We want to find the Fourier series representation of the function $f(x) = x$ on the interval $-\pi < x < \pi$.

1. Problem Description

We want to find the Fourier series representation of the function $f(x) = x$ on the interval $-\pi < x < \pi$.

2. Solution Steps

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

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

where the coefficients are:

$a_0 = \frac{1}{L} \int_{-L}^{L} f(x) dx$

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

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

In our case, $f(x) = x$ and $L = \pi$.

First, we compute $a_0$:

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x dx = \frac{1}{\pi} [\frac{x^2}{2}]_{-\pi}^{\pi} = \frac{1}{\pi} (\frac{\pi^2}{2} - \frac{\pi^2}{2}) = 0$

Next, we compute $a_n$:

$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos(nx) dx$

Since $x\cos(nx)$ is an odd function and we are integrating over a symmetric interval, the integral is

0. Therefore, $a_n = 0$ for all $n$.

Finally, we compute $b_n$:

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

Since $x\sin(nx)$ is an even function, we can write:

$b_n = \frac{2}{\pi} \int_{0}^{\pi} x \sin(nx) dx$

Using integration by parts, let $u = x$ and $dv = \sin(nx) dx$. Then $du = dx$ and $v = -\frac{1}{n} \cos(nx)$.

$\int x \sin(nx) dx = -\frac{x}{n} \cos(nx) + \int \frac{1}{n} \cos(nx) dx = -\frac{x}{n} \cos(nx) + \frac{1}{n^2} \sin(nx)$

Thus,

$b_n = \frac{2}{\pi} [-\frac{x}{n} \cos(nx) + \frac{1}{n^2} \sin(nx)]_0^{\pi} = \frac{2}{\pi} [-\frac{\pi}{n} \cos(n\pi) + \frac{1}{n^2} \sin(n\pi) - (0 + 0)]$

Since $\sin(n\pi) = 0$ and $\cos(n\pi) = (-1)^n$, we have:

$b_n = \frac{2}{\pi} [-\frac{\pi}{n} (-1)^n] = \frac{2}{n} (-1)^{n+1}$

Therefore, the Fourier series is:

$f(x) = \sum_{n=1}^{\infty} b_n \sin(nx) = \sum_{n=1}^{\infty} \frac{2}{n} (-1)^{n+1} \sin(nx)$

$f(x) = 2 [\sin(x) - \frac{\sin(2x)}{2} + \frac{\sin(3x)}{3} - \frac{\sin(4x)}{4} + \dots]$

3. Final Answer

$f(x) = 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx)$