The problem asks to prove the formula for the Fourier coefficient $b_n$ given the Fourier series expansion of a function $f(x)$ defined on the interval $(-L, L)$ with period $2L$: $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(\frac{n \pi x}{L}) + b_n \sin(\frac{n \pi x}{L}))$. We need to prove that $b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin(\frac{n \pi x}{L}) dx$.

1. Problem Description

The problem asks to prove the formula for the Fourier coefficient $b_n$ given the Fourier series expansion of a function $f(x)$ defined on the interval $(-L, L)$ with period $2L$:

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(\frac{n \pi x}{L}) + b_n \sin(\frac{n \pi x}{L}))$.

We need to prove that $b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin(\frac{n \pi x}{L}) dx$.

2. Solution Steps

To find the formula for $b_n$, we multiply both sides of the Fourier series expansion by $\sin(\frac{m \pi x}{L})$ and integrate from $-L$ to $L$, where $m$ is an integer.

$\int_{-L}^{L} f(x) \sin(\frac{m \pi x}{L}) dx = \int_{-L}^{L} (\frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(\frac{n \pi x}{L}) + b_n \sin(\frac{n \pi x}{L}))) \sin(\frac{m \pi x}{L}) dx$

Now, we can interchange the integral and summation (under appropriate conditions which we assume hold). Then,

$\int_{-L}^{L} f(x) \sin(\frac{m \pi x}{L}) dx = \int_{-L}^{L} \frac{a_0}{2} \sin(\frac{m \pi x}{L}) dx + \sum_{n=1}^{\infty} \int_{-L}^{L} (a_n \cos(\frac{n \pi x}{L}) + b_n \sin(\frac{n \pi x}{L})) \sin(\frac{m \pi x}{L}) dx$

$\int_{-L}^{L} f(x) \sin(\frac{m \pi x}{L}) dx = \frac{a_0}{2} \int_{-L}^{L} \underbrace{\sin(\frac{m \pi x}{L})}_{\text{odd function}} dx + \sum_{n=1}^{\infty} (a_n \int_{-L}^{L} \underbrace{\cos(\frac{n \pi x}{L}) \sin(\frac{m \pi x}{L})}_{\text{odd function}} dx + b_n \int_{-L}^{L} \sin(\frac{n \pi x}{L}) \sin(\frac{m \pi x}{L}) dx)$

Since the integral of an odd function over a symmetric interval is zero, the first two terms on the right-hand side are zero. Thus,

$\int_{-L}^{L} f(x) \sin(\frac{m \pi x}{L}) dx = \sum_{n=1}^{\infty} b_n \int_{-L}^{L} \sin(\frac{n \pi x}{L}) \sin(\frac{m \pi x}{L}) dx$

Now, we use the orthogonality property of sine functions:

$\int_{-L}^{L} \sin(\frac{n \pi x}{L}) \sin(\frac{m \pi x}{L}) dx =

\begin{cases}

0, & \text{if } n \neq m \\

L, & \text{if } n = m

\end{cases}$

Therefore,

$\int_{-L}^{L} f(x) \sin(\frac{m \pi x}{L}) dx = b_m L$

Dividing by $L$, we get

$b_m = \frac{1}{L} \int_{-L}^{L} f(x) \sin(\frac{m \pi x}{L}) dx$

Replacing $m$ with $n$, we obtain the desired formula:

$b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin(\frac{n \pi x}{L}) dx$

3. Final Answer

$b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin(\frac{n \pi x}{L}) dx$