Proving $\int_{0}^{1} \frac{\tanh^{-1}\sqrt{x(1-x)}}{\sqrt{x(1-x)}}dx=\frac{1}{3}(8C-\pi\ln(2+\sqrt{3}))$ for an identity of Srinivasa Ramanujan

Question

Ramanujan is supposed to have given more than five thousand elegant results, a good number of them are yet to be proved or disproved.

Yesterday in the comment section of

Proving that $ \sum_{k=0}^\infty\frac1{2k+1}{2k \choose k}^{-1}=\frac {2\pi}{3\sqrt{3}} $

A wonderful Ramanujan identity $$S=\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}{2k \choose k}^{-1}=\frac{1}{3}(8C-\pi\ln(2+\sqrt{3}))~~~~(1)$$ was showcased, Mathematica also gives this out.

My Attempt to prove (1) by hand:

Note the integral representation of the reciprocal of the binomial co-efficient: $${n \choose j}^{-1}=(n+1)\int_{0}^{1} x^j (1-x)^{n-j}~ dx~~~~(2)$$ $$S=\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}{2k \choose k}^{-1}= \int_{0}^{1} \sum_{k=0}^{\infty} \frac{[x(1-x)]^{k}}{(2k+1)} dx= \int_{0}^{1} \frac{\tanh^{-1}\sqrt{x(1-x)}}{\sqrt{x(1-x)}} dx~~~~(3)$$

The question is: How to get this integral (3) by hand ?

Answer 1

You can give Feynman's trick a shot. \begin{align*} I&=\int _0^1\frac{\operatorname{arctanh} \left(\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx\\[3mm] I\left(a\right)&=\int _0^1\frac{\operatorname{arctanh} \left(a\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx\\[3mm] I'\left(a\right)&=\int _0^1\frac{1}{1-a^2x\left(1-x\right)}\:dx=\frac{4}{a\sqrt{4-a^2}}\arctan \left(\frac{a}{\sqrt{4-a^2}}\right)\\[3mm] \int _0^1I'\left(a\right)da&=4\underbrace{\int _0^1\frac{1}{a\sqrt{4-a^2}}\arctan \left(\frac{a}{\sqrt{4-a^2}}\right)\:da}_{t=\frac{a}{\sqrt{4-a^2}}}\\[3mm] I&=8\underbrace{\int _0^{\frac{1}{\sqrt{3}}}\frac{\arctan \left(t\right)}{4t\sqrt{1+t^2}}\:dt}_{t=\tan\left(x\right)}=2\int _0^{\frac{\pi }{6}}\frac{x\sec \left(x\right)}{\tan \left(x\right)}\:dx\\[3mm] &=2\int _0^{\frac{\pi }{6}}\frac{x}{\sin \left(x\right)}\:dx \end{align*} That integral has been evaluated here by Zacky, using its result yields $$\boxed{I=\int _0^1\frac{\operatorname{arctanh} \left(\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx=\frac{\pi}{3}\ln(2-\sqrt 3) +\frac{8}{3}G}$$

---

One can also find the last integral by using the Weierstrass substitution. \begin{align*} 2\int _0^{\frac{\pi }{6}}\frac{x}{\sin \left(x\right)}\:dx&=4\underbrace{\int _0^{2-\sqrt{3}}\frac{\arctan \left(t\right)}{t}\:\:dt}_{\operatorname{IBP}}\\[3mm] &=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)-4\underbrace{\int _0^{2-\sqrt{3}}\frac{\ln \left(t\right)}{1+t^2}\:dt}_{t=\tan\left(x\right)}\\[2mm] &=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)-4\int _0^{\frac{\pi }{12}}\ln \left(\tan \left(x\right)\right)\:dx\\[3mm] &=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+8\sum _{k=1}^{\infty }\frac{1}{2k-1}\int _0^{\frac{\pi }{12}}\cos \left(2\left(2k-1\right)x\right)\:dx\\[3mm] &=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+4\sum _{k=1}^{\infty }\frac{\sin \left(\frac{\pi }{6}\left(2k-1\right)\right)}{\left(2k-1\right)^2}\\[3mm] &=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+\frac{8}{3}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{\left(2k-1\right)^2}\\[3mm] &=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+\frac{8}{3}G \end{align*}

Answer 2

Note

$\int_{0}^{1} \frac{\tanh^{-1}\sqrt{x(1-x)}}{\sqrt{x(1-x)}}dx \overset{x=\sin^2t} =\int_0^{\pi/2} 2 \tanh^{-1}\frac{\sin 2t}2dt = \int_0^{\pi/2}\ln\left(\frac{1+\frac{\sin2t}2}{1-\frac{\sin2t}2}\right)dt $

Solving the integral $\int_0^{\pi/2}\log\left(\frac{2+\sin2x}{2-\sin2x}\right)\mathrm dx$ $=\frac{1}{3}[8C-\pi\ln(2+\sqrt{3})]$

Answer 3

If we denote the sum by $S$, we have the short proof

$$S=\sum_{k=0}^{\infty} \int_0^1 \frac{x^{2k}}{\displaystyle (2k+1){2k \choose k}}\textrm{d}x=4\int_0^1\frac{\arcsin(x/2)}{x\sqrt{4-x^2}}\textrm{d}x=2\int_0^{\pi/6}\frac{x}{\sin(x)}\textrm{d}x$$ $$=4\int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\textrm{d}x=4\operatorname{Ti}_2(2-\sqrt{3})=\frac{8}{3}G+\frac{\pi}{3}\log(2-\sqrt{3}).$$ Q.E.D.

$\operatorname{Ti}_2(2-\sqrt{3})$ is a special value of the inverse tangent integral that is extracted immediately by (also) using a famous result by Ramanujan,
$$\sum_{n=1}^{\infty} \frac{\sin(2(2n-1)x)}{(2n-1)^2}=\operatorname{Ti}_2(\tan(x))-x \log(\tan(x)), \ 0<x<\frac{\pi}{2},$$ and these details may be found in the book, (Almost) Impossible Integrals, Sums, and Series, pages $215$-$216$.

A note: To have a clear picture of the arcsine series used one can express $\displaystyle {2k \choose k}$ in terms of $\displaystyle {2k+2 \choose k+1}$ and then reindex the series. That's all.