Hi I am trying to calculate this integral: $$ I=\frac{1}{\pi}\int_0^{\pi/3}\log\left( \frac{1+2\cos\theta}{2}+\sqrt{\bigg( \frac{1+2\cos\theta}{2} \bigg)^2-1} \right)\ d\theta. $$ The integral evaluation is related to Mahler measures. You may also recognize that it is related to the evaluation of log-sine integrals at $\dfrac{\pi}{3}$.
This integral $I$ is somewhat related to $$ \int_0^1\log\big|2a+2b\cos (2\pi \theta)\big|\ d\theta=\log \big( |a|+\sqrt{a^2-b^2} \big), $$ for $a,b\in\mathbb{R}$ with $|a|\geq|b|> 0$. This can be seen in Gradstein and Ryzhik's tables of integrals, but I am not sure how to use that to help me to solve $I$.


  • $\begingroup$ This may help: $$\int_1^{(1+2\cos \theta)/2} \frac{du}{\sqrt{u^2-1}} = \left.\log\left(u+\sqrt{u^2-1}\right)\right|_1^{(1+2\cos \theta)/2}=\log\left(\frac{1+2\cos \theta}{2}+\sqrt{\left(\frac{1+2\cos \theta}{2}\right)^2-1}\right)$$ so your integral is really $$\frac{1}{\pi}\int_0^{\pi/3}\left(\int_1^{(1+2\cos\theta)/2}\frac{du}{\sqrt{u^2-1}}\right)d\theta$$ $\endgroup$ Apr 15 '14 at 23:01
  • $\begingroup$ thats the same integral we started with after performing the definite integral over u, you will be back with a logarithmic integral in terms of $\theta$ $\endgroup$ Apr 15 '14 at 23:04
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    $\begingroup$ Ah. I just wanted to throw that out there in case someone does know how to solve it through the double integral. $\endgroup$ Apr 15 '14 at 23:05
  • $\begingroup$ @EugeneBulkin Well Thank you! I mean, I am not sure how to work with it because we cannot switch the order of integration $\endgroup$ Apr 15 '14 at 23:06
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    $\begingroup$ I'll try my luck to evaluate this integral Jeff. :D $\endgroup$
    – Tunk-Fey
    Apr 21 '14 at 19:21

The integral can be expressed in terms of a series of Gauss hypergeometric functions. It is doubtfull that it would be possible to go further on this way.

$I = \frac{1}{\pi}\int_0^{\pi/3}\ln\left(\mu(\theta)+\sqrt{\mu^2(\theta)-1}\right)d\theta = \frac{1}{\pi}\int_0^{\pi/3}\cosh^{-1}(\mu(\theta))$, where $\mu(\theta) = \frac{1}{2}+\cos(\theta)$.

$$\cosh^{-1}(\mu)=\sum_{k=0}^\infty \frac{(-1)^k\Gamma(k+1/2)}{2^{k-1/2}(2k+1)k!\sqrt{\pi}}(\mu-1)^{k+1/2}$$ so $$\mu(\theta)=\frac{1}{2}+\cos(\theta)\to I = \frac{1}{\pi^{3/2}}\sum_{k=0}^\infty \frac{(-1)^k\Gamma(k+1/2)}{2^{k-1/2}(2k+1)k!}(\cos(\theta)-1/2)^{k+1/2}$$

Let $I_k = \int(cos(\theta) - 1/2)^{k+1/2}d\theta$. Then $$I_k = -\frac{1}{(2k+3)2^k}\sqrt{\frac{2}{3}}(2\cos(\theta)-1)^{k+3/2}F_1\left(k+\frac{3}{2};\frac{1}{2},\frac{1}{2};k+\frac{5}{2};\frac{1}{3}(1-2\cos(\theta)),2\cos(\theta)-1\right)+C$$ where $C$ is some constant and $F_1$ is the Appell hypergeometric function of two variables.

$2\cos(0)-1=1$, $2\cos(\pi/3)-1=0$, and $F_1(a;b,b;c;-1/3,1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}_2F_1(a,b;c-b;-1/3)$, so $$I_k=\frac{1}{(2k+3)2^k}\sqrt{\frac{2}{3}}\frac{\Gamma(k+5/2)\Gamma(1/2)}{\Gamma(1)\Gamma(k+2)} {}_2F_1\left(k+\frac{3}{2},\frac{1}{2};k+2;-\frac{1}{3}\right)$$ $$I_k=\sqrt{\frac{2\pi}{3}}\frac{(2k+1)\Gamma(k+1/2)}{2^{k+2}(k+1)!} {}_2F_1\left(k+\frac{3}{2},\frac{1}{2};k+2;-\frac{1}{3}\right)$$

Finally, from $I = \frac{1}{\pi^{3/2}}\sum_{k=0}^\infty \frac{(-1)^k\Gamma(k+1/2)}{2^{k-1/2}(2k+1)k!}I_k$, we have $$\frac{1}{\pi}\int_0^{\pi/3}\ln\left(\mu(\theta)+\sqrt{\mu^2(\theta)-1}\right)d\theta=\boxed{\frac{1}{\pi\sqrt{3}}\sum_{k=0}^\infty \frac{(-1)^k(\Gamma(k+1/2))^2}{2^{2k+1}(2k+1)k!(k+1)!} {}_2F_1\left(k+\frac{3}{2},\frac{1}{2};k+2;-\frac{1}{3}\right)}$$

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    $\begingroup$ What is this? Why did you not use LaTeX? Your answer seems correct although it's so difficult to me to understand it. $\endgroup$ Apr 24 '14 at 11:43
  • $\begingroup$ Did you check this numerically? I will also wait for other solutions in the meantime. Thanks $\endgroup$ Apr 24 '14 at 16:36
  • $\begingroup$ @ Integrals : Yes, I checked it numerically. The result is approximately 0.251330433713252231 The series is fast convergent. This allows an accurate computation insofar the hypergeometric function can be computed with a sufficient accurency. Of course, my contribution is not a full answer to the question because it is not a closed form such as expected. You are right to wait for a more complete answer. $\endgroup$
    – JJacquelin
    Apr 24 '14 at 21:09
  • $\begingroup$ @JJacquelin Since this is the only answer posted, and is correct despite not being a closed form. I still award you +400. Thanks very much, however I am still looking for a closed form. But for now, this is great. Thanks again $\endgroup$ Apr 27 '14 at 16:15

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