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Hi this will soon end my posts on Log Sine integrals, and we can progress into other classes of integrals.

Given the generalized log sine integral,

$$\rm{Ls}_n^{(k)}(\sigma) = -\int_0^{\sigma}\theta^k \Big(\ln\big(2\sin\tfrac{\theta}{2}\big)\Big)^{n-1-k}\,d\theta$$

The log sine integral I am trying to calculate is given by $k=1$, $n=4$,

$$ \rm{Ls}_4^{(1)}\big(\tfrac{\pi}{3}\big)=-\int_0^{\pi/3}\theta\, \ln^2\big(2\sin\tfrac{\theta}{2}\big)\,d\theta=-\,\frac{17\pi^4}{6480}. $$ What a beautiful closed form! This integral is known as one of many "Log Sine integrals at $\pi/3$." It also seems that many people have not been responding to these log sine integral posts. This is quite puzzling as these integrals are fundamental to mathematical analysis and are quite old..

If anybody wants literature on these integrals, please let me know.

I am not quite sure how to evaluate this. Perhaps a change of variables or logarithmic expansion may work. Thanks for your help.

EDIT: This is also of interest when discussing Mahler measures. http://en.wikipedia.org/wiki/Mahler_measure

Tito Piezas III
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Jeff Faraci
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1 Answers1

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Hint:

$$\int_0^\frac{\pi}{3}\theta\ln^2\left(2\sin\dfrac{\theta}{2}\right)d\theta$$

$$=\int_0^\frac{\pi}{3}\theta\ln^2\sqrt{2(1-\cos\theta)}~d\theta$$

$$=\int_0^\frac{\pi}{3}\dfrac{\theta}{4}\ln^2(2(1-\cos\theta))~d\theta$$

$$=\int_0^\frac{\pi}{3}\dfrac{\theta}{4}(\ln2+\ln(1-\cos\theta))^2~d\theta$$

$$=\int_0^\frac{\pi}{3}\dfrac{\theta}{4}(\ln^22+2\ln2\ln(1-\cos\theta)+\ln^2(1-\cos\theta))~d\theta$$

Harry Peter
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