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Hi I am trying to calculate $$ I:=\int_0^\pi \theta^2 \ln^2\big(2\sin\frac{\theta}{2}\big)d \theta. $$ Here is a related Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$.. This paper may also be of interest to people here : http://www.math.uwo.ca/~dborwein/cv/zeta4.pdf.

We can expand the log in the integral to obtain three interals, one trivial, the other 2 are not so easy, any ideas? I tried doing the following $$ \left( \ln 2 +\ln \sin \frac{\theta}{2} \right)^2=\ln^2(2)+\ln^2\sin\frac{\theta}{2}+2\ln (2)\ln \sin\big(\frac{\theta}{2}\big). $$ We can write I as $$ I=\ln^2(2)\int_0^\pi \theta^2d\theta +\int_0^\pi\theta^2 \ln^2 \sin \frac{\theta}{2}d\theta+2\ln 2 \int_0^\pi\theta^2 \ln \sin{\frac{\theta}{2}}d\theta. $$ Change of variables $x=\theta/2$ and performing the trivial integral we obtain $$ I=\frac{\pi^3\ln^2 2}{3}+8\int_0^{\pi/2} x^2 \ln^2 \sin x\, dx+16\ln 2\int_0^{\pi/2} x^2 \ln \sin x \, dx. $$ I am stuck at this point, I was trying to somehow work these two integrals into the form of $$ \int_0^{\pi/2} \ln \sin x dx= \frac{-\pi\ln(2)}{2}\approx -1.08879 $$ but couldn't do so. Thanks.

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  • $\begingroup$ For what it may be worth, an equivalent form of the integral is $I=2\int_{0}^{\infty}\arctan^2{t}\log^2{\left(\frac{4t^2}{1+t^2}\right)}\frac{dt}{1+t^2}$. $\endgroup$ – David H May 2 '14 at 5:58
  • $\begingroup$ Oh great, this again... $\endgroup$ – IAmNoOne May 6 '14 at 9:52
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Hint:

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

$$=\int_\pi^0(\pi-\theta)^2\ln^2\left(2\sin\dfrac{\pi-\theta}{2}\right)d(\pi-\theta)$$

$$=\int_0^\pi\theta^2\ln^2\left(2\cos\dfrac{\theta}{2}\right)d\theta-2\pi\int_0^\pi\theta\ln^2\left(2\cos\dfrac{\theta}{2}\right)d\theta+\pi^2\int_0^\pi\ln^2\left(2\cos\dfrac{\theta}{2}\right)d\theta$$

Which may use the result in Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$.

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