Calculating the Fourier Coefficient a₁ for f(x) = x sin x

The question asks us to find the value of the Fourier coefficient \(a_1\) for the function \(f(x) = x \sin x\) over the interval \([-\pi, \pi]\).

For a function \(f(x)\) defined on the interval \([-L, L]\), the Fourier series is given by:

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

The coefficients \(a_n\) and \(b_n\) are calculated using the Euler-Fourier formulas:

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

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

In this problem, the interval is \([-\pi, \pi]\), so \(L = \pi\). The formulas become:

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

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

We need to find the coefficient \(a_1\). We set \(n=1\) in the formula for \(a_n\):

\(a_1 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(1x) dx\)

Substitute the function \(f(x) = x \sin x\):

\(a_1 = \frac{1}{\pi} \int_{-\pi}^{\pi} (x \sin x) \cos x dx\)

We can simplify the integrand using the trigonometric identity \(\sin(2\theta) = 2 \sin \theta \cos \theta\), which means \(\sin \theta \cos \theta = \frac{1}{2} \sin(2\theta)\). Let \(\theta = x\):

\(\sin x \cos x = \frac{1}{2} \sin(2x)\)

So the integral for \(a_1\) becomes:

\(a_1 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \left(\frac{1}{2} \sin(2x)\right) dx = \frac{1}{2\pi} \int_{-\pi}^{\pi} x \sin(2x) dx\)

Evaluating the Integral for Fourier Coefficient a₁

Now, let's evaluate the integral \(\int_{-\pi}^{\pi} x \sin(2x) dx\). We can check the parity of the function \(g(x) = x \sin(2x)\):

\(g(-x) = (-x) \sin(2(-x)) = -x \sin(-2x) = -x (-\sin(2x)) = x \sin(2x) = g(x)\)

Since \(g(-x) = g(x)\), the function \(g(x) = x \sin(2x)\) is an even function. For an even function \(g(x)\) integrated over a symmetric interval \([-a, a]\), we have \(\int_{-a}^{a} g(x) dx = 2 \int_{0}^{a} g(x) dx\). Here, \(a = \pi\).

So, the integral becomes:

\(\int_{-\pi}^{\pi} x \sin(2x) dx = 2 \int_{0}^{\pi} x \sin(2x) dx\)

Now we need to evaluate \(\int_{0}^{\pi} x \sin(2x) dx\). We can use integration by parts, which states \(\int u dv = uv - \int v du\). Let:

• \(u = x\)
• \(dv = \sin(2x) dx\)

Then:

• \(du = dx\)
• \(v = \int \sin(2x) dx = -\frac{1}{2} \cos(2x)\)

Applying the integration by parts formula:

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

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

Evaluate the first part:

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

Since \(\cos(2\pi) = 1\) and \(\cos(0) = 1\):

\(= \left(-\frac{\pi}{2} \cdot 1\right) - (0) = -\frac{\pi}{2}\)

Evaluate the second part (the integral):

\(\frac{1}{2} \int_{0}^{\pi} \cos(2x) dx = \frac{1}{2} \left[\frac{1}{2} \sin(2x)\right]_{0}^{\pi}\)

\(= \frac{1}{4} [\sin(2x)]_{0}^{\pi} = \frac{1}{4} (\sin(2\pi) - \sin(0))\)

Since \(\sin(2\pi) = 0\) and \(\sin(0) = 0\):

\(= \frac{1}{4} (0 - 0) = 0\)

So, the definite integral is:

\(\int_{0}^{\pi} x \sin(2x) dx = -\frac{\pi}{2} + 0 = -\frac{\pi}{2}\)

Now substitute this back into the formula for \(a_1\):

\(a_1 = \frac{1}{2\pi} \int_{-\pi}^{\pi} x \sin(2x) dx = \frac{1}{2\pi} \left(2 \int_{0}^{\pi} x \sin(2x) dx\right)\)

\(a_1 = \frac{1}{\pi} \int_{0}^{\pi} x \sin(2x) dx = \frac{1}{\pi} \left(-\frac{\pi}{2}\right)\)

\(a_1 = -\frac{1}{2}\)

Based on the standard calculation for the Fourier coefficient \(a_1\) of \(f(x) = x \sin x\) on \([-\pi, \pi]\), the value obtained is \(-1/2\).

The provided options are:

1. -3/2
2. -1/2
3. -2/3
4. -2/5

The result of our calculation is \(-1/2\).

The provided correct answer is Option 3: -2/3.

Revision Table: Key Concepts for Fourier Series

Additional Information on Fourier Series Expansion

Fourier series expansion is a powerful tool in mathematics and engineering for analyzing periodic functions. By breaking down a complex periodic function into a sum of simpler sine and cosine functions (harmonics), we can study its properties more easily. The coefficients \(a_n\) and \(b_n\) represent the contribution of each harmonic frequency to the overall function. The term \(a_0/2\) represents the DC component or the average value of the function over the interval.

For functions defined on \([-\pi, \pi]\), the basis functions are \(\{1, \cos(x), \cos(2x), \dots, \sin(x), \sin(2x), \dots\}\). These functions are orthogonal over this interval, which is why the integral formulas for the coefficients work. The process involves projecting the function \(f(x)\) onto each of these basis functions.

Understanding the parity of the function \(f(x)\) or the integrand \(f(x) \cos(nx)\) or \(f(x) \sin(nx)\) can significantly simplify the calculation of Fourier coefficients. For instance, if \(f(x)\) is even, all \(b_n\) coefficients are zero. If \(f(x)\) is odd, all \(a_n\) coefficients (including \(a_0\)) are zero.

In our calculation for \(a_1\) of \(f(x) = x \sin x\), the integrand was \(x \sin x \cos x = \frac{1}{2} x \sin(2x)\). The function \(f(x) = x \sin x\) is even because \(f(-x) = (-x) \sin(-x) = (-x)(-\sin x) = x \sin x = f(x)\). When an even function is multiplied by \(\cos(nx)\), the result is also even (\(even \times even = even\)), hence \(a_n\) integrals are non-zero in general. When an even function is multiplied by \(\sin(nx)\), the result is odd (\(even \times odd = odd\)), hence all \(b_n\) coefficients for \(f(x)=x \sin x\) are zero.