The problem asks to prove the formula for the Fourier sine series coefficient $b_n$ for a function $f(x)$ defined on the interval $[-L, L]$. The formula is given as: $b_n = \frac{1}{L}\int_{-L}^{L} f(x) \sin(\frac{n\pi x}{L}) dx$.

1. Problem Description

The problem asks to prove the formula for the Fourier sine series coefficient $b_n$ for a function $f(x)$ defined on the interval $[-L, L]$. The formula is given as:

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

2. Solution Steps

The formula for the Fourier series of a function $f(x)$ defined on the interval $[-L, L]$ is given by:

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

To find the coefficient $b_n$, we multiply both sides of the equation by $\sin(\frac{m\pi x}{L})$, where $m$ is an integer, and integrate from $-L$ to $L$:

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

We can split the integral on the right-hand side:

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

Now we use the following orthogonality properties of sine and cosine functions:

$\int_{-L}^{L} \sin(\frac{n\pi x}{L}) dx = 0$

$\int_{-L}^{L} \cos(\frac{n\pi x}{L}) \sin(\frac{m\pi x}{L}) dx = 0$

$\int_{-L}^{L} \sin(\frac{n\pi x}{L}) \sin(\frac{m\pi x}{L}) dx = \begin{cases} 0, & \text{if } n \neq m \\ L, & \text{if } n = m \end{cases}$

Using these properties, the first two integrals on the right-hand side become zero.

So we have:

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

When $n \neq m$, the integral is zero. When $n = m$, the integral is $L$. Therefore, the summation reduces to a single term where $n = m$:

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

Dividing both sides 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

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