I am computing the Fourier series of
$$f(x)=\sin\frac{\pi x}{L}.$$
The Fourier series of a piecewise smooth function $f(x)$ defined on the interval $-L\leq x\leq L$ is given by
$$f(x)\sim a_0+\sum_{n=1}^{\infty}a_n\cos\frac{n\pi x}{L}+\sum_{n=1}^{\infty}b_n\sin\frac{n\pi x}{L},$$
where
$$\begin{align}
a_0&=\frac{1}{2L}\int_{-L}^{L}f(x)\,dx,\\
a_n&=\frac{1}{L}\int_{-L}^{L}f(x)\cos\frac{n\pi x}{L}\,dx,\\
b_n&=\frac{1}{L}\int_{-L}^{L}f(x)\sin\frac{n\pi x}{L}\,dx.\\
\end{align}$$
Since $f$ is an odd function, we have that both $a_0$ and $a_n$ are equal to zero. However, $b_n$ is the integral of an even function. Hence
$$
b_n=\frac{2}{L}\int_{0}^{L}f(x)\sin\frac{n\pi x}{L}\,dx.\tag{1}
$$
But it turns out that $(1)$ is equal to zero, because
$$
b_n=\frac{2L\sin(n\pi)}{\pi(1-n^2)}=0,\qquad n\in\mathbb{N}.
$$
On the other hand, my book claims that $b_n=1$. Why is this so?
Answer
A trigonometric polynomial is equal to its own fourier expansion. So f(x)=sin(x) has a fourier expansion of sin(x) only (from $[-\pi,\pi]$ I mean). The series is finite just like how the taylor expansion of a polynomial is itself (and hence finite). In addition, $b_n=0$ IF $n\neq1$ because your expression is undefined for $n=1$. For all other values of $n$, you are correct and $b_n=0$. And you can see that from equation (1) too. $n=1$ must be handled separately. And when you do compute the integral with $n=1$, you will get $b_1=1$.
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