We are asked to construct the Fourier series for the function $f(x) = x$ on the interval $-\pi < x < \pi$.

1. Problem Description

We are asked to construct the Fourier series for 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 $-L < x < L$ is given by

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

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 this problem, $f(x) = x$ and $L = \pi$. Thus, we have

$$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \, dx$$

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

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

Since $x$ is an odd function, the integral of $x$ from $-\pi$ to $\pi$ is

0. Similarly, $x \cos(nx)$ is an odd function, so its integral from $-\pi$ to $\pi$ is also

0. Therefore, $a_0 = 0$ and $a_n = 0$ for all $n$.

Since $x \sin(nx)$ is an even function, we have

$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx = \frac{2}{\pi} \int_{0}^{\pi} x \sin(nx) \, dx$$

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

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

Therefore,

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

Thus,

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

$$f(x) = 2 \left( \frac{\sin x}{1} - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \frac{\sin 4x}{4} + \cdots \right)$$

3. Final Answer

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