PlanetPhysics/Determination of Fourier Coefficients

Suppose that the real function $$f$$ may be presented as sum of the Fourier series: $$\begin{matrix} f(x) \;=\; \frac{a_0}{2}+\sum_{m=0}^\infty(a_m\cos{mx}+b_m\sin{mx}) \end{matrix}$$ Therefore, $$f$$ is periodic with period $$2\pi$$.\, For expressing the Fourier coefficients $$a_m$$ and $$b_m$$ with the function itself, we first multiply the series (1) by $$\cos{nx}$$ ($$n \in \mathbb{Z}$$) and integrate from $$-\pi$$ to $$\pi$$.\, Supposing that we can integrate termwise, we may write $$\begin{matrix} \int_{-\pi}^\pi\!f(x)\cos{nx}\,dx \,=\, \frac{a_0}{2}\!\int_{-\pi}^\pi\!\cos{nx}\,dx +\!\sum_{m=0}^\infty\!\left(a_m\!\int_{-\pi}^\pi\!\cos{mx}\cos{nx}\,dx+b_m\!\int_{-\pi}^\pi\!\sin{mx}\cos{nx}\,dx\right)\!. \end{matrix}$$ When\, $$n = 0$$,\, the equation (2) reads $$\begin{matrix} \int_{-\pi}^\pi f(x)\,dx = \frac{a_0}{2}\cdot2\pi = \pi a_0, \end{matrix}$$ since in the sum of the right hand side, only the first addend is distinct from zero.

When $$n$$ is a positive integer, we use the product formulas of the trigonometric identities, getting $$\int_{-\pi}^\pi\cos{mx}\cos{nx}\,dx = \frac{1}{2}\int_{-\pi}^\pi[\cos(m-n)x+\cos(m+n)x]\,dx,$$ $$\int_{-\pi}^\pi\sin{mx}\cos{nx}\,dx = \frac{1}{2}\int_{-\pi}^\pi[\sin(m-n)x+\sin(m+n)x]\,dx.$$ The latter expression vanishes always, since the sine is an odd function.\, If\, $$m \neq n$$,\, the former equals zero because the antiderivative consists of sine terms which vanish at multiples of $$\pi$$; only in the case\, $$m = n$$\, we obtain from it a non-zero result $$\pi$$.\, Then (2) reads $$\begin{matrix} \int_{-\pi}^\pi f(x)\cos{nx}\,dx = \pi a_n \end{matrix}$$ to which we can include as a special case the equation (3).

By multiplying (1) by $$\sin{nx}$$ and integrating termwise, one obtains similarly $$\begin{matrix} \int_{-\pi}^\pi f(x)\sin{nx}\,dx = \pi b_n. \end{matrix}$$ The equations (4) and (5) imply the formulas $$a_n \;=\; \frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos{nx}\,dx \quad (n = 0,\,1,\,2,\,\ldots)$$ and $$b_n \;=\; \frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin{nx}\,dx \quad (n = 1,\,2,\,3,\,\ldots)$$ for finding the values of the Fourier coefficients of $$f$$.