 \pi }$

Proof

For the first four entries in the table, we use Theorem 4. The singular values $k_p$, as well as $K(k_p)$, are well tabulated in [2] and [21]. For instance, for the third entry, we need

$$\begin{aligned} k_7 = \frac{\sqrt{2}(3-\sqrt{7})}{8}, \quad K(k_7) = \frac{\Gamma (\frac{1}{7})\Gamma (\frac{2}{7})\Gamma (\frac{4}{7})}{4\root 4 \of {7}\pi }. \end{aligned}$$

In general, $k_p$ is algebraic and $K(k_p)$ involves only algebraic numbers and Gamma functions.

The last three entries are weight 3 cusp forms in Martin’s list [10]. For the third last one, we convert it to the following integral:

$$\begin{aligned} L(f,2)=\frac{1}{4\sqrt{2}}\int _{0}^{1}\frac{K(k)}{\sqrt{k(1-k)}}{\mathrm d}k, \end{aligned}$$

which can be evaluated using the series representation of $K$ and interchanging the order of integration and summation.

For the second last entry, we have

$$\begin{aligned} L(f,2)=\frac{1}{8}\int _{0}^{1}\frac{K(k)}{\sqrt{k}(1-k^2)^{3/4}}{\mathrm d}k. \end{aligned}$$

The evaluation of this integral follows from [17, Eq. (8)]. Indeed, many integrals over a single $K$ have closed forms, and the two integrals we just evaluated can also be done by a computer algebra system.

For the last entry in the table, we obtain

$$\begin{aligned} L(f,2)=\frac{1}{2\sqrt{2}}\int _{0}^{1}\frac{K(k)}{\sqrt{1+k}}{\mathrm d}k. \end{aligned}$$

Denoting the integral by $I_1$, we use the moments of $K'$ found in [17] and a quadratic transformation to produce

$$\begin{aligned} I_1 = \frac{1}{\sqrt{2}} \int _0^1 \frac{K'(x)}{\sqrt{1+x}}\mathrm dx = \frac{1}{2\sqrt{2}} \biggl [\frac{\Gamma ^2(\frac{1}{8})\Gamma ^2(\frac{3}{8})}{16\pi }-{}_4F_3\biggl (\begin{array}{c} \frac{3}{4},1,1,\frac{5}{4} \\ \frac{3}{2},\frac{3}{2},\frac{3}{2} \end{array};1\biggr )\biggr ]. \end{aligned}$$

(23)

Similarly, with the auxiliary integral $I_2 :=\int _0^1 K(x)/\sqrt{x(1+x)}\, \mathrm dx$, we have

$$\begin{aligned} I_2 = \frac{1}{\sqrt{2}} \int _0^1 \frac{K'(x)}{\sqrt{1-x}}\mathrm dx = \frac{1}{2\sqrt{2}} \biggl [\frac{\Gamma ^2(\frac{1}{8})\Gamma ^2(\frac{3}{8})}{16\pi }+{}_4F_3\biggl (\begin{array}{c} \frac{3}{4},1,1,\frac{5}{4} \\ \frac{3}{2},\frac{3}{2},\frac{3}{2} \end{array};1\biggr )\biggr ]. \end{aligned}$$

(24)

Experimentally, it is observed that $I_2=2I_1$, which can be shown as follows.

We know that $I_1/(2\sqrt{2}) = L(f,2)$. On the other hand, it is readily verifiable that $I_2/(2\sqrt{2})=L(f_0,2)$, where $f_0(q):=-f(-q)$. Consequently, by looking at the $q$-expansion of $f_0-f$ and using the fact that the coefficients of $f$ are multiplicative, we deduce that $I_2-I_1=I_1$. The desired evaluation of $I_1$ now follows by combining (23) and (24), and inter alia we also obtain a closed form for the $_4F_3$. $\square$

Remark 3

Here, we retain the notation used in the last part of the proof above. First, with $f(\tau ) = \eta ^2(\tau )\eta (2\tau )\eta (4\tau )\eta ^2(8\tau )$, Somos showed that

$$\begin{aligned} f(\tau ) = \frac{1}{2} \sum _{(n,m) \ne (0,0) } (m^2-2n^2)q^{m^2+2n^2}. \end{aligned}$$

Second, it is known that the series (3) for the $L$-value of a weight $k$ cusp form converges conditionally for some $s < (k+1)/2$.

In particular, the series for $L(f,2)$ converges, and so by looking at the partial sums, we have

$$\begin{aligned} I_1 = \sqrt{2} \, \sum _{(m, n) \ne (0,0)} \frac{m^2-2n^2}{(m^2+2n^2)^2}, \end{aligned}$$

where we sum over expanding ellipses $m^2+2n^2 \le M, \, M \rightarrow \infty$. Similarly,

$$\begin{aligned} I_2 = \sqrt{2} \, \sum _{(m, n) \ne (0,0)} (-1)^{m+1} \frac{m^2-2n^2}{(m^2+2n^2)^2}. \end{aligned}$$

Subtracting the sums gives another proof that $I_2-I_1=I_1$. Since the sum for $I_2$ has better convergence properties (it can be summed over expanding rectangles), we will deal exclusively with alternating versions of the lattice sums we encounter. Note that we can decompose $f$ into weight $3/2$ theta functions $(\eta ^2(\tau )\eta ^2(4\tau )/\eta (2\tau )) \cdot (\eta ^2(2\tau )\eta ^2(8\tau )/\eta (4\tau ))$, and therefore obtain a different series expansion. Such expansions imply the alternating sum identities

$$\begin{aligned} \sum _{(m, n) \ne (0,0)} \frac{(-1)^{m+1} (m^2-2n^2)}{(m^2+2n^2)^2} = \sum _{m,n} \frac{18(-1)^m(3m+1)(3n+1)}{((3m+1)^2+2(3n+1)^2)^2}= \frac{\Gamma ^2(\frac{1}{8})\Gamma ^2(\frac{3}{8})}{48\pi }. \end{aligned}$$

(25)

$\Diamond$

Example 2

The first entry in Theorem 5 ($f(\tau ) = \eta ^6(4\tau )$) corresponds to the result

$$\begin{aligned} L(f,2) = \frac{1}{4} \int _0^1 \frac{K(k)}{\sqrt{1-k^2}}\mathrm {d}k = \frac{\Gamma ^4(\frac{1}{4})}{64\pi }, \end{aligned}$$

(26)

and may be expressed as a lattice sum using a binary theta expansion of $f$:

$$\begin{aligned} \sum _{(m, n) \ne (0,0)} (-1)^{m+1}\frac{m^2-4n^2}{(m^2+4n^2)^2} = \frac{\Gamma ^4(\frac{1}{4})}{32\pi }. \end{aligned}$$

(27)

Combined with well-known lattice sum evaluations [20], consequences of (27) include

$$\begin{aligned}&\sum _{(m, n) \ne (0,0)} \frac{(-1)^m n^2}{(m^2+4n^2)^2} = \frac{\Gamma ^4(\frac{1}{4})}{256\pi }-\frac{3\pi \log 2}{32}, \\&\sum _{(m, n) \ne (0,0)} \frac{(-1)^{m+1} m^2}{(m^2+4n^2)^2} = \frac{\Gamma ^4(\frac{1}{4})}{64\pi }+\frac{3\pi \log 2}{8}. \end{aligned}$$

On the other hand,

$$\begin{aligned} L(f,1) = \frac{1}{\pi } \int _0^1 \frac{K(k)}{\sqrt{1-k^2}}\mathrm {d}k = \frac{\Gamma ^4(\frac{1}{4})}{16\pi ^2}, \end{aligned}$$

which is consistent with the functional equation satisfied by $L(f,s)$.

We note that (26) actually is a specialization of [6, Equation after (5.14)], which states

$$\begin{aligned} \sum _{n=1}^\infty \frac{a_n}{n^2} q^{n/4} = \frac{\pi \sqrt{k}}{4K(k)}\,_3F_2\biggl (\begin{array}{c} \frac{3}{4},\frac{3}{4},1 \\ \frac{5}{4},\frac{5}{4} \end{array};k^2\biggr ), \end{aligned}$$

where $a_n$ are the coefficients in the $q$-expansion of $f$, and as usual $k=\theta _2^2(q)/\theta _3^2(q)$; taking the limit $k \rightarrow 1^-$ and appealing to the Stolz-Cesàro theorem recovers (26). $\Diamond$

Example 3

The second entry in Theorem 5 ($f(\tau ) = \eta ^3(2\tau )\eta ^3(6\tau )$) gives the non-trivial integral evaluation

$$\begin{aligned} L(f,2) = \int _0^1 (3+6p)^{-\frac{1}{2}} K\biggl (\frac{p^{\frac{3}{2}}(2+p)^{\frac{1}{2}}}{(1+2p)^{\frac{1}{2}}}\biggr )\mathrm {d}p = \frac{\Gamma ^6(\frac{1}{3})}{2^{\frac{17}{3}} \pi ^2}, \end{aligned}$$

(28)

where we have used the parametrization of the degree 3 modular equation and multiplier [2, Ch. 19]. We also have $L(f,1) = \sqrt{3}/\pi \, L(f,2)$, which can be shown from (28) either by the functional equation or by a cubic transformation. We note that (28) (after a change of variable) appears in a very different context in [4, Sect. 3].

The lattice sum associated with (28) is

$$\begin{aligned} \sum _{(m, n) \ne (0,0)} (-1)^{m+n+1}\frac{m^2-3n^2}{(m^2+3n^2)^2} = \frac{\Gamma ^6(\frac{1}{3})}{2^{\frac{14}{3}} \pi ^2}. \end{aligned}$$

(29)

For the third entry in Theorem 5, we do not seem to obtain a reasonably concise integral involving $K$ (due to the apparent lack of a parametrization for the degree 7 modular equation). Using the binary theta function for $\eta ^3(\tau )\eta ^3(7\tau )$ [5], we obtain the sum

$$\begin{aligned} \sum _{(m, n) \ne (0,0)} \frac{(-1)^{m}(2n^2-m^2)}{(m^2+mn+2n^2)^2} = \frac{\Gamma ^2(\frac{1}{7})\Gamma ^2(\frac{2}{7})\Gamma ^2(\frac{4}{7})}{56\pi ^2}. \end{aligned}$$

(30)

$\Diamond$

Example 4

An $L$-value of the function $g$ used in the proof of (1) also gives some interesting lattice sum evaluations. Using the closed form for $L(g,3)$, the multiplicativity of the coefficients of $g$, and results in [20], we deduce

$$\begin{aligned} \sum _{(m, n) \ne (0,0)}\frac{(-1)^{m+n}m^2n^2}{(m^2+n^2)^3}&= \frac{\Gamma ^8(\frac{1}{4})}{2^9 \,3\, \pi ^3}-\frac{\pi \log 2}{8}, \nonumber \\ \sum _{(m, n) \ne (0,0)}\frac{(-1)^{m+n}m^4}{(m^2+n^2)^3}&= -\frac{\Gamma ^8(\frac{1}{4})}{2^9 \,3\, \pi ^3}-\frac{3\pi \log 2}{8}, \\ \sum _{(m, n) \ne (0,0)}\frac{(-1)^{m}m^2n^2}{(m^2+n^2)^3}&= -\frac{\Gamma ^8(\frac{1}{4})}{2^{10} \,3\, \pi ^3}-\frac{\pi \log 2}{16}.\nonumber \end{aligned}$$

(31)

$\Diamond$

5 Even weight cases and conclusion

For even weight cusp forms, we do not seem to obtain formulas for the $L$-values in terms of Gamma functions; instead hypergeometric functions are involved. Consider the following pair of weight $4$ examples:

$$\begin{aligned} f_1(\tau )=\frac{\eta ^{16}(4\tau )}{\eta ^4(2\tau )\eta ^4(8\tau )},\qquad \qquad f_2(\tau )=\eta ^4(2\tau )\eta ^4(4\tau ). \end{aligned}$$

All of the critical $L$-values of $f_1$ and $f_2$ reduce to special values of hypergeometric functions. Furthermore, there are some very curious relations between the $L$-values of both cusp forms:

$$\begin{aligned} L(f_1,3)=\frac{\pi }{2}L(f_{2},2)= \frac{\pi ^2}{8} L(f_1,1)=\frac{1}{8}\int _{0}^{1}\frac{K(k)^2}{\sqrt{1-k^2}}{\mathrm d}k = \frac{\pi ^3}{32}\,{_4F_3}\biggl (\begin{array}{c} \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2} \\ 1,1,1 \end{array};1\biggr ).\nonumber \\ \end{aligned}$$

(32)

The last equality follows from [17, Eq. 35]. Similarly, we have

$$\begin{aligned} L(f_2,3)=\frac{\pi }{4}L(f_1,2)=\frac{\pi ^2}{4}L(f_2,1)&= \frac{1}{8} \int _0^1 \frac{K(k)K'(k)}{\sqrt{1-k^2}}{\mathrm d}k=\frac{1}{4} \int _{0}^{1}K(k)^2 {\mathrm d}k \nonumber \\&= \frac{\pi ^4}{128}\,{_7F_6}\biggl (\begin{array}{c} \frac{5}{4},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2} \\ \frac{1}{4},1,1,1,1,1 \end{array};1\biggr ). \end{aligned}$$

(33)

The second last equality comes from [17, Sect. 5.4], while the hypergeometric evaluation follows from [17, Eqs. 3 and 18]. Curiously, this last integral also appears in connection with random walks [4, Sect. 3].

5.1 Conclusion

As developed by Manin and Shimura (see e. g. [15, Thm. 1]), and also recorded in [9, Sect. 3.4], the critical $L$-values of a modular form $f$ satisfy the following property: the ratios $L(f,2k)/L(f,2k-2)$ and $L(f,2k+1)/L(f,2k-1)$ can all be expressed in terms of algebraic (often rational) numbers and powers of $\pi$, where $s$ is the weight and $k=1,2,\ldots$. When the weight $s$ is odd, the functional equation relating $L(f,k)$ to $L(f,s-k)$ then implies that all the critical $L$-values of $f$ are related by constants of said type. The explicit algebraic numbers involved may be found using Rankin’s method, as explained in [14], though the computation is highly non-trivial and tedious.

It was simple to directly verify this property for $f_1$ and $f_2$ studied above. On the other hand, more effort and ad hoc strategies were needed to show that for the weight 5 cusp forms $g$ and $h$, all the critical $L$-values are related by rational multiples and powers of $\pi$. (Nevertheless, we believe that these approaches are still easier than Rankin’s method for the cusp forms concerned.) It would be valuable to find a general and more approachable method for computing the ratio of two critical $L$-values.

We conclude with two more observations and directions for further research:

• (1) It seems that the critical $L$-values of some even weight cusp forms can be expressed as hypergeometric functions, while those of odd weight cusp forms can often be expressed in terms of Gamma functions. It would be interesting to explain this discrepancy, because the Ramanujan zeta function and various other interesting zeta functions are attached to such cusp forms (see e. g. [12]). It would also be illuminating to see if many other critical $L$-values of odd weight cusp forms evaluate in terms of Gamma functions.

• (2) As shown above, $L(f_1,s)/L(f_2,s-1)/\pi$ is a rational number. (Note that $f_2$ is $f_1$ twisted by the non-trivial Dirichlet character of conductor 4.) Are there other pairs of even weight modular forms with the same property? If so, is there a method to find, given one function in the pair, the other function?