## bookmark_borderIntroduction to Theta Functions II

Infinite product representations of theta functions

Let $f(z) = \prod_{n=1}^{\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})$. Each of these two products converge absolutely and uniformly in any bounded domain of values of $z$. Hence $f(z)$ is analytic throughout the finite part of the $z$ plane. The zeros of $f(z)$ are simple zeros at the points $z=\frac{2n+1}{2}\pi \tau i + m\pi i$ where $m,n\in \mathbb{Z}$. So, the zeros of $f(z)$ coincide with the zeros of $\vartheta_4(z)$. Therefore, the function $\frac{\vartheta_4(z)}{f(z)}$ has no poles or zeros in the finite part of the plane. Also, it is easy to see that $f(z+\pi) = f(z)$ and \begin{aligned} f(z+\pi \tau) &= \prod_{n=1}^\infty (1-q^{2n+1}e^{2iz})(1-q^{2n-3}e^{-2iz}) \\ &= f(z) \frac{1-q^{-1}e^{-2iz}}{1-qe^{2iz}} \\ &= -q^{-1}e^{-2iz} f(z) \end{aligned} This is, $\frac{\vartheta_4(z)}{f(z)}$ is a doubly periodic function with periods $\pi,\pi\tau$ and has no poles and zeros. By section 20.12 of [2], it is simply a constant, say $G$. We have $$\vartheta_4(z) = G \prod_{n=1}^{\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})\quad (1)$$ Incrementing $z$ by the half periods $\frac{\pi}{2}, \frac{\pi \tau}{2}$ and $\frac{\pi +\pi\tau}{2}$, we get the product representations for the other theta functions: \begin{aligned} \vartheta_3(z) &= G \prod_{n=1}^{\infty} (1+q^{2n-1}e^{2iz})(1+q^{2n-1}e^{-2iz})\quad (2) \\ \vartheta_1(z) &= 2G q^{\frac{1}{4}}\sin(z) \prod_{n=1}^{\infty} (1-q^{2n}e^{2iz})(1-q^{2n}e^{-2iz}) \quad (3) \\ \vartheta_2(z) &= 2G q^{\frac{1}{4}}\cos(z) \prod_{n=1}^{\infty} (1+q^{2n}e^{2iz})(1+q^{2n}e^{-2iz}) \quad (4) \end{aligned} Now, all that left is to find the constant $G$. From identity (3), it easy to see that $$\vartheta_1'(0) = 2G q^{\frac{1}{4}}\prod_{n=1}^{\infty} (1-q^{2n})^2$$ Now, using the identity $\vartheta_1'(0)=\vartheta_2(0)\vartheta_3(0)\vartheta_4(0)$, we get: \begin{aligned} 2G q^{\frac{1}{4}}\prod_{n=1}^{\infty} (1-q^{2n})^2 &= \left\{2G q^{\frac{1}{4}}\prod_{n=1}^{\infty}(1+q^{2n})^2 \right\} \times \left\{G\prod_{n=1}^{\infty}(1+q^{2n-1})^2 \right\} \\ &\quad \times \left\{G \prod_{n=1}^{\infty}(1-q^{2n-1})^2 \right\} \\ \implies G &= \prod_{n=1}^\infty (1-q^{2n}) \end{aligned} Note that the rearrangements are justified since all products converge absolutely. Finally, we have \begin{aligned} \vartheta_1(z) &= 2 q^{\frac{1}{4}}\sin(z) \prod_{n=1}^{\infty} (1-q^{2n})(1-2q^{2n} \cos(2z)+ q^{4n}) \quad (5) \\ \vartheta_2(z) &= 2 q^{\frac{1}{4}}\cos(z) \prod_{n=1}^{\infty} (1-q^{2n})(1+2q^{2n} \cos(2z) + q^{4n}) \quad (6) \\ \vartheta_3(z) &= \prod_{n=1}^{\infty} (1-q^{2n})(1+2q^{2n-1} \cos(2z) + q^{4n-2}) \quad (7) \\ \vartheta_4(z) &= \prod_{n=1}^{\infty} (1-q^{2n})(1-2q^{2n-1} \cos(2z) + q^{4n-2}) \quad (8) \end{aligned}

Derivatives of Ratios of Theta Functions

Consider the function $\phi(z) = \frac{\vartheta_1'(z)\vartheta_4(z) - \vartheta_1(z)\vartheta_4'(z)}{\vartheta_2(z)\vartheta_3(z)}$. Now, $\phi(z)$ is doubly periodic with periods $\pi$ and $\frac{\pi\tau}{2}$. Relative to these periods, the only poles of $\phi(z)$ are at points congruent to $\pi\over 2$. It follows that $\phi(z)$ is a constant. Letting $z\to 0$, we see that $\phi(z)=\vartheta_4^2(0)$. It is therefore established that $$\frac{d}{dz}\left(\frac{\vartheta_1(z)}{\vartheta_4(z)} \right)=\vartheta_4^2(0) \frac{\vartheta_2(z)\vartheta_3(z)}{\vartheta_4^2(z)} \quad (9)$$ Two other identities of this kind can be derived using the same technique: \begin{aligned} \frac{d}{dz}\left(\frac{\vartheta_2(z)}{\vartheta_4(z)} \right) &=-\vartheta_3^2(0) \frac{\vartheta_1(z)\vartheta_3(z)}{\vartheta_4^2(z)} \quad (10) \\ \frac{d}{dz}\left(\frac{\vartheta_3(z)}{\vartheta_4(z)} \right) &=-\vartheta_2^2(0) \frac{\vartheta_1(z)\vartheta_2(z)}{\vartheta_4^2(z)} \quad (11) \end{aligned} By writing $\xi = \frac{\vartheta_1(z)}{\vartheta_4(z)}$ and making use of identities (13) and (14) from part 1, we obtain: $$\left(\frac{d\xi}{dz}\right)^2 = (\vartheta_2^2(0) - \xi^2 \vartheta_3^2(0))(\vartheta_3^2(0)-\xi^2 \vartheta_2^2(0))$$ Using the change of variables $y = \frac{\vartheta_3(0)}{\vartheta_2(0)}\xi$ and $u = \vartheta_3^2(0) z$, the above differential equation can be written as: $$\left(\frac{dy}{du}\right)^2 = (1-y^2)(1-k^2 y^2)$$ where $k = \frac{\vartheta_2^2(0)}{\vartheta_3^2(0)}$. This differential equation has the particular solution: $$y = \frac{\vartheta_3(0)}{\vartheta_2(0)} \frac{\vartheta_1(u \vartheta_3^{-2}(0))}{\vartheta_4(u \vartheta_3^{-2}(0))}$$ We can write $y$ as a function of $u$ and $k$ as $y = \text{sn}(u,k)$ or simply $\text{sn}(u)$. Clearly, $\text{sn}(u,k)$ has periods $2\pi \vartheta^2_3(0)$ and $\pi \tau \vartheta_3^2(0)$. It has two simple poles congruent to $\frac{1}{2}\pi\tau \vartheta_3^2(0)$ and $\pi \vartheta_3^2(0) + \frac{1}{2}\pi\tau \vartheta_3^2(0)$. Also, it is easy to see that $$\int_0^1 \frac{dy}{\sqrt{(1-y^2)(1-k^2 y^2)}} = \text{sn}^{-1}(1) = \frac{\pi}{2}\vartheta_3^2(0)$$ The integral on the left is denoted by complete elliptic integral of the first kind and is denoted by $K(k)$.

References

1. Derek F. Lawden (1989). Elliptic Functions and Applications. Springer-Verlag New York
2. E. T. Whittaker, G. N. Watson (1927). A Course of Modern Analysis. Cambridge University Press

## bookmark_borderIntroduction to Theta Functions I

The Jacobi theta functions are defined for all complex variables of $z$ and $q$ such that $|q| < 1$, as follows: \begin{aligned} \vartheta_1 (z,q) &= -i \sum_{n=-\infty}^{\infty} (-1)^n q^{(n+1/2)^2} e^{i(2n+1)z} \\ \vartheta_2 (z,q) &= \sum_{n=-\infty}^{\infty} q^{(n+1/2)^2} e^{i(2n+1)z} \\ \vartheta_3 (z,q) &= \sum_{n=-\infty}^{\infty} q^{n^2} e^{2inz} \\ \vartheta_4 (z,q) &= \sum_{n=-\infty}^{\infty} (-1)^n q^{n^2} e^{2inz} \\ \end{aligned} The parameter $q$ is called the nome. Let $\tau$ be a complex number whose imaginary part is positive and write $q=e^{i\pi \tau}$ so that $|q| < 1$. Sometimes, $q$ will not be specified, so that $\vartheta_r (z)$ is written for $\vartheta_r(z,q)$ where $r=1,2,3,4$. It is easy to see that \begin{aligned} \vartheta_1(z+\pi) &= -i e^{i \pi}\sum_{n=-\infty}^{\infty}(-1)^n q^{\left(n+\frac{1}{2}\right)^2}e^{i(2n+1)z} = -\vartheta_1(z) \\ \vartheta_1(z+\pi \tau) &= i e^{-i\pi \tau -2iz} \sum_{n=-\infty}^{\infty} (-1)^{n+1} e^{i\pi \tau \left( n+\frac{3}{2}\right)^2 + (2n+3) iz} = - (q e^{2iz})^{-1} \vartheta_1(z) \end{aligned} Similarly, the periodicity laws for the other theta functions can be derived as well. The multipliers of the theta functions associated with the periods $\pi$ and $\pi \tau$ are summarized in the following table: $$\begin{array}{|c|c|c|c|c|} \hline \; & \vartheta_1(z) & \vartheta_2(z) & \vartheta_3(z) & \vartheta_4(z) \\ \hline \hline \pi & -1 & -1 & 1 & 1 \\ \hline \pi \tau & -\lambda & \lambda & \lambda & -\lambda \\ \hline \end{array}$$ where $\lambda = (q e^{2iz})^{-1}$. Furthermore, incrementation of $z$ by the half periods $\frac{\pi}{2}, \frac{\pi \tau}{2}, \frac{\pi +\pi \tau}{2}$ yields the following identities: $$\begin{array}{|c|c|c|c|c|} \hline \omega & \vartheta_1(z+\omega) & \vartheta_2(z+\omega) & \vartheta_3(z+\omega) & \vartheta_4(z+\omega) \\ \hline \hline \frac{\pi}{2} & \vartheta_2(z) & -\vartheta_1(z) & \vartheta_4(z) & \vartheta_3(z) \\ \hline \frac{\pi \tau}{2} & i \mu \vartheta_4(z) & \mu \vartheta_3(z) & \mu \vartheta_2(z) & i\mu \vartheta_1(z) \\ \hline \frac{\pi + \pi \tau}{2} & \mu \vartheta_3(z) & -i\mu \vartheta_4(z) & i\mu \vartheta_1(z) & \mu \vartheta_2(z) \\ \hline \end{array}$$ where $\mu = (q^{\frac{1}{4}}e^{iz})^{-1}$.

## Identities involving products of theta functions

Many theta function identities can be derived by the multiplication of two of their series and rearrangement of the terms in the product series. This is justified since the series are absolutely convergent. We have \begin{aligned} \vartheta_3(x,q)\vartheta_3(y,q) &= \left(\sum_{n=-\infty}^{\infty} q^{n^2} e^{2inx} \right)\left(\sum_{m=-\infty}^{\infty} q^{m^2} e^{2imy} \right) \\ &= \sum_{n=-\infty}^\infty \sum_{m=-\infty}^\infty q^{n^2+m^2}e^{2i(nx+my)} \end{aligned} Now, we change the summation indices from $(m,n)$ to $(r,s)$ by the following equations: \begin{aligned} r &= m+n \\ s &= m-n \end{aligned} If $(m,n)$ are both even then $(r,s)$ will both be even and if $(m,n)$ have opposite parity then $(r,s)$ will both be odd. Therefore, we can rearrange the series as follows: \begin{aligned} \vartheta_3(x,q)\vartheta_3(y,q) &= \sum_{r=-\infty}^{\infty} \sum_{s=-\infty}^{\infty} q^{2(r^2 +s^2)}e^{2i(r(x+y)+s(x-y))} \\ &\quad + \sum_{r=-\infty}^{\infty} \sum_{s=-\infty}^{\infty}q^{2\left[\left(r+\frac{1}{2} \right)^2 + \left(s+\frac{1}{2} \right)^2\right]} e^{i(2r+1)(x+y) + i(2s+1)(x-y)} \\ &= \vartheta_3(x+y,q^2)\vartheta_3(x-y,q^2) + \vartheta_2(x+y,q^2)\vartheta_2(x-y,q^2) \quad (1) \end{aligned} Some similar identities that can be derived using this method are: \begin{aligned} \vartheta_1(x,q) \vartheta_1(y,q) &= \vartheta_3(x+y,q^2)\vartheta_2(x-y,q^2) - \vartheta_2(x+y,q^2)\vartheta_3(x-y,q^2) \quad (2) \\ \vartheta_2(x,q) \vartheta_2(y,q) &= \vartheta_2(x+y,q^2)\vartheta_3(x-y,q^2) + \vartheta_3(x+y,q^2)\vartheta_2(x-y,q^2) \quad (3) \\ \vartheta_4(x,q) \vartheta_4(y,q) &= \vartheta_3(x+y,q^2)\vartheta_3(x-y,q^2) - \vartheta_2(x+y,q^2)\vartheta_2(x-y,q^2) \quad (4) \\ \vartheta_1(x,q) \vartheta_2(y,q) &= \vartheta_1(x+y,q^2)\vartheta_4(x-y,q^2) + \vartheta_4(x+y,q^2)\vartheta_1(x-y,q^2) \quad (5) \\ \vartheta_3(x,q) \vartheta_4(y,q) &= \vartheta_4(x+y,q^2)\vartheta_4(x-y,q^2) - \vartheta_1(x+y,q^2)\vartheta_1(x-y,q^2) \quad (6) \end{aligned} Squaring and subtracting identities (4) and (2) gives us: \begin{aligned} &\; \vartheta_4^2(x,q) \vartheta_4^2(y,q) - \vartheta_1^2(x,q) \vartheta_1^2(y,q)\\ &= \left[\vartheta_3^2(x+y,q^2)-\vartheta_2^2(x+y,q^2) \right]\times \left[\vartheta_3^2(x-y,q^2)-\vartheta_2^2(x-y,q^2) \right] \quad (7) \end{aligned} Putting $y=0$ in the above equation gives the result: $$\vartheta_3^2(x,q^2)-\vartheta_2^2(x,q^2) = \vartheta_4(x,q)\vartheta_4(0,q)$$ Now, using the above result in equation (7) gives us: $$\vartheta_4(x+y,q)\vartheta_4(x-y,q)\vartheta_4^2(0,q) = \vartheta_4^2(x,q) \vartheta_4^2(y,q) - \vartheta_1^2(x,q) \vartheta_1^2(y,q) \quad (8)$$ Note that all theta functions in this equation have the same nome $q$. More identities of this type can be derived by incrementing $x$ and/or $y$ by the half periods $\frac{\pi}{2},\frac{\pi \tau}{2},\frac{\pi+\pi\tau}{2}$: \begin{aligned} \vartheta_1(x+y)\vartheta_1(x-y)\vartheta_4^2(0) &= \vartheta_3^2(x) \vartheta_2^2(y) - \vartheta_2^2(x) \vartheta_3^2(y) \\ &= \vartheta_1^2(x) \vartheta_4^2(y) - \vartheta_4^2(x) \vartheta_1^2(y)\quad (9) \\ \vartheta_2(x+y)\vartheta_2(x-y)\vartheta_4^2(0) &= \vartheta_4^2(x) \vartheta_2^2(y) - \vartheta_1^2(x) \vartheta_3^2(y) \\ &= \vartheta_2^2(x) \vartheta_4^2(y) - \vartheta_3^2(x) \vartheta_1^2(y)\quad (10) \\ \vartheta_3(x+y)\vartheta_3(x-y)\vartheta_4^2(0) &= \vartheta_4^2(x) \vartheta_3^2(y) - \vartheta_1^2(x) \vartheta_2^2(y) \\ &= \vartheta_3^2(x) \vartheta_4^2(y) - \vartheta_2^2(x) \vartheta_1^2(y)\quad (11) \\ \vartheta_4(x+y)\vartheta_4(x-y)\vartheta_4^2(0) &= \vartheta_3^2(x) \vartheta_3^2(y) - \vartheta_2^2(x) \vartheta_2^2(y) \\ &= \vartheta_4^2(x) \vartheta_4^2(y) - \vartheta_1^2(x) \vartheta_1^2(y)\quad (12) \\ \end{aligned} One can keep on deriving similar identities by squaring and adding equations (1) and (2) and squaring and subtracting equations (3) and (2). For a complete list of such identities, see section 1.4 of [1]. Another approach to derive theta function identities is to utilize properties of doubly periodic functions. This method is used extensively by [2].

Consider the function: $$\frac{a \vartheta_1^2(z) + b \vartheta_4^2(z)}{\vartheta_2^2(z)}$$ This is a doubly periodic function with periods $\pi$ and $\pi \tau$. Furthermore, we can choose the constants $a$ and $b$ such that there is at most one simple pole in every cell. By section 20.13 of [2], such a function is a constant. By appropriately scaling the constants $a$ and $b$, we can make the function equal to 1. Therefore, there exists a relationship of the form: $$a \vartheta_1^2(z) + b \vartheta_4^2(z) = \vartheta_2^2(z)$$ To find $a$ and $b$, we can put $z = 0, \frac{\pi\tau}{2}$: \begin{aligned} b \vartheta_4^2(0) &= \vartheta_2^2(0) \; \implies b = \frac{\vartheta_2^2(0)}{\vartheta_4^2(0)} \\ a (i \mu\vartheta_4(0))^2 &= (\mu \vartheta_3(0))^2 \; \implies a = -\frac{\vartheta_3^2(0)}{\vartheta_4^2(0)} \end{aligned} Thus, we have obtained the identity: $$\vartheta_2^2(0) \vartheta_4^2(z) -\vartheta_3^2(0) \vartheta_1^2(z) = \vartheta_4^2(0) \vartheta_2^2(z) \quad (13)$$ A similar technique yields the identity: $$\vartheta_3^2(0) \vartheta_4^2(z) - \vartheta_2^2(0) \vartheta_1^2(z) = \vartheta_4^2(0) \vartheta_3^2(z) \quad (14)$$ Incrementing $z$ by $\frac{\pi}{2}$ in (11) and (12) gives two additional identities: \begin{aligned} \vartheta_4^2(0) \vartheta_1^2(z) &= \vartheta_2^2(0) \vartheta_3^2(z) - \vartheta_3^2(0) \vartheta_2^2(z) \quad (15) \\ \vartheta_4^2(0) \vartheta_4^2(z) &= \vartheta_3^2(0) \vartheta_3^2(z) - \vartheta_2^2(0) \vartheta_2^2(z) \quad (16) \end{aligned} With these relations one can express any theta function in terms of any other pair of theta functions. Putting $z=0$ in (16) gives the identity: $$\vartheta_4^4(0)+\vartheta_2^4(0) = \vartheta_3^4(0)$$

## The identity $\vartheta_1'(0)=\vartheta_2(0)\vartheta_3(0)\vartheta_4(0)$

This proof is taken from [1]. Differentiating equation (5) with respect to $x$ and substituting $x=y=0$, we get: $$\vartheta_1'(0,q) \vartheta_2(0,q) = 2 \vartheta_1'(0,q^2)\vartheta_4(0,q^2) \quad (17)$$ Substituting $x=y=0$ in equations (3) and (6) gives: \begin{aligned} \vartheta_2^2(0,q) &= 2\vartheta_2(0,q^2) \vartheta_3(0,q^2) \quad (18) \\ \vartheta_3(0,q)\vartheta_4(0,q) &= \vartheta_4^2(0,q^2) \quad (19) \end{aligned} Now, dividing (17) by both (18) and (19) gives $$\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)} = \frac{\vartheta_1'(0,q^2)}{\vartheta_2(0,q^2)\vartheta_3(0,q^2)\vartheta_4(0,q^2)}$$ The repeated application of this result gives: $$\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)} = \frac{\vartheta_1'(0,q^{2^n})}{\vartheta_2(0,q^{2^n})\vartheta_3(0,q^{2^n})\vartheta_4(0,q^{2^n})}$$ for all positive integers $n$. Letting $n\to\infty$ in the above equation, we find that: $$\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)} = \lim_{q\to 0}\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)} \quad (19)$$ From the definitions of the theta functions, it is evident that \begin{aligned} \vartheta_1'(0,q) &= 2q^{\frac{1}{4}} + \mathcal{O}(q^{\frac{9}{4}}) \\ \vartheta_2(0,q) &= 2q^{\frac{1}{4}} + \mathcal{O}(q^{\frac{9}{4}}) \\ \vartheta_3(0,q) &= 1 + \mathcal{O}(q) \\ \vartheta_4(0,q) &= 1 + \mathcal{O}(q) \\ \end{aligned} Therefore, the limit in equation (19) equals 1 and $$\vartheta_1'(0)=\vartheta_2(0)\vartheta_3(0)\vartheta_4(0)$$ is established as desired.

References

1. Derek F. Lawden (1989). Elliptic Functions and Applications. Springer-Verlag New York
2. E. T. Whittaker, G. N. Watson (1927). A Course of Modern Analysis. Cambridge University Press

## bookmark_borderSimplification of Incomplete Beta Functions

In this post, we will derive closed form solutions for two incomplete beta integrals. \begin{aligned} \int_0^1\frac{dx}{\sqrt{1-x}\sqrt[6]{9-x}\sqrt[3]x}&=\frac\pi{\sqrt3} \quad\quad (1)\\ \int_0^1\frac{dx}{\sqrt{1-x}\sqrt[4]x\sqrt[4]{2-x\sqrt3}} &= \frac{2\sqrt{2}\pi}{3\sqrt[8]{3}} \quad\quad (2) \end{aligned} These problems are from a math.stackexchange.com question posted by Vladimir Reshetnikov.

Alternate forms

The equations (1) and (2) can be expressed in terms of the incomplete beta function using some elementary hypergeometric identities. This makes our problems equivalent to proving that \begin{aligned} B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) &= \frac{\sqrt{3}}{2^{\frac{4}{3}}\pi}\Gamma\left(\frac{1}{3}\right)^3 \quad\quad (3)\\ B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) &= \frac{2}{3\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^2 \quad\quad (4) \end{aligned} Proof of (3) $$B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) = \int_0^{\frac{1}{9}}x^{-\frac{5}{6}}\left( 1-x\right)^{-\frac{2}{3}} dx$$ Substitute $x=\frac{y^3}{1+y^3}$ in the above integral to get $$B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) =3\int_0^{\frac{1}{2}}\frac{1}{\sqrt{y+y^4}}dy$$ We can now transform this integral into a beta integral using the substitution $y=\frac{1-z}{2+z}$. $$B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) =3\int_0^1\frac{1}{\sqrt{1-z^3}}dz = B\left( \frac{1}{3},\frac{1}{2}\right)=\frac{\sqrt{3}}{2^{\frac{4}{3}}\pi}\Gamma\left(\frac{1}{3}\right)^3$$ This proves equations (1) and (3).

Proof of (4)

The proof of integrals (2) and (4) is more involved. Using the substitution $x=\frac{4t^2}{1+4t^2}$, the integral transforms into the Weierstrass form: $$B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) =\int_0^{\frac{\sqrt{3}}{2}}x^{-\frac{3}{4}}(1-x)^{-\frac{3}{4}}dx=2\sqrt{2} \int_0^\alpha \frac{dt}{\sqrt{4t^3+t}}$$ where $\alpha=\frac{\sqrt{3+2\sqrt{3}}}{2}$. Let $\wp(z)$ denote the Weierstrass Elliptic function with invariants $g_2=-1$ and $g_3=0$. $\wp(z)$ satisfies the differential equation: $$\wp'(z)^2=4\wp(z)^3+\wp(z)$$ The half periods are $\omega_1=\frac{1+i}{4}L$ and $\omega_2=\frac{-1+i}{4}L$ where $L=\frac{1}{\sqrt{2\pi}}\Gamma\left(\frac{1}{4}\right)^2$ is the Lemniscate constant. Now, we can express our integral as follows: $$B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) =2\sqrt{2}\left(\wp^{-1}(0)-\wp^{-1}(\alpha) \right)$$Note that$$\int_0^\infty \frac{dt}{\sqrt{4t^3+t}}=\frac{1}{2\sqrt{2\pi}}\Gamma\left(\frac{1}{4}\right)^2=\frac{L}{2}$$ Therefore, $\wp^{-1}(0)=\frac{L}{2}$. To prove the claim we only need to prove that $\wp\left(\frac{L}{6}\right)=\alpha$. Using the addition theorem of $\wp(z)$ and the fact that $\wp\left(\frac{L}{2}\right)=0$ we get: \begin{aligned} \wp\left(\frac{L}{6}\right) &= \wp\left(\frac{L}{2}-\frac{L}{3}\right)\\ &= \frac{1}{4}\left(\frac{\wp'\left(\frac{L}{3}\right)}{\wp\left(\frac{L}{3}\right)} \right)^2-\wp\left(\frac{L}{3}\right) \\ &= \frac{4\wp\left( \frac{L}{3}\right)^3+\wp\left(\frac{L}{3}\right)}{4\wp\left(\frac{L}{3}\right)^2}-\wp\left(\frac{L}{3}\right)\\ &= \frac{1}{4\wp\left(\frac{L}{3}\right)}\quad\quad\quad (5) \end{aligned} On the other hand, by the duplication theorem: \begin{aligned} \wp\left(\frac{L}{3}\right)&= \frac{1}{4}\frac{\left(6\wp\left(\frac{L}{6}\right)^2+\frac{1}{2}\right)^2}{4\wp\left(\frac{L}{6}\right)^3+\wp\left(\frac{L}{6}\right)}-2\wp\left(\frac{L}{6}\right)\quad\quad\quad ({6}) \end{aligned} Combine equations (5) and (6) to get: $$\frac{1+4\wp\left(\frac{L}{6}\right)^2}{\left(6\wp\left(\frac{L}{6}\right)^2+\frac{1}{2} \right)^2-32\wp\left(\frac{L}{6}\right)^4-8\wp\left(\frac{L}{6}\right)^2}=1 \\ \implies 16\wp\left(\frac{L}{6}\right)^4-24\wp\left(\frac{L}{6}\right)^2-3=0$$ Since $\wp\left( \frac{L}{6}\right)$ is positive, we conclude that $$\wp\left(\frac{L}{6}\right)=\frac{\sqrt{3+2\sqrt{3}}}{2}$$ This completes our proof of equation (4) and (2).