# Simplification 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).