HARD INTEGRAL - PART I

Today´s blog we will evaluate a hard integral, namely:

$\int_0^{\pi/4}\ln^2\left( \sin(x) \right)\,dx=\frac{\pi^3}{192}+\frac{3\pi \ln^2(2)}{16}+\frac{\beta(2)\ln(2)}{2}- \Im \operatorname{Li}_3\left(1-i\right)$

Where $\beta(2)$ is Catalan´s constant (G=0.915965594177219015054603514932384110774) and $\operatorname{Li}_3\left(x\right)$ is the Polylogarithm function.

Recall the expansion (see here)

$\ln^2\left(2 \sin(x) \right)=\frac{\pi^2}4-\pi x+x^2+2 \sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}\cos(2nx)}{n}$ (1)

Therefore

$\ln^2\left( \sin(x) \right)=-\ln^2(2)-2\ln(2)\ln(\sin(x))+\frac{\pi^2}4-\pi x+x^2+2 \sum_{n=1}^\infty \frac{\operatorname{H}_{n}\cos(2nx)}{n}-2 \sum_{n=1}^\infty \frac{\cos(2nx)}{n^2}$ (2)

Integrating both sides of (2) from 0 to $\pi/4$

$\begin{aligned} \int_0^{\pi/4}\ln^2\left( \sin(x) \right)\,dx&=-\ln^2(2)\int_0^{\pi/4}\,dx-2\ln(2)\int_0^{\pi/4}\ln(\sin(x))\,dx+\frac{\pi^2}4\int_0^{\pi/4}\,dx-\pi\int_0^{\pi/4} x\,dx+\int_0^{\pi/4}x^2\,dx+2 \sum_{n=1}^\infty \frac{\operatorname{H}_{n}}{n}\int_0^{\pi/4}\cos(2nx)\,dx-2 \sum_{n=1}^\infty \frac{1}{n^2}\int_0^{\pi/4}\cos(2nx)\,dx\\ &=-\frac{\pi \ln^2(2)}{4}+\ln(2)\beta(2)+\frac{\pi \ln^2 (2)}{2}+\frac{\pi^3}{16}-\frac{\pi^3}{32}+\frac{\pi^3}{192}+\sum_{n=1}^\infty \frac{\operatorname{H}_{n}\sin\left( \frac{n \pi}{2}\right)}{n^2}-\sum_{n=1}^\infty \frac{\sin\left( \frac{n \pi}{2}\right)}{n^3}\\ &=-\frac{\pi \ln^2(2)}{4}+\ln(2)\beta(2)+\frac{\pi \ln^2 (2)}{2}+\frac{7\pi^3}{192}+\left(- \Im \operatorname{Li}_3\left(1-i\right)-\frac{\beta(2)\ln(2)}{2}-\frac{\pi}{16}\ln^2(2) \right)-\frac{\pi^3}{32}\\ &=\frac{\pi^3}{192}+\frac{3\pi \ln^2(2)}{16}+\frac{\beta(2)\ln(2)}{2}- \Im \operatorname{Li}_3\left(1-i\right) \qquad \blacksquare \end{aligned}$

We used that (see Appendix below):

$\sum_{n=1}^\infty \frac{\operatorname{H}_{n}\sin\left( \frac{n \pi}{2}\right)}{n^2}=- \Im \operatorname{Li}_3\left(1-i\right)-\frac{\beta(2)\ln(2)}{2}-\frac{\pi}{16}\ln^2(2)$

And

$\sum_{n=1}^\infty \frac{\sin\left( \frac{n \pi}{2}\right)}{n^3}&=\frac{\pi^3}{32}$

Appendix

Recall the generator function (see this post)

$\sum_{n=1}^\infty \frac{\operatorname{H}_{n}x^n}{n^2}=\operatorname{Li}_3\left( x\right)-\operatorname{Li}_3\left(1-x \right)+\ln\left( 1-x\right)\operatorname{Li}_2\left(1-x \right)+\frac12\ln\left(x\right)\ln^2\left(1-x\right)+\zeta(3)$ (A.1)

Letting $x=e^{i \pi/2}$ in (A.1) we obtain

$\begin{aligned} \sum_{n=1}^\infty \frac{\operatorname{H}_{n}\cos\left( \frac{n \pi}{2}\right)}{n^2}+i\sum_{n=1}^\infty \frac{\operatorname{H}_{n}\sin\left( \frac{n \pi}{2}\right)}{n^2}&=\operatorname{Li}_3\left( i\right)-\operatorname{Li}_3\left(1-i\right)+\ln\left( 1-i\right)\operatorname{Li}_2\left(1-i \right)+\frac12\ln\left(e^{i\pi/2}\right)\ln^2\left(1-e^{i\pi/2}\right)+\zeta(3)\\ &=\left(-\frac{3 \zeta(3)}{32}+i\frac{\pi^3}{32} \right)-\operatorname{Li}_3\left(1-i\right)+\left(\frac{\pi^2}{16}-i\beta(2)-i\frac{\pi \ln(2)}{4} \right)\left(\frac{\ln(2)}{2}-i\frac{\pi}{4} \right)+\frac12 \left(\frac{i \pi}{2}\right)\left(\frac{\ln^2(2)}{4}-\frac{\pi^2}{16}-\frac{i \pi}{4}\ln(2) \right)+\zeta(3)\\ \end{aligned}$

We are looking for the Imaginary part of the equation above, hence

$\begin{aligned} \sum_{n=1}^\infty \frac{\operatorname{H}_{n}\sin\left( \frac{n \pi}{2}\right)}{n^2}&=\frac{\pi^3}{32}- \Im \operatorname{Li}_3\left(1-i\right)-\frac{\pi^3}{64}-\frac{\beta(2)\ln(2)}{2}-\frac{\pi \ln^2(2)}{8}+\frac{\pi\ln^2(2)}{16}-\frac{\pi^3}{64}\\ &=- \Im \operatorname{Li}_3\left(1-i\right)-\frac{\beta(2)\ln(2)}{2}-\frac{\pi}{16}\ln^2(2) \end{aligned}$