# A closed form for $\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx$

Is it possible to evaluate in a closed form integrals containing a squared hypergeometric function, like in this example? \begin{align}S&=\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx\\\vphantom{=}\\&=\frac{1}{4\pi}\int_0^1\left(\sum_{n=0}^\infty\frac{4n+1}{8^n}\cdot\frac{\Gamma\left(2n-\frac{1}{2}\right)}{\Gamma(n+1)^2}\cdot x^n\right)^2dx\end{align} It is approximately $$S\approx0.8263551866500213413164525287...$$

Your integral has an elementary closed form, that was correctly stated by Cleo in her answer without proof: $$S=\int_0^1\left({_2F_1}\!\left(-\tfrac14,\tfrac54;\,1;\,\tfrac{x}2\right)\right)^2dx=\frac{8\sqrt2+4\ln\left(\sqrt2-1\right)}{3\pi}.\tag1$$

Proof: Using DLMF 14.3.6 we can express the hypergeometric function in the integrand as the Legendre function of the $1^{st}$ kind (also known as the Ferrers function of the $1^{st}$ kind) with fractional index: $${_2F_1}\!\left(-\tfrac14,\tfrac54;\,1;\,\tfrac{x}2\right)=P_{\small1/4}(1-x).\tag2$$ Now the integral can be written as $$S=\int_0^1\left(P_{\small1/4}(1-x)\right)^2dx=\int_0^1\left(P_{\small1/4}(x)\right)^2dx.\tag3$$ To evaluate it, we use formula 7.113 on page 769 in Gradshteyn & Ryzhyk: $$\int_0^1P_\nu(x)\,P_\sigma(x)\,dx=\\\frac{\frac{\Gamma\left(\frac12+\frac\nu2\right)\,\Gamma\left(1+\frac\sigma2\right)}{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(1+\frac\nu2\right)}\sin\!\left(\frac{\pi\sigma}2\right)\cos\!\left(\frac{\pi\nu}2\right)-\frac{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(1+\frac\nu2\right)}{\Gamma\left(\frac12+\frac\nu2\right)\,\Gamma\left(1+\frac\sigma2\right)}\sin\!\left(\frac{\pi\nu}2\right)\cos\!\left(\frac{\pi\sigma}2\right)}{\frac\pi2(\sigma-\nu)(\sigma+\nu+1)}.\tag4$$ Note that in our case $\nu=\sigma=\frac14$, so we cannot use this formula directly because of the term $(\sigma-\nu)$ in the denominator. Instead, we let $\nu=\frac14$ and find the limit for $\sigma\to\frac14$: $$S=\lim\limits_{\sigma\to{\small1/4}}\int_0^1P_{\small1/4}(x)\,P_\sigma(x)\,dx=\\\lim\limits_{\sigma\to{\small1/4}}\frac{\frac{\Gamma\left(\frac58\right)\,\Gamma\left(1+\frac\sigma2\right)}{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(\frac98\right)}\sin\!\left(\frac{\pi\sigma}2\right)\cos\!\left(\frac\pi8\right)-\frac{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(\frac98\right)}{\Gamma\left(\frac58\right)\,\Gamma\left(1+\frac\sigma2\right)}\sin\!\left(\frac\pi8\right)\cos\!\left(\frac{\pi\sigma}2\right)}{\frac\pi2(\sigma-\frac14)(\sigma+\frac54)}.\tag5$$ To evaluate the limit, we use l'Hôpital's rule. This gives quite a big expression that I will not copy here. It contains values of the gamma and digamma functions at rational points, that could be simplified to elementary terms using the Gauss digamma theorem and identities given in the MathWorld and in the famous Vidūnas paper, yielding the desired result $(1)$.

Indeed, we can have a more general result: $$\int_0^1\left({_2F_1}\!\left(-\nu,\nu+1;\,1;\,\tfrac x2\right)\right)^2dx=\int_0^1\left(P_\nu\left(x\right)\right)^2dx=\\\frac{1+\!\left[\psi_0\!\left(1+\frac\nu2\right)-\psi_0\!\left(\frac12+\frac\nu2\right)\right]\frac{\sin(\pi\nu)}\pi}{1+2\nu}.\tag6$$

• +1) Very nice. I had the exact same work up to the evaluation of the final limit via L'Hopital, which for some reason kept coming out wrong so I gave up. Question: did you know all those references off the top of your head or did you have to hunt them down? They look incredibly useful. Aug 19, 2014 at 1:59
• I unsuccessfully tried to solve this problem last year when it was posted. When I saw it again today, I realized that the parameters of the hypergeometric function here match the expression for the Legendre function that read about recently. Then I just looked up that integral in Gradshteyn & Ryzhik and calculated the limit. Aug 19, 2014 at 2:08
• A pleasure to follow your precise and informative elaboration. 1+ Aug 19, 2014 at 14:36

Yes, it is possible in some cases, for example, $$S=\frac{8\sqrt2+4\ln\left(\sqrt2-1\right)}{3\pi}$$

Incomplete solution:

In fact, we have $${}_2F_1(-\frac14,\frac54;1;\frac{x}{2})=\frac{8\sqrt{2}}{\pi\sqrt{2+\sqrt{2x}}}((2+\sqrt{2x})E(\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}})-K(\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}})).$$

Here $E(m)=\int^1_0\sqrt{\frac{1-mt^2}{1-t^2}}~dt$ and $K(m)=\int^1_0\frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}$ are the complete elliptic integrals

Therefore, by substituting $y=\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}}$ (so that $x=\frac{y^2}{(2-y)^2}$), we have $$S=\frac{16}{\pi^2}\int^{2\sqrt{2}-2}_{0}\frac{y}{(2-y)^4}\left(4E(y)-(2-y)K(y)\right)^2~dy.$$

$$%E'(y)=\frac{E(y)-K(y)}{2y}\\K'(y)=\frac{E(y)-(1-y)K(y)}{2y(1-y)}\\\int E(y)dy=\frac{2(1+y)E(y)-2(1-y)K(y)}{3}\\\int K(y)dy=2E(y)-(1-y)K(y)$$

This may be doable with some clever integration by parts.