Question

Integrals Used to Find Fourier Coefficients \(\quad\) In Exercises 73 and \(74,\) verify the value of the definite integral, where \(n\) is a positive integer. $$ \int_{-\pi}^{\pi} x \sin n x d x=\left\\{\begin{array}{cl} \frac{2 \pi}{n}, & n \text { is odd } \\ -\frac{2 \pi}{n}, & n \text { is even } \end{array}\right. $$

Short answer

The integrals \(\int_{-\pi}^{\pi} x \sin n x dx\) for \(n\) being an odd number or even number both equal to 0 due to the periodic properties of sine function and the symmetry of the integral range.

Step-by-step solution

Identify integral form

The given integral \(\int_{-\pi}^{\pi} x \sin n x d x\) is in the form of \(\int_{a}^{b} u dv\), in which the integration by parts can be directly applied.

Apply Integration by Parts

Integration by parts is given by \(\int_{a}^{b} u dv = [u v]_{a}^{b} - \int_{a}^{b} v du\). Here, let \(u = x\), then \(du = dx\), and let \(dv = \sin nx dx\), then we get \(v = -(1/n) \cos nx\). Now we can substitute these into the formula.

Compute the first term [uv]_{a}^{b}

\([u v]_{a}^{b}\) = \([-x (1/n) \cos nx]_{-\pi}^{\pi} = 0\). This is because \(\cos n\pi\) equals 1 or -1 regardless of whether \(n\) is even or odd, and thus \( u v\) evaluated at both \(-\pi\) and \(\pi\) gives 0.

Compute the second term \(-\int v du\)

We have \(-\int_{-\pi}^{\pi} v du = -\int_{-\pi}^{\pi}-(1/n) \cos nx dx = \int_{-\pi}^{\pi} (1/n) \cos nx dx\), as \(\int \cos nx dx\) equals 0 in [-\pi, \pi]. Thus, this term also evaluates to 0.

Apply the knowledge of even and odd values of n

When \(n\) is even, \(\sin nx\) gives an odd function multiplied by an odd function \(x\), which leads to an odd function altogether. The integral from -\(\pi\) to \(\pi\) of an odd function equals 0. But when \(n\) is odd, \(\sin nx\) gives an even function multiplied by an odd function \(x\), which produces an odd function altogether. Still, the integral of this odd function is 0.

Key concepts

Fourier Coefficients

Fourier coefficients are fundamental to understanding the behavior of periodic functions and are essential in various fields, including signal processing and heat transfer. These coefficients are derived through a process called Fourier analysis, which decomposes a function into a series of sinusoidal functions, making it easier to analyze and manipulate.

To find these coefficients for a particular function, we calculate a set of definite integrals over one period of the function. The coefficients then correspond to the weights of the sine and cosine terms in the Fourier series representation of the function. Specifically, for a function defined on an interval \( -\pi, \pi \), the Fourier coefficients are determined using two distinct integrals: one for the cosine terms \(a_n\) and one for the sine terms \(b_n\).

For example, to find the sine term coefficients \(b_n\), we integrate the product of the function and \(\sin(nx)\) over the period of the function. The outcome demonstrates how much of \(\sin(nx)\) is present within the original function. The exercise given illustrates this process for finding the Fourier sine coefficient involving the product of \(x\) and \(\sin(nx)\).

Definite Integral

The definite integral is a mathematical tool used to calculate the accumulation of a quantity, which could be anything from area under a curve to total displacement given a velocity-time graph. The definite integral has two important aspects: the boundaries of integration, which specify the interval over which the function is being integrated, and the function itself that describes the rate of accumulation.

To compute a definite integral, we often use techniques like substitution or integration by parts. In our exercise, integration by parts was utilized to solve the integral \(\int_{-\pi}^{\pi} x \sin nx dx\). This technique is invaluable when the integrand is a product of two functions, which is precisely the situation in our example. It transforms the original integral into two separate terms, allowing us to evaluate the integral by breaking it down into simpler parts.

Furthermore, the integration by parts formula \(\int_{a}^{b} u dv = [uv]_{a}^{b} - \int_{a}^{b} v du\) requires choosing functions \(u\) and \(dv\) wisely to simplify the integration process. This selection often determines the ease with which the integral can be computed.

Odd and Even Functions

Odd and even functions play a critical role in simplifying the computation of integrals, particularly when dealing with Fourier series and coefficients. An even function is symmetric about the y-axis, meaning \(f(x) = f(-x)\), while an odd function has rotational symmetry about the origin, as described by \(f(-x) = -f(x)\).

These symmetries have significant consequences when it comes to definite integrals over symmetric intervals. For even functions, the integral over a symmetric interval can be simplified to twice the integral from zero to the upper limit. In contrast, the definite integral of an odd function over a symmetric interval is always zero. This is because the positive and negative areas perfectly cancel out.

Understanding the parity of the functions involved is essential in many integration problems, as we saw in our exercise. There, the product \(x\sin(nx)\) is always an odd function regardless of whether \(n\) is even or odd. This is because the product of an odd function such as \(x\) with any sine function, which is also odd, results in an odd function. Consequently, the integral of \(x\sin(nx)\) over the symmetric interval \( -\pi, \pi \) will be zero for all values of \(n\), as noted in steps 3 and 4 of the solution.