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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: $$K(x)={_2F_1}\left(\frac12,\frac12;\ 1;\ x^2\right)\cdot\frac\pi2.\tag2$$

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  • $\begingroup$ $$I(n)=\int_0^\pi(2+\cos\phi)^nd\phi\quad\iff\quad\int_0^\pi\dfrac{\ln(2+\cos x)}{\sqrt{2+\cos x}}dx=I'\Big(-\tfrac12\Big)$$ $\endgroup$
    – Lucian
    Feb 27, 2014 at 8:06
  • $\begingroup$ With Mathematica I got $\mbox{LHS} - \mbox{RHS} = -0.169648$ $\endgroup$ Mar 7, 2014 at 23:12
  • $\begingroup$ @FelixMarin You should use EllipticK[x^2] to represent $K(x)$ in Mathematica. See the MathWorld link in my question. $\endgroup$ Mar 8, 2014 at 0:36
  • $\begingroup$ @VladimirReshetnikov $0$ k. I just checked it: $-9.15712 \times 10^{-13}$. $\endgroup$ Mar 8, 2014 at 4:41
  • $\begingroup$ @Felix Marin that's small enough of a margin of error to be caused by numerical instability/rounding errors $\endgroup$ Jul 19, 2014 at 4:25

2 Answers 2

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$\int^{\pi}_0\cos^{2k+1}\phi d\phi=0$, 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 $$ \begin{align*} I'(-1/2) &=\left.\frac{d}{dn}3^{n/2}J(n)\right|_{n=-1/2}\\ &=\left.\left(3^{n/2}J'(n)+3^{n/2}\frac{\log 3}{2}J(n)\right)\right|_{n=-1/2}\\ &=3^{-1/4}\frac{\log 3}{2}J(-1/2)\\ &=\frac{\log 3}{2}I(-1/2)\\ &=\frac{\log 3}{\sqrt{3}}K(\sqrt{2/3}). \end{align*} $$

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    $\begingroup$ Here are some formulas. $\endgroup$
    – Lucian
    Mar 1, 2014 at 14:50
  • $\begingroup$ very nice solution +1 $\endgroup$ Mar 2, 2014 at 7:30
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$$I=\int_0^\pi \frac{\ln(2+\cos x)}{\sqrt{2+\cos x}}dx\overset{2+\cos x=t}=\int_1^3\frac{\color{blue}{\ln t}}{\sqrt{t}}\frac{dt}{\sqrt{1-(2-t)^2}}\overset{t\to \frac{3}{t}}=\int_1^3\frac{\color{red}{\ln\left(\frac{3}{t}\right)}}{\sqrt t}\frac{dt}{\sqrt{1-(2-t)^2}}$$ $$\require{cancel}\Rightarrow 2I= \int_1^3\frac{\color{blue}{\cancel{\ln 3}}+\color{red}{\ln 3-\cancel{\ln t}}}{\sqrt t}\frac{dt}{\sqrt{1-(2-t)^2}}\overset{t=2+\cos x}=\ln 3\int_0^\pi\frac{dx}{\sqrt{2+\cos x}}$$ $$\Rightarrow I\overset{x\to 2x}=\frac{\ln 3}{\sqrt 3}\int_0^\frac{\pi}{2}\frac{dx}{\sqrt {\frac{2+\cos(2x)}{3}}}=\frac{\ln 3}{\sqrt 3}\int_0^\frac{\pi}{2} \frac{dx}{\sqrt{1-\frac{2}{3}\sin^2 x}}=\frac{\ln 3}{\sqrt 3}K\left(\sqrt{\frac23}\right)$$

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    $\begingroup$ Very clever use of integral substitutions. +1 $\endgroup$
    – Paramanand Singh
    Mar 13, 2020 at 13:08

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