# Closed-form of an integral involving a Jacobi theta function, $\int\limits_0^\infty \frac{\theta_4^{n}\left(e^{-\pi x}\right)}{1+x^2} dx$

The Jacobi theta function $$\theta_4$$ is defined by $$\displaystyle \theta_4(q)=\sum_{n \in \mathbb{Z}} (-1)^n q^{n^2} \tag{1}$$ For this question, set $$q=\large e^{-\pi x}$$ and $$\theta_4 \equiv \theta_4(q)$$. Define $$\theta_3(q)=\theta_4(-q)$$.

Using Lambert-Series representation for powers of $$\theta_4$$ (which I will describe in a moment) and integrating term by term, I have obtained a family of neat identities: $$\int_0^{\infty} \frac{\theta_4^2}{1+x^2} dx=1 \tag{2}$$ $$\int_0^{\infty} \frac{\theta_4^4}{1+x^2} dx=\frac{4 \ln2}{\pi} \tag{3}$$ $$\int_0^{\infty} \frac{\theta_4^6}{1+x^2} dx=\frac{16 G}{\pi^2}-\frac23 \tag{4}$$ $$\int_0^{\infty} \frac{\theta_4^8}{1+x^2} dx=\frac{20 \zeta(3)}{\pi^3} \tag{5}$$ Here, $$G$$ is Catalan's constant. These identities check out to a lot of digits, according to Mathematica.

As I mentioned, in order to drive these identities I used Lambert series representations for powers of $$\theta_4$$ which I have found online. For instance:

$$\theta_4^2 = 1+4\sum_{n=1}^{\infty} \frac{(-1)^n q^n}{1+q^{2n}} \tag{6}$$ (See e.g. [1], or use $$r_2(n)=4\sum_{d \mid n} \sin(\frac{\pi}{2} d)$$ ([2]), and switch order of summation)

$$\theta_4^6=1+16\sum_{n=0}^{\infty}\frac{(-1)^n n^2 q^n}{1+q^{2n}}+4\sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)^2 q^{2n+1}}{1+q^{2n+1}} \tag{7}$$ (A proof is given in [3], together with proofs for similar formulas for the $$4$$th power of $$\theta_4$$, and the $$8$$th power.)

I am interested in a closed form for the integral $$I(n)= \int_0^{\infty} \frac{\theta_4(e^{-\pi x})^n}{1+x^2} dx.\tag{8}$$

1st Question

I have searched the web for a similar Lambert series combination for $$\theta_4^{10}$$, but all my efforts remain fruitless. Also, I wasn't able to derive one using the approach taken in [3]. Can we find a formula, possibly in the form of $$(6)$$ or $$(7)$$, for $$\theta_4^{10}(q)$$? Such a formula may be used to find the closed form of $$I(10)$$.

2nd Question

Can we find closed forms for $$I(n)$$ for other values of $$n$$?

In general, can we find a Lambert series representation of $$\theta_4^n$$ for each even $$n$$? If not, for which $$n$$ we can find one and for which we cannot?

References

[1]: Proving $\left(\sum_{n=-\infty}^\infty q^{n^2} \right)^2 = \sum_{n=-\infty}^\infty \frac{1}{\cos(n \pi \tau)}$ ,(Note that $$\theta_4(q)=\theta_3(-q)$$)

[2]: Eric W. Weisstein, Sum of Squares Function,Mathworld.(25),(The line below references proofs). Link

[3]: George E. Andrews, Richard Lewis and Zhi-Guo Liu, An identity relating a theta function to a sum of Lambert series, (7)-(9).

This question has been edited. For more information about how I obtained these results, check the original version of this post.

$$\displaystyle \large \mathbf{Progress}\,\mathbf{Report}$$

Thanks to Paramanand Singh (see his answer below) I've been able to make some progress. Again, we set $$q=e^{-\pi x}$$ and $$\theta_2 \equiv \theta_2(q)= \sum_{n \in \mathbb{Z}} q^{(n+\frac12)^2}\\\theta_3\equiv \theta_3(q)=\sum_{n \in \mathbb{Z}} q^{n^2}\\\theta_4\equiv \theta_4(q)=\sum_{n \in \mathbb{Z}}(-1)^n q^{n^2}$$ We also have the relations $$\theta_2^4+\theta_4^4=\theta_3^4 \\ \theta_2(e^{-\pi/x})=\frac1{\sqrt{x}}\theta_4(e^{-\pi x}) \\ \theta_3(e^{-\pi/x})=\frac1{\sqrt{x}}\theta_3(e^{-\pi x})\tag{9}$$ I will omit all the intermediate steps and just state what I've found so far, and also some other related identities which I've not proven, but match numerically. So, using Singh's formulas (changing $$q$$ to $$-q$$ and rewriting in terms of hyperbolic functions), I get that $$\int_0^{\infty} \frac{\theta_4^{10}}{1+x^2}dx=\frac{768 \beta(4)}{5 \pi^4}-\frac{32}{75}-2\int_0^{\infty} \frac{\theta_2^4\theta_4^6}{1+x^2}dx\tag{10}$$ $$\int_0^{\infty} \frac{\theta_4^{12}}{1+x^2}dx=\frac{450 \zeta(5)}{\pi^5}-\int_0^{\infty} \frac{\theta_2^4\theta_3^4\theta_4^4}{1+x^2}dx\tag{11}$$ Notice that $$(10)$$ and $$(11)$$ are a bit different from the expressions I wrote in the comments on Singh's answer. Numerical evidence suggests that: $$\int_0^{\infty} \frac{\theta_2^4\theta_3^4\theta_4^4}{1+x^2}dx=\frac23\int_0^{\infty} \theta_2^4\theta_3^4\theta_4^4 dx\tag{12}$$ $$\int_0^{\infty} \frac{\theta_2^4\theta_3^4\theta_4^2}{1+x^2}dx=\frac83\int_0^{\infty} \theta_2^2\theta_3^4\theta_4^4 dx\tag{13}$$ $$\int_0^{\infty} \theta_2^4\theta_4^2dx=\int_0^{\infty} \theta_2^2\theta_4^2dx=1\tag{14}$$ $$\int_0^{\infty} \frac{\theta_2^4\theta_4^4}{1+x^2}dx=\frac{8\zeta(3)}{\pi^3}\tag{15}$$ $$\int_0^{\infty} \frac{\theta_2^4\theta_4^2}{1+x^2}dx=\frac23\tag{16}$$ All of these just make me more confident that an ultimate closed form for both $$(10)$$, $$(11)$$ and higher power combinations of theta functions exist. As a side note, these integrals identities can be translated into identities about lattice sums, by expanding the theta functions into their series definition.

• Expressing the function $\vartheta_{3}^{k}(q)$ in the form of a Lambert series is related to finding a formula for expressing a positive integer $n$ as sum of $k$ squares in terms of divisor functions of $n$. I checked the book "Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work" by G. H. Hardy and on page 135 Hardy mentions that Liouville and Eisentein found formulas for sum of $k$ squares for $k = 10$ and $k = 12$ respectively. This means that Lambert series for $\vartheta_{3}^{10}(q), \vartheta_{3}^{12}(q)$ are available. Contd.. May 29, 2016 at 10:55
• And by changing $q$ into $-q$ we can get similar series for powers of $\vartheta_{4}(q)$. Unfortunately Hardy's book gives only the formulas for expressing $n$ as sum of $10$ squares. It does not give the corresponding Lambert series. May 29, 2016 at 10:57
• @ParamanandSingh Thank you, that was very helpful. I assumed that finding these lambert series was the convenient way to prove theorems about the sum of squares functions, and not the other way around. I found this article (MIilne) which includes a very interesting discussion on the sum of sqaures functions and gives explicit closed forms for $\theta_4^{16}$ and $\theta_4^{24}$ (along with dozens other formulas). May 31, 2016 at 14:49
• For example: with $$U_k=\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^k q^n}{1+q^n},$$ we have $$\theta_4^{16}=1-\frac{32}{3}(U_1+U_3+U_5)+\frac{256}{3}(U_1 U_5 -U_3^2)$$ May 31, 2016 at 14:54
• Thanks man for the link to Milne's article! And yes for general $k$ the problem of representing a number as sum of $k$ squares is solved via Lambert series expression for $\vartheta_{3}^{k}(q)$. For small values of $k$ (for example $k = 2, 4$) there are approaches from elementary number theory without the use of theta functions. May 31, 2016 at 16:28

$$||$$ Suppose for non-negative $$a,b,c$$, $$L(2a,2b,2c)= \left ( \frac\pi2 \right )^{a+b+c-1} \int_{0}^{\infty}\frac{ \vartheta_2(e^{-\pi x})^{2a}\vartheta_3(e^{-\pi x})^{2b}\vartheta_4(e^{-\pi x})^{2c}}{ 1+x^2}\text{d}x.\tag{1}$$ By making a modular parameterization $$x=K^\prime(k)/K(k)$$, and noticing that $$\vartheta_2^2=\frac{2kK}{\pi}, \vartheta_3^2=\frac{2K}{\pi}, \vartheta_4^2=\frac{2k^\prime K}{\pi}, \frac{\mathrm{d}x}{\mathrm{d}k} =-\frac{\pi}{2kk^\prime{}^2K^2},$$we have $$L(2a,2b,2c)=\int_{0}^{1} \frac{k^{a-1}k^\prime{}^{c-2} K^{a+b+c}}{K^2+K^\prime{}^2}\text{d}k.\tag{2}$$ To evaluate $$(2)$$, I introduce a variation of Hrilbert transformation.

For two integrable functions $$f_{1}(x),f_2(x)$$ for $$x\in[0,1]$$, define $$\frac{2}{\pi}\int_{0}^{1} \frac{f_{1,2}(x)}{1-k^2x^2}\text{d}x =g_{1,2}(k),$$ obviously we have $$\int_{0}^{1} f_1(x)g_2(x)\text{d}x =\int_{0}^{1}f_2(x)g_1(x)\text{d}x.\tag{3}$$ What's special of the transformation is that for certain functions(containing $$K$$), they act invariants alike.

Some transform pairs can be computed explicitly by residues, e.g. $$\begin{array}{|l|l|}\hline f(x)&g(k)\\ \hline 1&\frac{1}{\pi k}\ln\left ( \frac{1+k}{1-k} \right )\\ K^\prime(x)&K(k)\\ xK(x)K^\prime(x)& \frac{1}{2} K(k)^2\\ 3K(x)^2K^\prime(x)-K^\prime(x)^3&k^2K(k)^3\\ x^2\left(3K(x)^2K^\prime(x)-K^\prime(x)^3\right)&K(k)^3\\ \frac{1}{\sqrt{1-x^2}}&\frac{1}{\sqrt{1-k^2}}\\ \frac{xK(x)}{\sqrt{1-x^2}}&\frac{K(k)}{\sqrt{1-k^2}}\\ x\sqrt{1-x^2}\left(3K(x)K^\prime(x)^2-K(x)^3\right)&\sqrt{1-k^2}K(k)^3\\ \sqrt{1-x^2}\left(K(x)^4-6K(x)^2K^\prime(x)^2+K^\prime(x)^4\right)& -k^2\sqrt{1-k^2}K(k)^4\\ x^2\sqrt{1-x^2}\left(K(x)^4-6K(x)^2K^\prime(x)^2+K^\prime(x)^4\right)& -\sqrt{1-k^2}K(k)^4\\ \frac{1}{x\sqrt{1-x^2} }\frac{K(x)}{ K(x)^2+K^\prime(x)^2}&\frac{1}{K(k)\sqrt{1-k^2} }\\ \frac{x}{\sqrt{1-x^2} }\frac{K(x)}{ K(x)^2+K^\prime(x)^2}&\frac{1}{k^2}\left(\frac{1}{K(k)\sqrt{1-k^2} }-\frac2\pi\right)\\ \frac{x^{2a-1}}{(1-x^2)^a}&\frac{1}{\sin(\pi a)}\frac1{(1-k^2)^a}\\ \hline \end{array}$$

Example Evaluation
Proposition: Show that $$L(4,0,4)=\left ( \frac{\pi}{2} \right )^3 \int_{0}^{\infty} \frac{\vartheta_2^4(e^{-\pi x}) \vartheta_4^4(e^{-\pi x})}{1+x^2}\text{d}x =\zeta(3),$$ $$L(0,4,4)=\left ( \frac{\pi}{2} \right )^3 \int_{0}^{\infty} \frac{\vartheta_3^4(e^{-\pi x}) \vartheta_4^4(e^{-\pi x})}{1+x^2}\text{d}x =\frac72\zeta(3).$$ Proof: On the one hand, from $$(2)$$, we have $$L(0,4,4)=\int_{0}^{1}\frac{K(k)^4}{k(K(k)^2+K^\prime(k)^2)} \text{d}k.$$ Setting $$f_1=\frac{1}{x\sqrt{1-x^2} }\frac{K(x)}{ K(x)^2+K^\prime(x)^2},f_2=x\sqrt{1-x^2}\left(3K(x)K^\prime(x)^2-K(x)^3\right)$$, using $$(3)$$ we have $$L(0,4,4)=\int_{0}^{1}x\left(3K^\prime(x)^2-K(x)^2\right) \text{d}x=\frac72\zeta(3).$$ The last equality is well-known.
On the other hand, due to Jacobi's functional relation $$\vartheta_2^4+\vartheta_4^4=\vartheta_3^4,$$ we have $$L(4,0,4)+L(0,0,8)=L(0,4,4).$$ $$L(0,0,8)=\frac{5}{2}\zeta(3)$$ can be found in the original post, and it then completes the proof.

Expressions obtained by this method: $$\begin{array}{l} L(0,0,2)=1\\ L(0,2,2)=2G\\ L(0,2,4)=\frac{7}{2}\zeta(3)-2G\\ L(0,4,2)=4G\\ L(4,2,2)=\frac{3\pi^4}{16} {}_7F_6\left ( \begin{array}{c|} \frac12,\frac12,\frac12,\frac12,\frac12,\frac12,\frac54\\ \frac14,1,1,1,1,1 \end{array}\text{ }1 \right )-\int_{0}^{1} \frac{K^\prime{}^3}{K} \frac{\text{d}k}{k^\prime}\\ L(4,2,4)=\frac{27\text{ }\Gamma\left ( \frac14 \right )^8 }{1280\pi^2} -\int_{0}^{1} \frac{K^\prime{}^4}{K}\text{d}k\\ L(4,4,2)=\frac{\text{}\Gamma\left ( \frac14 \right )^8 }{160\pi^2}\\ L(0,8,4)=-\int_{0}^{1}k^3\left ( 5K^\prime{}^4-10K^2K^\prime{}^2 +K^4 \right )\text{d}k \end{array}$$ where here and below $$G$$ denotes Catalan's constant $$G=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}$$. And then: $$\begin{array}{l} L(0,0,4)=2\ln(2)\\ L(0,0,6)=4G-\frac{\pi^2}{6}\\ L(4,0,2)=\frac{\pi^2}{6}\\ L(0,0,10)=\frac{48}{5}\beta(4)+\frac{2\pi^4}{75} -\frac{\Gamma\left ( \frac14 \right )^8 }{400\pi^2}\\ L(0,8,4)=\frac{93}{8}\zeta(5)+\frac{\pi^6}{256} {}_9F_8\left ( \frac12,\frac12 ,\frac12,\frac12, \frac12,\frac12,\frac12,\frac12,\frac54; \frac14,1,1,1,1,1,1,1;1\right ). \end{array}$$ And we also follow OP's $$q$$-expansion method, to obtain more results. Along with some L-series like $$L_8(s)=\sum_{n=0}^{\infty}\left ( \frac{1}{(8n+1)^s} - \frac{1}{(8n+3)^s}- \frac{1}{(8n+5)^s}+ \frac{1}{(8n+7)^s}\right ),$$$$L_{-8}(s)=\sum_{n=0}^{\infty}\left ( \frac{1}{(8n+1)^s}+ \frac{1}{(8n+3)^s}- \frac{1}{(8n+5)^s}-\frac{1}{(8n+7)^s}\right ),$$ here is the summary: $$\begin{array}{|l} \color{red}{L(0,0,1)} =\sqrt{\frac\pi2}+2\sqrt{\frac2\pi} \sum_{n=1}^{\infty}(-1)^n\left ( \frac\pi2-\operatorname{Si}\left ( \pi n^2 \right ) \right )\\ \color{red}{L(0,0,2)}=1\\ L(1,0,1)=\frac{1}{\sqrt{2}}\\ L(0,1,1)=\sqrt{2}\ln(1+\sqrt{2})\\ \color{red}{L(0,0,4)}=2\ln(2)\\ L(2,0,2)=\frac\pi4\\ L(0,2,2)=2G\\ \color{magenta}{L(2,1,1)} =\frac{\Gamma\left ( \frac14 \right )^2 }{2\sqrt{2\pi} }- \sqrt{2}\pi\,_3F_2\left ( \frac12,\frac12,\frac12;1,\frac32;\frac12 \right ) -\frac{\pi}{16\sqrt{2} }\,_4F_3\left ( 1,1,\frac32,\frac32;2,2,2;\frac12 \right ) +\frac{5\pi}{2\sqrt{2} }\ln(2)\\ L(1,2,1)=\frac{\Gamma\left ( \frac14 \right )^2 }{4\sqrt{\pi} }\\ \color{magenta}{L(1,1,2)} =\sqrt{2}\pi\,_3F_2\left ( \frac12,\frac12,\frac12;1,\frac32;\frac12 \right ) +\frac{\pi}{16\sqrt{2} }\,_4F_3\left ( 1,1,\frac32,\frac32;2,2,2;\frac12 \right ) -\frac{5\pi}{2\sqrt{2} }\ln(2)\\ L(1,0,3)=\frac{\pi-2G+\ln(2)}{2\sqrt{2} } \\ L(3,0,1)=\frac{\pi+2G-\ln(2)}{2\sqrt{2} }\\ L(0,1,3)=2L_{-8}(2)+\frac{\pi}{2\sqrt{2} } \ln\left ( 1+\sqrt{2} \right ) -\frac\pi2\\ L(0,3,1)=2L_{-8}(2)-\frac{\pi}{2\sqrt{2} } \ln\left ( 1+\sqrt{2} \right ) +\frac\pi2\\ \color{red}{L(0,0,6)}=4G-\frac{\pi^2}{6}\\ L(2,0,4)=2G-\frac78\zeta(3)\\ L(0,2,4)=\frac72\zeta(3)-2G\\ L(0,4,2)=4G\\ L(4,0,2)=\frac{\pi^2}{6}\\ \color{magenta}{L(2,2,2)} =\frac{\Gamma\left ( \frac14 \right )^4 }{8\pi} -\frac{3\pi^2}{4}\ln(2) +\frac{\pi^2}{64}\,_5F_4\left ( 1,1,\frac32,\frac32,\frac32;2,2,2,2;1 \right )\\ L(3,0,3)=\frac{11\pi^2}{48\sqrt{2} } -\frac{G}{\sqrt{2} }\\ L(0,3,3)=4L_8(3)-\frac{3\pi^2}{8\sqrt{2} }\ln\left ( 1+\sqrt{2} \right ) +\frac{\pi^2}{8}\\ \color{red}{L(0,0,8)}=\frac52\zeta(3)\\ L(4,0,4)=\zeta(3)\\ L(0,4,4)=\frac72\zeta(3)\\ L(4,2,2)=\frac{3\pi^4}{16} {}_7F_6\left ( \frac12,\frac12,\frac12,\frac12,\frac12,\frac12,\frac54; \frac14,1,1,1,1,1;1\right )-\int_{0}^{1} \frac{K^\prime{}^3}{K} \frac{\text{d}k}{k^\prime}\\ L(2,2,4)=\frac{5\pi^4}{32} {}_7F_6\left ( \frac12,\frac12,\frac12,\frac12,\frac12,\frac12,\frac54; \frac14,1,1,1,1,1;1\right )-\int_{0}^{1} \frac{K^\prime{}^3}{K} \frac{\text{d}k}{k^\prime}\\ \color{magenta}{L(2,4,2)} =\frac{3\pi^3}{8} {}_4F_3 \left ( \frac12,\frac12,\frac12,\frac12;1,1,1;1 \right ) -\frac{\pi^3}{8} {}_5F_4 \left ( \frac12,\frac12,\frac12,\frac12,\frac12;1,1,1,\frac32;1 \right ) -\frac{7\pi}{8}\zeta(3)\\ \color{magenta}{L(2,0,6)} =\frac{3\pi^3}{16} {}_4F_3 \left ( \frac12,\frac12,\frac12,\frac12;1,1,1;1 \right ) -\frac{\pi^3}{16} {}_5F_4 \left ( \frac12,\frac12,\frac12,\frac12,\frac12;1,1,1,\frac32;1 \right ) -\frac{7\pi}{16}\zeta(3)-\frac{\pi^3}{16}\\ \color{magenta}{L(6,0,2)} =\frac{3\pi^3}{16} {}_4F_3 \left ( \frac12,\frac12,\frac12,\frac12;1,1,1;1 \right ) -\frac{\pi^3}{16} {}_5F_4 \left ( \frac12,\frac12,\frac12,\frac12,\frac12;1,1,1,\frac32;1 \right ) -\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\\ \color{red}{L(0,0,10)}=\frac{48}{5}\beta(4)+\frac{2\pi^4}{75} -\frac{\Gamma\left ( \frac14 \right )^8 }{400\pi^2}\\ L(4,4,2)=\frac{\text{}\Gamma\left ( \frac14 \right )^8 }{160\pi^2}\\ L(4,2,4)=\frac{27\text{ }\Gamma\left ( \frac14 \right )^8 }{1280\pi^2} -\int_{0}^{1} \frac{K^\prime{}^4}{K}\text{d}k\\ L(0,6,4) =\frac{93}{20}\zeta(5) -\frac{12}{5}\beta(4)+ \frac{27\,\Gamma\left ( \frac14 \right )^8 }{6400\pi^2} -\frac15\int_{0}^{1} \frac{K^\prime{}^4}{K}\text{d}k\\ L(0,2,8) =\frac{93}{20}\zeta(5) -\frac{12}{5}\beta(4)- \frac{27\,\Gamma\left ( \frac14 \right )^8 }{1600\pi^2} +\frac45\int_{0}^{1} \frac{K^\prime{}^4}{K}\text{d}k\\ L(0,8,2)=\frac{48}{5}\beta(4)+\frac{\Gamma\left ( \frac14 \right )^8 }{200\pi^2}\\ L(0,4,6)=\frac{48}{5}\beta(4)-\frac{\Gamma\left ( \frac14 \right )^8 }{800\pi^2}\\ L(4,0,6)=\frac{\Gamma\left ( \frac14 \right )^8 }{800\pi^2} -\frac{2\pi^4}{75}\\ L(8,0,2)=\frac{\Gamma\left ( \frac14 \right )^8 }{200\pi^2} +\frac{2\pi^4}{75}\\ \color{red}{L(0,0,12)}=\frac{225}{16}\zeta(5)-\frac{\pi^6}{128} {}_9F_8\left ( \frac12,\frac12 ,\frac12,\frac12, \frac12,\frac12,\frac12,\frac12,\frac54; \frac14,1,1,1,1,1,1,1;1\right )\\ L(0,8,4)=\frac{93}{8}\zeta(5)+\frac{\pi^6}{256} {}_9F_8\left ( \frac12,\frac12 ,\frac12,\frac12, \frac12,\frac12,\frac12,\frac12,\frac54; \frac14,1,1,1,1,1,1,1;1\right )\\ L(0,4,8)=\frac{93}{8}\zeta(5)-\frac{\pi^6}{256} {}_9F_8\left ( \frac12,\frac12 ,\frac12,\frac12, \frac12,\frac12,\frac12,\frac12,\frac54; \frac14,1,1,1,1,1,1,1;1\right )\\ L(8,0,4)=\frac{39}{16}\zeta(5)+\frac{\pi^6}{256} {}_9F_8\left ( \frac12,\frac12 ,\frac12,\frac12, \frac12,\frac12,\frac12,\frac12,\frac54; \frac14,1,1,1,1,1,1,1;1\right )\\ L(4,0,8)=-\frac{39}{16}\zeta(5)+\frac{\pi^6}{256} {}_9F_8\left ( \frac12,\frac12 ,\frac12,\frac12, \frac12,\frac12,\frac12,\frac12,\frac54; \frac14,1,1,1,1,1,1,1;1\right )\\ L(4,4,4)=\frac{\pi^6}{128} {}_9F_8\left ( \frac12,\frac12 ,\frac12,\frac12, \frac12,\frac12,\frac12,\frac12,\frac54; \frac14,1,1,1,1,1,1,1;1\right ) . \end{array}$$ Some of them are from a different background and we highlight them in magenta. The red means that they match the form in the original post.

Something hasn't to do with the question
If we set $$f_1=\frac{x}{\sqrt{1-x^2} }\frac{K(x)}{ K(x)^2+K^\prime(x)^2},f_2=x\sqrt{1-x^2}\left(3K(x)K^\prime(x)^2-K(x)^3\right)$$, an integral identity is given: $$\int_{0}^{1} \frac{\sqrt{1-x^2}}{x} \left(3K^\prime(x)^2-K(x)^2\right)\left ( \frac{1}{\sqrt{1-x^2}} -\frac{2}{\pi}K(x) \right )\text{d}x=\int_{0}^{1} \frac{kK(k)^4}{K(k)^2+K^\prime(k)^2}\text{d}k= L(4,0,4)=\zeta(3).$$ Reformulate it in an elegant way: $$\int_{0}^{1} \frac{ \left (3K^\prime(x)^2-K(x)^2\right ) \left ( \frac{\pi}{2}-K(x)\sqrt{1-x^2} \right )}{x} \text{d}x=\frac{\pi}{2}\zeta(3).$$ Here are more applications, $$\int_{0}^{1}\left(1-x^4\right) K(x)^3\left ( 3K(x)^2K^\prime(x)-K^\prime(x)^3\right )\text{d}x =0,$$ $$\int_{0}^{1} \frac{K(x)\ln(1+x)}{\sqrt{1-x^2} }\text{d}x =\frac{\Gamma\left ( \frac14 \right )^4}{96}, \int_{0}^{1} \frac{K(x)\ln(1-x)}{\sqrt{1-x^2} }\text{d}x =-\frac{5\text{ }\Gamma\left ( \frac14 \right )^4}{96}.$$ And a remarkable one, $$\int_{0}^{1}\frac{x\ln\left ( \frac{1+x^2}{2} \right ) K(x)K^\prime(x)}{ 1-x^2}\text{d}x=-\frac18\pi^2G.$$

A corresponding and remarkable integral(only obtainable via $$q$$-expansions): $$\int_{0}^{1}\frac{K(k)^3}{ 9K(k)^2+K^\prime(k)^2}\text{d}k =\frac{2G}{9}-\frac{19\zeta(3)}{648}.$$ With more possible applications, we add $$\int_{0}^{1} \frac{(2k^2-1)K(k)^2\arctan\left ( \frac{K^\prime(k)}{K(k)} \right ) }{ \sqrt{1-k^2} }\text{d}k =-4\operatorname{Li}_4\left ( \frac12 \right ) -\frac72\zeta(3)\ln(2)+\frac{\pi^2}{6}\ln(2)^2-\frac{1}{6}\ln(2)^4+\frac{151}{2880}\pi^4.$$

• Remember MSE is not a place for you to store your results and observations. Your results are new and potentially interesting, I strongly suggest you put them on arXiv instead. Jan 9 at 3:32

This is a sequel to my comments to the question which was too long to fit in another comment.

We have the formulas for $\vartheta_{3}^{10}(q), \vartheta_{3}^{12}(q)$ from Topics in Analytic Number Theory by Rademacher (famous for proving an infinite series formula to calculate the number of partitions of a positive integer) on page 198: \begin{align} \vartheta_{3}^{10}(q) &= 1 + \frac{4}{5}\left\{\sum_{n = 1}^{\infty}\frac{2n^{4}q^{n}}{1 + q^{2n}} + \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{(2n - 1)^{4}q^{2n - 1}}{1 - q^{2n - 1}}\right\} + \frac{2}{5}\vartheta_{3}^{2}(q)\vartheta_{2}^{4}(q)\vartheta_{4}^{4}(q)\tag{1}\\ \vartheta_{3}^{12}(q) &= 1 + 8\sum_{n = 1}^{\infty}\frac{n^{5}q^{n}}{1 - q^{2n}} - 8\sum_{n = 1}^{\infty}(-1)^{n}\frac{n^{5}q^{2n}}{1 - q^{2n}} + \vartheta_{2}^{4}(q)\vartheta_{3}^{4}(q)\vartheta_{4}^{4}(q)\tag{2} \end{align}

Finding a general formula for $\vartheta_{3}^{k}(q)$ for even positive integer $k$ is a difficult problem but using the methods given in Rademacher's book it looks like it is possible to obtain such formulas at the cost of heavy symbolic manipulation for a specific $k$.

Update: I found one pattern in your formulas by using the substitution $x = K'(k)/K(k)$ so that when $x = 0$ then $k = 1$ and when $x = \infty$ then $k = 0$ and moreover $$\frac{dx}{dk} = -\frac{\pi}{2kk'^{2}K^{2}}$$ so that the integral of $\vartheta_{4}^{2n}(e^{-\pi x})/(1 + x^{2})$ is transformed into $$\int_{0}^{1}\left(\frac{2k'K}{\pi}\right)^{n}\frac{1}{K^{2} + K'^{2}}\frac{\pi}{2kk'^{2}}\,dk = \left(\frac{2}{\pi}\right)^{n - 1}\int_{0}^{1}\frac{k^{-1}k^{'(n - 2)}K^{n}}{K^{2} + K'^{2}}\,dk$$ and that explains (at least to some extent) the occurrence of $\dfrac{1}{\pi^{n - 1}}$ in your formulas.

Next it is easy to prove one of the formulas in $(14)$. We have $$\vartheta_{2}^{2}\vartheta_{4}^{2} = kk'(2K/\pi)^{2}$$ and hence $$\int_{0}^{\infty}\vartheta_{2}^{2}\vartheta_{4}^{2}\,dx = \int_{0}^{1}kk'\cdot\frac{4K^{2}}{\pi^{2}}\cdot\frac{\pi}{2kk'^{2}K^{2}}\,dk = \frac{2}{\pi}\int_{0}^{1}\frac{dk}{\sqrt{1 - k^{2}}} = 1$$ I wonder if similar technique can be applied to prove other formulas.

If $q = e^{-\pi x}$ then $dx = -\dfrac{dq}{\pi q}$ and interval $(0, \infty)$ changes to $(0, 1)$ and hence we can express the first integral of $(14)$ as $$\frac{1}{\pi}\int_{0}^{1}\vartheta_{2}^{4}(q)\vartheta_{4}^{2}(q)\,\frac{dq}{q} = \frac{16}{\pi}\int_{0}^{1}\psi^{4}(q^{2})\phi^{2}(-q)\,dq$$ Next $$\psi^{4}(q^{2}) = \sum_{n = 0}^{\infty}\frac{(2n + 1)q^{2n}}{1 - q^{4n + 2}}, \phi^{2}(-q) = 1 + 4\sum_{n = 1}^{\infty}\frac{(-1)^{n}q^{n}}{1 + q^{2n}}$$ I wonder if you can utilize the above Lambert series to prove that the desired integral is equal to $1$. It appears that if we express the integrand as a Lambert series then it can also be expressed as the logarithmic derivative of some product of theta functions and the integral can be evaluated. See this paper regarding some integrals related to theta functions (all of it was given by Ramanujan in his lost notebook).

• Thank you very much for that formula for $\theta_4^{12}$! Using this, I obtained $$\int_0^{\infty} \frac{\theta_4^{12}(e^{-\pi x})}{1+x^2} dx=\frac{450 \zeta(5)}{\pi^5}-\frac23 \int_0^{\infty} \theta_2^4(e^{-\pi x}) \theta_3^4(e^{-\pi x}) \theta_4^4(e^{-\pi x})dx,$$ which kinda reminds me of equation $(4)$ in the OP. A bit stuck on that last integral, but that was certainly helpful, so I am accepting this answer. May 31, 2016 at 19:38
• After some more work and some guesses, using your formula for $\theta_4^{10}$ I obtain $$\int_0^{\infty} \frac{\theta_4^{10}(e^{-\pi x})}{1+x^2} dx=\frac{768 \beta(4)}{5 \pi^4}+\frac{32}{75}-\frac{16}{15} \int_0^{\infty} \theta_2^2(e^{-\pi x}) \theta_3^4(e^{-\pi x}) \theta_4^4(e^{-\pi x})dx,$$ where $\beta$ is Dirichlet's. Jun 1, 2016 at 11:35
• Wow! That's such a natural substitution to do and I completely missed it. That certainly sheds some light on this nature of these closed forms. Also, notice that $\theta_4^2(e^{-\pi x}) \theta_2^2(e^{-\pi x}) dx$ is invariant under $x \mapsto 1/x$, so expanding the theta functions into series identity $(14)$ that you've prove can be restated as: $$\sum_{ (a,b,c,d) \in \mathbb{Z}^4} \frac{(-1)^{a+b} e^{-\pi(a^2+b^2+(c+1/2)^2 + (d+1/2)^2)}}{a^2 +b^2 + (c+1/2)^2+(d+1/2)^2} = \frac{ \pi}{2}$$ which is one beautiful series. Jun 2, 2016 at 11:23
• @nospoon: I must say that you have done a remarkable work here in this post. You should perhaps organize all your proofs and publish it somewhere (say a blog or a journal if you have access). And yes you should post these integral identities (the ones whose proof you are still searching) as separate questions on MSE. Maybe some on the giants (you are also one of them) of integrals fame would solve it. Jun 2, 2016 at 15:11
• @NoamShalev-nospoon Your $\int_{0}^{\infty} \vartheta_2(e^{-\pi x})^2 \vartheta_3(e^{-\pi x})^4\vartheta_4(e^{-\pi x})^4\text{d}x$ seems to be very close to $\int_{0}^{\infty} \frac{x}{1+x^2}\vartheta_2(e^{-\pi x})^2 \vartheta_3(e^{-\pi x})^4\vartheta_4(e^{-\pi x})^4\text{d}x =\frac{\Gamma\left ( \frac14 \right )^8 }{60\pi^6}.$ Jan 14, 2023 at 15:02

Are you still interested on a solution of that problem. I found a solution for $$\int_0^{\infty} \frac{\theta_4^2}{1+x^2} dx=1 \tag{2}$$ and it seems to be a reasonable way to solve the general problem .