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

1. Problem Description

The problem asks 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

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

where the Fourier coefficients are given by

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

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

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

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} \left[ \frac{x^2}{2} \right]_{-\pi}^{\pi} = \frac{1}{\pi} \left( \frac{\pi^2}{2} - \frac{(-\pi)^2}{2} \right) = \frac{1}{\pi} (0) =

0. $$

Next, we compute $a_n$:

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

Since $x$ is an odd function and $\cos(nx)$ is an even function, their product $x \cos(nx)$ is an odd function. The integral of an odd function over a symmetric interval $(-\pi, \pi)$ is zero.

$$$$

a_n =

0. $$

Now, we compute $b_n$:

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

Since $x$ is an odd function and $\sin(nx)$ is also an odd function, their product $x \sin(nx)$ is an even function. Therefore,

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

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

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

So,

\begin{align*} b_n &= \frac{2}{\pi} \left[ -\frac{x}{n} \cos(nx) + \frac{1}{n^2} \sin(nx) \right]_{0}^{\pi} \\ &= \frac{2}{\pi} \left( -\frac{\pi}{n} \cos(n\pi) + \frac{1}{n^2} \sin(n\pi) - (0 + 0) \right) \\ &= \frac{2}{\pi} \left( -\frac{\pi}{n} (-1)^n \right) = -\frac{2}{n} (-1)^n = \frac{2}{n} (-1)^{n+1}. \end{align*}

Therefore, the Fourier series is

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

3. Final Answer

The Fourier series of $f(x) = x$ on $-\pi < x < \pi$ is

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