Integral evaluated with the Guassian function

Today we show the proof of this integral posted by \[ \int_{0}^{1} \frac{dx}{(x^2-x+1)\sqrt[3]{x(1-x)}} = \frac{4\pi}{3\sqrt{3}}\] The proof relies on a special case of the Gaussian function involving one of the roots of the unity, $e^{\frac{i\pi}{3}}$

Proof:

First note that \[ \Im \left(\frac{4}{\sqrt{3}(1-i\sqrt{3}-2x)}\right) = \frac{1}{x^2-x+1 }\] Hence \begin{align*} I = \int_{0}^{1} \frac{dx}{(x^2-x+1)\sqrt[3]{x(1-x)}} = & \Im \left( \frac{4}{\sqrt{3}}\int_{0}^{1} \frac{dx}{(1-i\sqrt{3}-2x)\sqrt[3]{x(1-x)}}\right)\\ =& \Im \left( \frac{4}{\sqrt{3}(1-i\sqrt{3})}\int_{0}^{1} \frac{dx}{\left(1-\frac{2}{1-i\sqrt{3}}x\right)\sqrt[3]{x(1-x)}}\right)\\ \end{align*} Note that \[ \frac{2}{1-i\sqrt{3}} = \frac{2}{1-i\sqrt{3}}\left(\frac{1+i\sqrt{3}}{1+i\sqrt{3}}\right) = \frac{1+i\sqrt{3}}{2} = e^{\frac{i\pi}{3}} \] Hence \[ I = \Im \left( \frac{2}{\sqrt{3}}e^{\frac{i\pi}{3}} \underbrace{\int_{0}^{1} \frac{dx}{\left(1-e^{\frac{i\pi}{3}}x\right)\sqrt[3]{x(1-x)}}}_{J}\right) \] Recall the integral representation of the Gaussian function: \[F(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_{0}^{1} x^{b-1}(1-x)^{c-b-1}(1-xz)^{-a} dx \] This is called the Pochhammer integral and converges whenever $\quad \Re(c)>\Re(b)>0 \quad \arg\left(|1-z|\right)\lt \pi $

Therefore if we put $\displaystyle a=1, b= \frac{2}{3}$ and $\displaystyle c=\frac{4}{3}$ we have \begin{align*} J = \int_{0}^{1}\left(1-e^{\frac{i\pi}{3}} x\right)^{-1} x^{-\frac{1}{3}}(1-x)^{-\frac{1}{3}} dx =& \frac{\Gamma^2\left(\frac{2}{3}\right)}{\Gamma\left(\frac{4}{3}\right)} F\left(1, \frac{2}{3};\frac{4}{3};e^{\frac{i\pi}{3}}\right) \end{align*} The Gaussian hypergeometric function satisfies the following property: \[F\left(3a,\tfrac{1}{3}+a;\tfrac{2}{3}+2a;{\mathrm{e}}^{\frac{\mathrm{i}\pi}{3% }}\right)=\sqrt{\pi}{\mathrm{e}}^{\frac{\mathrm{i}\pi a}{2}}\left(\frac{16}{2% 7}\right)^{(3a+1)/6}\frac{\Gamma\left(\frac{5}{6}+a\right)}{\Gamma\left(\frac{% 2}{3}+a\right)\Gamma\left(\frac{2}{3}\right)},\] If we put $a = \frac{1}{3}$ \[F\left(1,\tfrac{2}{3};\tfrac{4}{3};{\mathrm{e}}^{\frac{\mathrm{i}\pi}{3% }}\right)=\sqrt{\pi}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}\left(\frac{16}{2 7}\right)^{\frac{1}{3}}\frac{\Gamma\left(\frac{7}{6}\right)}{\Gamma\left(\frac{2}{3}\right)} = \sqrt{\pi}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}\frac{2\sqrt[3]{2}}{3}\frac{\Gamma\left(\frac{7}{6}\right)}{\Gamma\left(\frac{2}{3}\right)} \] Therefore \begin{align*} J = \int_{0}^{1}\left(1-e^{\frac{i\pi}{3}} x\right)^{-1} x^{-\frac{1}{3}}(1-x)^{-\frac{1}{3}} dx =& \sqrt{\pi}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}\frac{2\sqrt[3]{2}}{3}\frac{\Gamma\left(\frac{7}{6}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{4}{3}\right)} \\ =& \pi{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}\frac{2}{3} \end{align*} Hence \begin{align*} I = \int_{0}^{1} \frac{dx}{(x^2-x+1)\sqrt[3]{x(1-x)}} = & \Im \left( \frac{2}{\sqrt{3}}e^{\frac{i\pi}{3}} \int_{0}^{1} \frac{dx}{\left(1-e^{\frac{i\pi}{3}}x\right)\sqrt[3]{x(1-x)}}\right)\\ =& \Im \left(\pi{\mathrm{e}}^{\frac{\mathrm{i}\pi }{2}}\frac{4}{3\sqrt{3}}\right)\\ =& \Im \left(i\pi\frac{4}{3\sqrt{3}}\right)\\ =& \frac{4\pi}{3\sqrt{3}} \end{align*} Therefore, we can conclude \[ \boxed{\int_{0}^{1} \frac{dx}{(x^2-x+1)\sqrt[3]{x(1-x)}} = \frac{4\pi}{3\sqrt{3}}}\]