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.

  • $\begingroup$ 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





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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.