数学家傅里叶猜想任何一个周期函数均可分解为一系列不同振幅、不同频率和不同相位的正弦函数的组合（由此，复杂的周期函数可分解为若干简单的三角函数，易于分析和处理）。即：
$$f(t)=C+\sum _{k=1}^{\infty }{A}_k\sin (k{\omega }_{0}t+{\phi }_k)$$

其中，$C$是常数项，也可理解为振幅为$0$的三角函数，其频率和相位任意；$A_k$、$k{\omega }_0$和${\phi }_k$表示正弦函数$k$的振幅、角频率和相位（${\omega }_0$为基频，即为角频率的最小单元）。$k$取值从$1$到正无穷。因此，傅里叶级数也就是一个无穷级数。

接着我们利用两角和公式$\sin (A+B)=\sin (A)\cos (B)+\cos (A)\sin (B)$对上式做分解，得到：
$$f(t)=C+\sum _{k=1}^{\infty }{A}_k\cos ({\phi }_k)\sin (k{\omega }_{0}t)+\sum _{k=1}^{\infty }{A}_k\sin ({\phi }_k)\cos (k{\omega }_{0}t)$$

令$A_k\cos ({\phi }_k)=a_k$和$A_k\sin ({\phi }_k)=b_k$，得到$f(t)$的傅里叶级数表达式：
$$f(t)=C+\sum _{k=1}^{\infty }{a}_k\sin (k{\omega }_{0}t)+\sum _{k=1}^{\infty }{b}_k\cos (k{\omega }_{0}t)$$

如果上式中的系数$C$、$a_n$和$b_n$可解，则$f(t)$可分解为一系列简单的三角函数。

我们选择正弦函数$k$中最大的周期$\max (\frac{2\pi }{k{\omega }_0})=\frac{2\pi }{{\omega }_0}$作为积分区间（保证所有的三角函数在此区间内积分为$0$）。对$f(t)$的傅里叶级数表达式在$[0,\frac{2\pi }{{\omega }_0}]$内积分得：
$$\begin{array}{rl}{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)dt& ={\int }_0^{\frac{2\pi }{{\omega }_0}}Cdt+\sum _{k=1}^{\infty }{a}_k{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin (k{\omega }_{0}t)dt+\sum _{k=1}^{\infty }{b}_k{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos (k{\omega }_{0}t)dt\\ & =C\frac{2\pi }{{\omega }_0}+\sum _{k=1}^{\infty }{a}_{k}0+\sum _{k=1}^{\infty }{b}_{k}0\\ & =C\frac{2\pi }{{\omega }_0}\end{array}$$

因此，系数$C$得解：
$$C=\frac{{\omega }_0}{2\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)dt$$

接着$f(t)$的傅里叶级数表达式左右两边同时乘以$\cos (n{\omega }_{0}t)$，得到：
$$\cos (n{\omega }_{0}t)f(t)=C\cos (n{\omega }_{0}t)+\sum _{k=1}^{\infty }{a}_k\sin (k{\omega }_{0}t)\cos (n{\omega }_{0}t)+\sum _{k=1}^{\infty }{b}_k\cos (k{\omega }_{0}t)\cos (n{\omega }_{0}t)$$

对上式在$[0,\frac{2\pi }{{\omega }_0}]$内积分得：
$$\begin{array}{rl}& {\int }_0^{\frac{2\pi }{{\omega }_0}}\cos (n{\omega }_{0}t)f(t)dt=C{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos (n{\omega }_{0}t)dt+\sum _{k=1}^{\infty }{a}_k{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin (k{\omega }_{0}t)\cos (n{\omega }_{0}t)dt+\sum _{k=1}^{\infty }{b}_k{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos (k{\omega }_{0}t)\cos (n{\omega }_{0}t)dt\\ & =C0+\sum _{k=1}^{\infty }{a}_k{\int }_0^{\frac{2\pi }{{\omega }_0}}\frac{1}{2}[\sin (k{\omega }_{0}t+n{\omega }_{0}t)+\sin (k{\omega }_{0}t-n{\omega }_{0}t)]dt\\ & +\sum _{k=1}^{\infty }{b}_k{\int }_0^{\frac{2\pi }{{\omega }_0}}\frac{1}{2}[\cos (k{\omega }_{0}t+n{\omega }_{0}t)+\cos (k{\omega }_{0}t-n{\omega }_{0}t)]dt\\ & =\sum _{k=1}^{\infty }\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin ((k+n){\omega }_{0}t)dt+\sum _{k=1}^{\infty }\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin ((k-n){\omega }_{0}t)dt\\ & +\sum _{k=1}^{\infty }\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((k+n){\omega }_{0}t)dt+\sum _{k=1}^{\infty }\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((k-n){\omega }_{0}t)dt\\ & =\sum _{k=1}^{\infty }\frac{a_k}{2}0+\sum _{k=1}^{\infty }\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin ((k-n){\omega }_{0}t)dt+\sum _{k=1}^{\infty }\frac{b_k}{2}0+\sum _{k=1}^{\infty }\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((k-n){\omega }_{0}t)dt\\ & =\sum _{k=1}^{n-1}\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}-\sin ((n-k){\omega }_{0}t)dt+\frac{a_n}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin ((n-n){\omega }_{0}t)dt+\sum _{k=n+1}^{\infty }\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin ((k-n){\omega }_{0}t)dt\\ & +\sum _{k=1}^{n-1}\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((n-k){\omega }_{0}t)dt+\frac{b_n}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((n-n){\omega }_{0}t)dt+\sum _{k=n+1}^{\infty }\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((k-n){\omega }_{0}t)dt\\ & =\sum _{k=1}^{n-1}\frac{a_k}{2}0+\frac{a_n}{2}0+\sum _{k=n+1}^{\infty }\frac{a_k}{2}0+\sum _{k=1}^{n-1}\frac{b_k}{2}0+\frac{b_n}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}1dt+\sum _{k=n+1}^{\infty }\frac{b_k}{2}0\\ & =\frac{b_n}{2}\frac{2\pi }{{\omega }_0}\end{array}$$

因此，系数$b_n$得解：
$$b_n=\frac{{\omega }_0}{\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos (n{\omega }_{0}t)f(t)dt$$

接着$f(t)$的傅里叶级数表达式左右两边同时乘以$\sin (n{\omega }_{0}t)$，得到：
$$\sin (n{\omega }_{0}t)f(t)=C\sin (n{\omega }_{0}t)+\sum _{k=1}^{\infty }{a}_k\sin (k{\omega }_{0}t)\sin (n{\omega }_{0}t)+\sum _{k=1}^{\infty }{b}_k\cos (k{\omega }_{0}t)\sin (n{\omega }_{0}t)$$

对上式在$[0,\frac{2\pi }{{\omega }_0}]$内积分得：
$$\begin{array}{rl}& {\int }_0^{\frac{2\pi }{{\omega }_0}}\sin (n{\omega }_{0}t)f(t)dt=C{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin (n{\omega }_{0}t)dt+\sum _{k=1}^{\infty }{a}_k{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin (k{\omega }_{0}t)\sin (n{\omega }_{0}t)dt+\sum _{k=1}^{\infty }{b}_k{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos (k{\omega }_{0}t)\sin (n{\omega }_{0}t)dt\\ & =C0+\sum _{k=1}^{\infty }{a}_k{\int }_0^{\frac{2\pi }{{\omega }_0}}\frac{1}{2}[\cos (k{\omega }_{0}t-n{\omega }_{0}t)-\cos (k{\omega }_{0}t+n{\omega }_{0}t)]dt\\ & +\sum _{k=1}^{\infty }{b}_k{\int }_0^{\frac{2\pi }{{\omega }_0}}\frac{1}{2}[\sin (k{\omega }_{0}t+n{\omega }_{0}t)-\sin (k{\omega }_{0}t-n{\omega }_{0}t)]dt\\ & =\sum _{k=1}^{\infty }\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((k-n){\omega }_{0}t)dt-\sum _{k=1}^{\infty }\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((k+n){\omega }_{0}t)dt\\ & +\sum _{k=1}^{\infty }\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin ((k+n){\omega }_{0}t)dt-\sum _{k=1}^{\infty }\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin ((k-n){\omega }_{0}t)dt\\ & =\sum _{k=1}^{\infty }\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((k-n){\omega }_{0}t)dt-\sum _{k=1}^{\infty }\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}0+\sum _{k=1}^{\infty }\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}0-\sum _{k=1}^{\infty }\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin ((k-n){\omega }_{0}t)dt\\ & =\sum _{k=1}^{n-1}\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((n-k){\omega }_{0}t)dt+\frac{a_n}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((n-n){\omega }_{0}t)dt+\sum _{k=n+1}^{\infty }\frac{a_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos ((k-n){\omega }_{0}t)dt\\ & -\sum _{k=1}^{n-1}\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}-\sin ((n-k){\omega }_{0}t)dt-\frac{b_n}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin ((n-n){\omega }_{0}t)dt-\sum _{k=n+1}^{\infty }\frac{b_k}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin ((k-n){\omega }_{0}t)dt\\ & =\sum _{k=1}^{n-1}\frac{a_k}{2}0+\frac{a_n}{2}{\int }_0^{\frac{2\pi }{{\omega }_0}}1dt+\sum _{k=n+1}^{\infty }\frac{a_k}{2}0-\sum _{k=1}^{n-1}\frac{b_k}{2}0-\frac{b_n}{2}0-\sum _{k=n+1}^{\infty }\frac{b_k}{2}0\\ & =\frac{a_n}{2}\frac{2\pi }{{\omega }_0}\end{array}$$

因此，系数$a_n$得解：
$$a_n=\frac{{\omega }_0}{\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin (n{\omega }_{0}t)f(t)dt$$

至此，已经得到傅里叶级数中各系数的表达式（如下），只要它们可积分，即$C$、$a_n$和$b_n$可解，那么就得到了函数$f(t)$的傅里叶级数。
$$\begin{gathered} f(t)=C+\sum _{k=1}^{\infty }{a}_k\sin (k{\omega }_{0}t)+\sum _{k=1}^{\infty }{b}_k\cos (k{\omega }_{0}t) \\ \{\begin{array}{l}C=\frac{{\omega }_0}{2\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)dt\\ a_n=\frac{{\omega }_0}{\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}\sin (n{\omega }_{0}t)f(t)dt\\ b_n=\frac{{\omega }_0}{\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}\cos (n{\omega }_{0}t)f(t)dt\end{array} \end{gathered}$$

傅里叶级数的复数形式

为此，我们需要引入欧拉公式。其构建了三角函数和复数之间的桥梁，如下：
$$e^{ix}=\cos x+i\sin x$$
其中，$i=\sqrt{-1}$。替换上式中的$x$为$-x$，得到$e^{-ix}=\cos -x+i\sin -x=\cos x-i\sin x$。得到：
$$\{\begin{array}{l}\cos x=\frac{e^{ix}+e^{-ix}}{2}\\ \sin x=\frac{e^{ix}-e^{-ix}}{2i}\end{array}$$
将上式代入三角函数形式的傅里叶级数中：
$$\begin{array}{rl}& f(t)=C+\sum _{k=1}^{\infty }{a}_k\sin (k{\omega }_{0}t)+\sum _{k=1}^{\infty }{b}_k\cos (k{\omega }_{0}t)\\ & =C+\sum _{k=1}^{\infty }(a_k\frac{e^{ik{\omega }_{0}t}-e^{-ik{\omega }_{0}t}}{2i}+b_k\frac{e^{ik{\omega }_{0}t}+e^{-ik{\omega }_{0}t}}{2})\\ & =C+\sum _{k=1}^{\infty }(\frac{a_k}{2i}{e}^{ik{\omega }_{0}t}-\frac{a_k}{2i}{e}^{-ik{\omega }_{0}t}+\frac{b_k}{2}{e}^{ik{\omega }_{0}t}+\frac{b_k}{2}{e}^{-ik{\omega }_{0}t})\\ & =C+\sum _{k=1}^{\infty }(-\frac{i{a}_k}{2}{e}^{ik{\omega }_{0}t}+\frac{i{a}_k}{2}{e}^{-ik{\omega }_{0}t}+\frac{b_k}{2}{e}^{ik{\omega }_{0}t}+\frac{b_k}{2}{e}^{-ik{\omega }_{0}t})\\ & =C+\sum _{k=1}^{\infty }\frac{b_k-i{a}_k}{2}{e}^{ik{\omega }_{0}t}+\sum _{k=1}^{\infty }\frac{b_k+i{a}_k}{2}{e}^{-ik{\omega }_{0}t}\\ & =C{e}^{i0{\omega }_{0}t}+\sum _{k=1}^{\infty }\frac{b_k-i{a}_k}{2}{e}^{ik{\omega }_{0}t}+\sum _{j=-1}^{-\infty }\frac{b_{-j}+i{a}_{-j}}{2}{e}^{ij{\omega }_{0}t}\end{array}$$

观察上式，可以得到傅里叶级数的复数形式：
$$f(t)=\sum _{k=-\infty }^{\infty }{X}_k{e}^{ik{\omega }_{0}t}$$

$k\ge 1$时：
$$X_k=\frac{b_k-i{a}_k}{2}=\frac{{\omega }_0}{2\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)(\cos (k{\omega }_{0}t)-i\sin (k{\omega }_{0}t))dt=\frac{{\omega }_0}{2\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)e^{-ik{\omega }_{0}t}dt$$
$k=0$时：
$$X_k=C=\frac{{\omega }_0}{2\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)dt=\frac{{\omega }_0}{2\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)e^{-i0{\omega }_{0}t}dt$$
$k\le -1$时，令$j=-k$：
$$\begin{array}{rl}& X_k=\frac{b_{-k}+i{a}_{-k}}{2}=\frac{b_j+i{a}_j}{2}=\frac{{\omega }_0}{2\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)(\cos (j{\omega }_{0}t)+i\sin (j{\omega }_{0}t))dt\\ & =\frac{{\omega }_0}{2\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)e^{ij{\omega }_{0}t}dt=\frac{{\omega }_0}{2\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)e^{-ik{\omega }_{0}t}dt\end{array}$$

至此，已经得到复数形式的傅里叶级数中系数的表达式（如下），只要它们可积分，即$X_k$可解，那么就得到了函数$f(t)$的复数形式的傅里叶级数。
$$f(t)=\sum _{k=-\infty }^{\infty }{X}_k{e}^{ik{\omega }_{0}t}$$
其中，$X_n=\frac{{\omega }_0}{2\pi }{\int }_0^{\frac{2\pi }{{\omega }_0}}f(t)e^{-in{\omega }_{0}t}dt$。