determination of Fourier coefficients

Suppose that the real function $f$ may be presented as sum of the Fourier series:

$f(x)={\displaystyle \frac{a_0}{2}}+{\displaystyle \sum _{m=0}^{\mathrm{\infty }}}(a_m\cos mx+b_m\sin mx)$(1)

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

${\displaystyle {\int }_{-\pi }^{\pi }}f(x)\cos nxdx={\displaystyle \frac{a_0}{2}}{\displaystyle {\int }_{-\pi }^{\pi }}\cos nxdx+{\displaystyle \sum _{m=0}^{\mathrm{\infty }}}\left(a_m{\displaystyle {\int }_{-\pi }^{\pi }}\cos mx\cos nxdx+b_m{\displaystyle {\int }_{-\pi }^{\pi }}\sin mx\cos nxdx\right).$(2)

When  $n=0$,  the equation (2) reads

${\displaystyle {\int }_{-\pi }^{\pi }}f(x)𝑑x={\displaystyle \frac{a_0}{2}}\cdot 2\pi =\pi a_0,$(3)

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

$${\int }_{-\pi }^{\pi }\cos mx\cos nxdx=\frac{1}{2}{\int }_{-\pi }^{\pi }[\cos (m-n)x+\cos (m+n)x]𝑑x,$$

$${\int }_{-\pi }^{\pi }\sin mx\cos nxdx=\frac{1}{2}{\int }_{-\pi }^{\pi }[\sin (m-n)x+\sin (m+n)x]𝑑x.$$

The latter expression vanishes always, since the sine is an odd function.  If  $m\ne n$,  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

${\displaystyle {\int }_{-\pi }^{\pi }}f(x)\cos nxdx=\pi a_n$(4)

to which we can include as a special case the equation (3).

By multiplying (1) by $\sin nx$ and integrating termwise, one obtains similarly

${\displaystyle {\int }_{-\pi }^{\pi }}f(x)\sin nxdx=\pi b_n.$(5)

The equations (4) and (5) imply the formulas

$$a_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\cos nxdx\mathit{\hspace{1em}}(n=0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots })$$

and

$$b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\sin nxdx\mathit{\hspace{1em}}(n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots })$$

for finding the values of the Fourier coefficients of $f$.