The problem states that a function $f(x)$ is defined on an open interval $(-L, L)$ and has period $2L$. Its Fourier series expansion is given by $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right)\right)$. We are asked to prove that the Fourier coefficient $b_n$ is given by $b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx$. Note that the original text in the image has $n\pi x/4$, but the correct form is $n\pi x/L$.

1. Problem Description

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

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

We are asked to prove that the Fourier coefficient $b_n$ is given by

$b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx$. Note that the original text in the image has $n\pi x/4$, but the correct form is $n\pi x/L$.

2. Solution Steps

To find the Fourier coefficient $b_n$, we can use the orthogonality properties of sine and cosine functions. Specifically, we will use the following:

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

$\int_{-L}^{L} \cos\left(\frac{n\pi x}{L}\right) \sin\left(\frac{m\pi x}{L}\right) dx = 0$ for all $n$ and $m$.

$\int_{-L}^{L} \sin\left(\frac{n\pi x}{L}\right) dx = 0$ for all $n$.

Multiply both sides of the Fourier series expansion by $\sin\left(\frac{m\pi x}{L}\right)$ and integrate from $-L$ to $L$:

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

We can interchange the summation and integration (assuming uniform convergence). Thus:

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

Using the orthogonality properties mentioned above, we have:

$\int_{-L}^{L} f(x) \sin\left(\frac{m\pi x}{L}\right) dx = 0 + \sum_{n=1}^{\infty} \left(a_n \cdot 0 + b_n \cdot L \delta_{nm}\right)$, where $\delta_{nm} = 1$ if $n=m$ and $0$ otherwise.

Thus:

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

Dividing by $L$, we get:

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

Replacing $m$ with $n$, we get:

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

3. Final Answer

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