# How to prove $\int_0^\pi\frac{\ln(2+\cos\phi)}{\sqrt{2+\cos\phi}}d\phi=\frac{\ln3}{\sqrt3}K\left(\sqrt{\frac23}\right)$?

How can I prove the following conjectured identity? $$\int_0^\pi\frac{\ln(2+\cos\phi)}{\sqrt{2+\cos\phi}}d\phi\stackrel?=\frac{\ln3}{\sqrt3}K\left(\sqrt{\frac23}\right),\tag1$$ where $K(x)$ is the complete elliptic integral of the 1ˢᵗ kind: $$K(x)={_2F_1}\left(\frac12,\frac12;\ 1;\ x^2\right)\cdot\frac\pi2.\tag2$$

• $$I(n)=\int_0^\pi(2+\cos\phi)^nd\phi\quad\iff\quad\int_0^\pi\dfrac{\ln(2+\cos x)}{\sqrt{2+\cos x}}dx=I'\Big(-\tfrac12\Big)$$ – Lucian Feb 27 '14 at 8:06
• With Mathematica I got $\mbox{LHS} - \mbox{RHS} = -0.169648$ – Felix Marin Mar 7 '14 at 23:12
• @FelixMarin You should use EllipticK[x^2] to represent $K(x)$ in Mathematica. See the MathWorld link in my question. – Vladimir Reshetnikov Mar 8 '14 at 0:36
• @VladimirReshetnikov $0$ k. I just checked it: $-9.15712 \times 10^{-13}$. – Felix Marin Mar 8 '14 at 4:41
• @Felix Marin that's small enough of a margin of error to be caused by numerical instability/rounding errors – frogeyedpeas Jul 19 '14 at 4:25

$\int^{\pi}_0\cos^{2k+1}\phi d\phi=0$, and $$\int^{\pi}_0\cos^{2k}\phi d\phi=\sqrt{\pi}\Gamma(k+1/2)/\Gamma(k+1)=2^{-2k}\pi\binom{2k}{k}.$$

Therefore \begin{align*} I(n) &=\int^{\pi}_0\sum_{m=0}^{\infty}2^{n-m}\binom{n}{m}\cos^m\phi~d\phi\\ &=2^n\sum_{m=0}^{\infty}2^{-m}\binom{n}{m}\int^{\pi}_0\cos^m\phi~d\phi\\ &=2^n\pi\sum_{k=0}^{\infty}2^{-2k}\binom{n}{2k}2^{-2k}\binom{2k}{k}\\ &=2^n\pi~{}_2F_1\left(\frac{1-n}2,-\frac{n}{2};1;\frac14 \right)\\ &=2^n\pi\left(\frac23\right)^{-n}~{}_2F_1\left(-n,\frac{1}{2};1;\frac23\right)\\ &=3^n\pi~{}_2F_1\left(-n,\frac{1}{2};1;\frac23\right)\\ \end{align*} Using DLMF 15.8.13 with $a=-n$, $b=1/2$ and $z=2/3$.

We note that $I(-1/2)=\frac{2}{\sqrt{3}}K(\sqrt{2/3})$.

Edit: We have $$I(n)=3^n\pi~{}_2F_1\left(-n,\frac{1}{2};1;\frac23\right)=\frac{\pi}{\sqrt{3}}~{}_2F_1\left(n+1,\frac{1}{2};1;\frac23\right)=3^{n+1/2}I(-n-1).$$

Therefore, If we write $J(n)=3^{-n/2}I(n)$, then $J(n)=J(-n-1)$, and consequently $J'(-1/2)=0$.

Thus \begin{align*} I'(-1/2) &=\left.\frac{d}{dn}3^{n/2}J(n)\right|_{n=-1/2}\\ &=\left.\left(3^{n/2}J'(n)+3^{n/2}\frac{\log 3}{2}J(n)\right)\right|_{n=-1/2}\\ &=3^{-1/4}\frac{\log 3}{2}J(-1/2)\\ &=\frac{\log 3}{2}I(-1/2)\\ &=\frac{\log 3}{\sqrt{3}}K(\sqrt{2/3}). \end{align*}

• Here are some formulas. – Lucian Mar 1 '14 at 14:50
• very nice solution +1 – Shobhit Bhatnagar Mar 2 '14 at 7:30