Does $\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\right)\,\ln\left(1-\frac{\cos^2\phi}{C^2}\right)$ have a closed form?

Question

I am wondering if anyone has a nice way of approaching the following definite integral $\newcommand{\dilog}{\operatorname{Li}_2}$

$$\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\right)\,\ln\left(1-\frac{\cos^2\phi}{C^2}\right)\,.$$

Here $C$ is a positive, real constant that satisfies the constraint $C>1$. So far I have tried a simple $u$ substitution, $u = \cos\left(\phi\right)/C$. However, this doesn't get me anywhere. I have also tried performing a series expansion in small $1/C$ in the second log, performing the integration, and then summing in powers of $1/C$. However, the sum does not simplify nicely.

I have also tried relating the expression to $\dilog\left(1-\frac{\cos^2\phi}{C^2}\right)$ and $\dilog\left(\frac{\cos^2\phi}{C^2}\right)$ and performing the integration of these polylog functions using their series representation. However I have trouble performing the summation for the $\dilog\left(1-\frac{\cos^2\phi}{C^2}\right)$ term.

Answer 1

This integral has a closed form in terms of dilogarithms; the idea is that the series $$\ell(r,\phi)=\log(1-2r\cos\phi+r^2)=-2\sum_{n=1}^\infty\frac{r^n}{n}\cos n\phi\qquad(|r|<1)$$ (obtained from $1-2r\cos\phi+r^2=(1-re^{i\phi})(1-re^{-i\phi})$ and the power series of $\log$) may be considered as a Fourier series, giving $\int_0^{2\pi}\ell(r,2\phi)\,d\phi=0$ for $|r|\leqslant 1$ and, by Parseval's identity, $$\int_0^{2\pi}\ell(a,2\phi)\ell(b,2\phi)\,d\phi=4\pi\sum_{n=1}^\infty\frac{a^n}{n}\frac{b^n}{n}=4\pi\operatorname{Li}_2(ab)$$ for $|a|,|b|\leqslant 1$ (all the boundary cases are attainable in the limit). Now $$ \log\left(\frac{\cos^2\phi}{C^2}\right)=\ell(-1,2\phi)-\log(4C^2),\\ \log\left(1-\frac{\cos^2\phi}{C^2}\right)=\ell(r,2\phi)-\log(4rC^2),\\ \color{blue}{r:=2C^2-1-2C\sqrt{C^2-1}}, $$ reducing the given integral to the above. The answer is $\color{blue}{2\operatorname{Li}_2(-r)+\log(4C^2)\log(4rC^2)}$.

Answer 2

Claim. Let $H_k$ denote the $k$th harmonic number. Then for any $0<c<1$, $$\frac1{2\pi}\int_0^{2\pi}\log(c\cos^2\phi)\log(1-c\cos^2\phi)\,d\phi=\small2\log c\log\frac{1+\sqrt{1-c}}{2}-\sum_{k\ge1}\frac{(2k-1)!!(H_{k-1/2}-H_k)}{k\cdot2^k\cdot k!}c^k.$$ Proof: Let $\psi=c\cos^2\phi$ so that $d\psi=-2c\sin\phi\cos\phi\,d\phi$. Then \begin{align}\frac1{2\pi}\int_0^{2\pi}\log(c\cos^2\phi)\log(1-c\cos^2\phi)\,d\phi&=\frac2\pi\int_0^{\pi/2}\log(c\cos^2\phi)\log(1-c\cos^2\phi)\,d\phi\\&=\frac1\pi\int_0^c\frac{\log\psi\log(1-\psi)}{\sqrt\psi\sqrt{c-\psi}}\,d\psi\\&=-\frac1\pi\frac d{ds}\int_0^c\frac{\log(1-\psi)}{\psi^s(c-\psi)^{1/2}}\,d\psi\bigg\vert_{s=1/2}.\end{align} Using the Taylor expansion of $\log(1-\psi)$, we obtain \begin{align}\int_0^c\frac{\log(1-\psi)}{\psi^s(c-\psi)^{1/2}}\,d\psi&=\sum_{k\ge1}\frac1k\int_0^c\psi^{k-s}(c-\psi)^{-1/2}\,d\psi\\&=\sum_{k\ge1}\frac1k\cdot\frac{\Gamma(k-s+1)\Gamma(1/2)c^{k-s+1/2}}{\Gamma(k-s+3/2)}\end{align} where the last equality is due to a modification of the standard Beta function. This gives \begin{align}\frac1{2\pi}\int_0^{2\pi}\log(c\cos^2\phi)\log(1-c\cos^2\phi)\,d\phi&=-\sqrt{\frac c\pi}\sum_{k\ge1}\frac{c^k}k\frac d{ds}\frac{\Gamma(k-s+1)}{c^s\Gamma(k-s+3/2)}\bigg\vert_{s=1/2}\end{align} and since \begin{align}\small\frac d{ds}\frac{\Gamma(k-s+1)}{c^s\Gamma(k-s+3/2)}\bigg\vert_{s=1/2}&\small=-\frac{\Gamma(k-s+1)}{c^s\Gamma(k-s+3/2)}\left(\log c+\psi^{(0)}(k-s+1)-\psi^{(0)}\left(k-s+\frac32\right)\right)\bigg\vert_{s=1/2}\\&=-\frac{\Gamma(k+1/2)}{\sqrt c\cdot\Gamma(k+1)}\left(\log c+\psi^{(0)}\left(k+\frac12\right)-\psi^{(0)}(k+1)\right)\\&=-\sqrt{\frac\pi c}\frac{(2k-1)!!}{2^k\cdot k!}(\log c+H_{k-1/2}-H_k),\end{align} the integral simplifies to \begin{align}\frac1{2\pi}\int_0^{2\pi}\log(c\cos^2\phi)\log(1-c\cos^2\phi)\,d\phi&=-\sum_{k\ge1}\frac{(2k-1)!!(\log c+H_{k-1/2}-H_k)}{k\cdot2^k\cdot k!}c^k.\end{align} The result follows by observing the Taylor expansion identity $$\sum_{k\ge1}\frac{(2k-1)!!}{k\cdot2^k\cdot k!}c^k=-2\log\frac{1+\sqrt{1-c}}2.\tag*{$\square$}$$ Note the interchanges of summations, derivatives and integrals are permitted due to absolute convergence.

Answer 3

It seems that a series solution is possible

$$I=\int_0^{2\pi} \log \left(\frac{\cos ^2(x)}{c^2}\right) \log \left(1-\frac{\cos ^2(x)}{c^2}\right)\,dx=4\int_0^{\frac \pi 2} \log \left(\frac{\cos ^2(x)}{c^2}\right) \log \left(1-\frac{\cos ^2(x)}{c^2}\right)\,dx$$ Expanding the second logarithms $$I= -4\sum_{n=1}^\infty \frac 1 {n\,c^{2n}}\int_0^{\frac \pi 2} \log \left(\frac{\cos ^2(x)}{c^2}\right)\cos ^{2n}(x) \,dx$$ The integrals $$J_n=\int\log \left(\frac{\cos ^2(x)}{c^2}\right)\cos ^{2n}(x) \,dx$$ are relatively simple. One integration by parts gives $$J_n=-\frac{\cos ^{2 n+1}(x) \log \left(\frac{\cos ^2(x)}{c^2}\right) \, _2F_1\left(\frac{1}{2},n+\frac{1}{2};n+\frac{3}{2};\cos ^2(x)\right)}{2 n+1}-$$ $$\frac{1}{2} \Gamma \left(n+\frac{1}{2}\right)^2 \cos ^{2 n+1}(x) \, _3\tilde{F}_2\left(\frac{1}{2},n+\frac{1}{2},n+\frac{1}{2};n+\frac{3}{2},n+\frac {3}{2};\cos ^2(x)\right)$$

Using the bounds $$K_n=\int_0^{\frac \pi 2}\log \left(\frac{\cos ^2(x)}{c^2}\right)\cos ^{2n}(x) \,dx$$ $$K_n=\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{1}{2}\right) \left(H_{n-\frac{1}{2}}-H_n-2 \log (c)\right)}{2 \Gamma (n+1)}$$ Then $$I=2 \sqrt{\pi }\sum_{n=1}^\infty \frac{ \Gamma \left(n+\frac{1}{2}\right) }{n \,c^{2 n}\,\Gamma (n+1)}\left(2 \log (c)+H_n-H_{n-\frac{1}{2}}\right)$$ $$\color{blue}{I=8 \pi \log (c) \left(\log (2 c)-\cosh ^{-1}(c)\right)+2 \sqrt{\pi }\sum_{n=1}^\infty \frac{ \Gamma \left(n+\frac{1}{2}\right) }{n \,c^{2 n}\,\Gamma (n+1)}\left(H_n-H_{n-\frac{1}{2}}\right)}$$

If we denote the summand by $a_n$ we have, for large values of $n$ $$\frac{a_{n+1}}{a_n}=\frac 1{c^2}\left(1-\frac{5}{2 n} \right)+O\left(\frac{1}{n^2}\right)$$ which leads to a quite fast convergence.

For $c=2$, the partial sums (from $n=0$ to $n=p$) $$\left( \begin{array}{cc} p & S_p \\ 0 & 1.20789 \\ 1 & 1.39219 \\ 2 & 1.51043 \\ 3 & 1.52617 \\ 4 & 1.52870 \\ 5 & 1.52914 \\ 6 & 1.52923 \\ 7 & 1.52925 \end{array} \right)$$

We ca have save calculation time using for large enough $n$ the expansion $$a_n=\frac{\sqrt \pi} {c^{2n}\, n^{\frac 52}}\left(1-\frac{3}{8 n}+\frac{5}{128 n^2}+\frac{35}{1024 n^3}+O\left(\frac{1}{n^4}\right)\right)$$