A question about derivation of Fourier coefficients As the textbook said:
Trigonometrical polynomials
$$
\begin{align}
f(x)=\frac{a_0}{2} + \sum_{v=1}^{n}(a_v\cos{vx}+b_v\sin{vx}) \tag{1}
\end{align}
$$
The Fourier coefficients can be expressed simply by the following formulas:
$$
a_u=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos{ux}\,dx, {\ } b_u=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin{ux}\,dx. \tag{2}
$$
The proof follows if we multiply Eq. (1) by $\cos{ux}$ or $\sin{ux}$ and then integrate.
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My derivation:
$$
\begin{align}
\cos{ux}\,f(x) & = \frac{a_0}{2} \cos{ux} + \sum_{v=1}^{n}(a_v\cos{vx} \cos{ux} +b_v\sin{vx} \cos{ux}) \\
\int_{-\pi}^{\pi}\cos{ux}\,f(x)\,dx & = \frac{a_0}{2} \int_{-\pi}^{\pi} \cos{ux}\,dx \\
& \phantom{={}} + \sum_{v=1}^{n}\left(a_v \int_{-\pi}^{\pi}\cos{vx} \cos{ux}\,dx +b_v \int_{-\pi}^{\pi}\sin{vx} \cos{ux}\,dx\right)
\end{align}
$$
According to the orthogonality relations of the trigonometric functions, we get:
$$
\begin{align}
& \int_{-\pi}^{\pi} \cos{ux}\,dx =0, \\
& \int_{-\pi}^{\pi}\cos{vx} \cos{ux}\,dx = \pi, \quad \text{if } u=v, \\
& \int_{-\pi}^{\pi}\sin{vx} \cos{ux}\,dx =0.
\end{align}
$$
So, if $u=v$
$$
\begin{align}
\int_{-\pi}^{\pi}\cos{ux}\,f(x)\,dx & =  \sum_{v=1}^{n} a_v \int_{-\pi}^{\pi}\cos{vx} \cos{ux}\,dx =  \sum_{v=1}^{n} a_v \pi  
\end{align}
$$
I can’t get the Eq. (2), where’s the mistake? Thanks.
my question's unique part is, derivate this equation:
$$
\begin{align}
\sum_{v=1}^{n} a_v \int_{-\pi}^{\pi}\cos{vx} \cos{ux}\,dx = a_v \pi  
\end{align}
$$
and it was solved by @Ak19.
 A: To generalize, the derivation is easier if you prove it for a complex variable; i.e: $$c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx:n\in\mathbb{Z}$$
where the Fourier expansion is of the form $$f(x)=\sum_{n\in\mathbb{Z}}c_n e^{inx}$$.
for $f:\mathbb{R}\rightarrow\mathbb{C}$; simply consider $\mathbb{R}$ as a subset of $\mathbb{C}$ and you can use this method for $f:\mathbb{R}\rightarrow\mathbb{R}$.
Proof:
Fix $m\in\mathbb{Z}$. Consider the integral:
$$\int_{-\pi}^{\pi}f(x)e^{-imx}dx=\int_{-\pi}^{\pi}\sum_{n\in\mathbb{Z}}c_n e^{inx}e^{-imx}dx$$
$$=\sum_{n\in\mathbb{Z}}\int_{-\pi}^{\pi}c_n e^{inx}e^{-imx}dx.$$
Clearly if there is any residual exponential term, integrating that term from $-\pi$ to $\pi$ will yield $0$, regardless of the magnitude of the coefficient of that term, as any exponential term refers simply to a rotation, which will be the same at $-\pi$ and $\pi$. We have:
$$=\cdots+0+\int_{-\pi}^{\pi}c_n dx + 0 + \cdots$$
$$=2\pi c_n,$$
divide by $2\pi$ to find $c_n$.
This proof comes with a nice visualization with epicycles: see here.
There's no mistake in your proof; you just need the last 'canceling-out' step.
