How to evaluate $\int_{0}^{2\pi}x^2\ln (1-\cos x)dx$?

Question

Wolfram Alpha shows that $$\int_{0}^{2\pi}x^2\ln (1-\cos x)dx = -\frac{8}{3} \pi (\pi^2 \ln(2) + 3 \zeta(3))$$ I tried to use the Fourier series $$\ln (1-\cos x)=-\sum_{n=1}^{\infty} \frac{\cos^nx}{n}.$$ I am not sure how to continue from this point. I need some help.

Answer 1

HINT

By using the basic trigonometric identity

$$1-\cos(x)=2\sin^2\left(\frac x2\right)$$

yor given integral becomes

$$\begin{align} \int_{0}^{2\pi}x^2\ln (1-\cos x)~dx &= \int_{0}^{2\pi}x^2 \ln\left(2\sin^2\left(\frac x2\right)\right)~dx\\ &=\int_{0}^{2\pi}x^2\ln(2)~dx+2\int_{0}^{2\pi}x^2 \ln\left(\sin\left(\frac x2\right)\right)~dx\\ &=\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\ln(\sin x)~dx \end{align}$$

where within the second integral the substitution $x=\frac x2$ was used. Now use the Fourier series expansion

$$\ln(\sin x)=-\ln(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}$$

to further get

$$\begin{align} \frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\ln(\sin x)~dx&=\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\left[-\ln(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}\right]~dx\\ &=\frac{8\pi^3}{3}\ln(2)-16\int_0^{\pi}x^2\ln(2)~dx-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx\\ &=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx \end{align}$$

The second integral can be evaluated by applying integration by parts. Can you finish it from hereon?

Answer 2

Thanks mrtaurho, I am able to finish it. $$\begin{align} \int_{0}^{2\pi}\pi^2\ln(1-\cos x)dx &=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx\\ &=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\left[-x^2\frac{\sin(2nx)}{2n}+\frac{2x\cos(2nx)}{4n^2}-\frac{2\sin(2nx)}{8n^3}\right]_{0}^{\pi}\\ &=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}{n}\frac{2\pi}{4n^2}\\ &= -\frac{8\pi^3}3\ln(2)-8\pi\sum_{n=1}^{\infty}\frac{1}{n^3}\\ &=-\frac{8}{3}\pi(\pi^2\ln2+3\zeta(3)) \end{align}$$