Hello I am trying to integrate $$ I=\int_0^\pi \theta^2 \ln^2\big(2\cosh\frac{\theta}{2}\big)d \theta $$ which is similar to Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$., however we have hyperbolic functions in this example. We can possibly write the integrand as $$ \theta^2 \ln^2\big(2\cosh\frac{\theta}{2}\big)=\theta^2\big(\ln 2+\ln \cosh \frac{\theta}{2}\big)^2=\theta^2\left(\ln^2(2)+\ln^2\cosh\frac{\theta}{2}+2\ln (2)\ln \cosh\big(\frac{\theta}{2}\big)\right). $$ We can now write $$ I=\ln^2(2)\int_0^\pi \theta^2d\theta +\int_0^\pi\theta^2 \ln^2 \cosh \frac{\theta}{2}d\theta+2\ln 2 \int_0^\pi\theta^2 \ln \cosh{\frac{\theta}{2}}d\theta. $$ Change of variables $y=\theta/2$ we obtain $$ I=\frac{\pi^3\ln^2(2)}{3}+16\ln(2)\int_0^{\pi/2} y^2 \ln \cosh (y) dy+8\int_0^{\pi/2} y^2 \ln^2 \cosh (y) dy. $$ I am stuck here, thanks.








$$=\left[\dfrac{\theta^5}{20}-\sum\limits_{n=1}^\infty\dfrac{(-1)^n(n^3\theta^3+3n^2\theta^2+6n\theta+6)e^{-n\theta}}{n^5}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^{n+k}((n+k)^2\theta^2+2(n+k)\theta+2)e^{-(n+k)\theta}}{nk(n+k)^3}\right]_0^\pi$$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$$=\dfrac{\pi^5}{20}-\sum\limits_{n=1}^\infty\dfrac{(-1)^n((\pi^3n^3+3\pi^2n^2+6\pi n+6)e^{-\pi n}-6)}{n^5}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^{n+k}((\pi^2(n+k)^2+2\pi(n+k)+2)e^{-\pi(n+k)}-2)}{nk(n+k)^3}$$

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    $\begingroup$ Are you sure the second line is valid? You dropped the square on the logarithm and I'm not sure where the $2$ came from. $\endgroup$ – Cameron Williams Apr 13 '14 at 0:17
  • $\begingroup$ @Harry How does this make the problem simpler though? Do you have suggestions how to sum these integrands/sum the result of the integrals? Luckily the integrals seem easy, but the sums... Thank you. $\endgroup$ – Jeff Faraci Apr 14 '14 at 17:37
  • $\begingroup$ @Integrals, integrator refuse to calculate $\int\theta^2(\ln(1+e^{-\theta}))^2~d\theta$ , so I think my calculation is already most simple. $\endgroup$ – Harry Peter Apr 15 '14 at 16:35
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    $\begingroup$ @Integrals, but if the integral expressing in terms of known special functions is much more complicated than expressing in infinite series forms, I will prefer for latter. $\endgroup$ – Harry Peter Apr 22 '14 at 13:20
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    $\begingroup$ @HarryPeter Likewise, thank you for this solution. I guess this is the best we can do. Nobody seems to come up with any other solutions. Although evaluating the sums isn't so clear, thanks $\endgroup$ – Jeff Faraci Apr 27 '14 at 0:12

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