The problem states that a function $f(x)$ is defined in an open interval $(-L, L)$ and has period $2L$. The Fourier series expansion of $f(x)$ is given as: $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(\frac{n \pi x}{L}) + b_n \sin(\frac{n \pi x}{L}))$ The task is to prove that the Fourier coefficient $b_n$ is given by: $b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin(\frac{n \pi x}{L}) dx$

1. Problem Description

The problem states that a function $f(x)$ is defined in an open interval $(-L, L)$ and has period $2L$. The Fourier series expansion of $f(x)$ is given as:

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

The task is to prove that the Fourier coefficient $b_n$ is given by:

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

2. Solution Steps

To find the coefficient $b_n$, we can multiply both sides of the Fourier series by $\sin(\frac{m \pi x}{L})$ and integrate from $-L$ to $L$.

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

Now, we can split the integral:

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

Since $\sin(\frac{m \pi x}{L})$ is an odd function and we are integrating from $-L$ to $L$, the first term is zero:

$\int_{-L}^{L} \frac{a_0}{2} \sin(\frac{m \pi x}{L}) dx = 0$

The second term is also zero. This is because the integral of the product of cosine and sine over a period is zero:

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

The third term simplifies due to the orthogonality 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, the equation becomes:

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

Dividing both sides by $L$:

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

Replacing $m$ with $n$:

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

3. Final Answer

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