1. ## Fourier-serie

Calculate :
$\begin{array}{l}
P = \frac{\pi }{8}\left( {\frac{{\sin (x)}}{{1^3 }} + \frac{{\sin (3x)}}{{3^3 }} + \frac{{\sin (5x)}}{{5^3 }} + .......} \right) \\
Q = \frac{\pi }{6} - \left( {\frac{{\cos (2x)}}{{1^2 }} + \frac{{\cos (4x)}}{{2^2 }} + \frac{{\cos (6x)}}{{3^2 }} + .....} \right) \\
\end{array}$

2. I think you are looking for the following trick:
$\int_{0}^{2\pi} \! P \sin m x \, dx = \frac{\pi}{8} \int _{0}^{2\pi} \! \left(\sum_{n=1}^{+\infty} \frac{\sin nx}{n^3}\right) \sin m x \, dx$
$P_m \int_{0}^{2\pi} \! \sin m x \, dx = \frac{\pi}{8} \int _{0}^{2\pi} \! \frac{\sin^2 mx}{m^3} \, dx$
$P_m = \frac{\frac{\pi}{8} \int _{0}^{2\pi} \! \frac{\sin^2 mx}{m^3} \, dx}{\int_{0}^{2\pi} \! \sin m x \, dx}$
$P = \sum_{m=1}^{\infty} P_m$.
Then use a similar trick for $Q$.