Question

There is a subtlety in the definition of \(\int_{-\infty}^{\infty} f(x) d x\) that is illustrated by the following: Show that (a) \(\int_{-\infty}^{\infty} \sin x d x\) diverges and (b) \(\lim _{a \rightarrow \infty} \int_{-a}^{a} \sin x d x=0\).

Short answer

(a) The integral of \(\sin x\) from \(-\infty\) to \(\infty\) diverges. (b) The limit of the symmetric integral is 0.

Step-by-step solution

Understanding the Problem

We need to find out if the integral of the sine function from negative to positive infinity converges or diverges, and then prove that the limit of the integral of sine over symmetric bounds tends towards zero.

Show Divergence of Integral

To show that \(\int_{-\infty}^{\infty} \sin x \, dx\) diverges, we can look at the behavior of the sine function over the real line. The function \(\sin x\) is oscillatory, alternating between -1 and 1, and does not settle towards any value. As a result, the integral cannot converge due to these continuous oscillations causing a 'cancelling out' effect without an actual sum reaching a specific value.

Evaluate Finite Bounds Integral

Consider the integral \(\int_{-a}^{a} \sin x \, dx\). Integrating sine gives \(\int \sin x \, dx = -\cos x\). Thus, we have:\[\int_{-a}^{a} \sin x \, dx = \left[ -\cos x \right]_{-a}^{a} = -\cos(a) + \cos(-a)\]Since \(\cos(-a) = \cos(a)\), this integral simplifies to \(-\cos(a) + \cos(a) = 0\).

Take the Limit of the Symmetric Bounds Integral

Take the limit as \(a\) approaches infinity of the integral from \(-a\) to \(a\):\[\lim_{{a \to \infty}} \int_{-a}^{a} \sin x \, dx = \lim_{{a \to \infty}} 0 = 0\]We see that for all finite limits \(a\), the integral results in zero, hence, even as \(a\) grows larger, the integral remains at zero, showing that the limit is indeed zero.

Key concepts

Divergence of Integrals

An improper integral like \(\int_{-\infty}^{\infty} f(x) \ dx\) diverges if the area under the curve does not converge to a finite value as the limits extend to infinity. In the case of \(\int_{-\infty}^{\infty} \sin x \, dx\), divergence occurs due to the oscillatory nature of the sine function. Because \(\sin x\) repeats its cycle every \(2\pi\), oscillating between -1 and 1, the integral does not settle towards a particular value. These perpetual oscillations never sum up to a finite number, as the positive and negative areas cancel each other out indefinitely. This characteristic ensures that this integral is divergent and does not result in a finite value regardless of attempting to evaluate it. Thus, improper integrals of oscillatory functions over infinite intervals often lead to divergence due to their nature and behavior.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in mathematics, with multiple applications ranging from physics to engineering. The function \(\sin x\) is a periodic function with a cycle repeating every \(2\pi\) radians. This means its values fluctuate between -1 and 1 consistently over its domain from negative to positive infinity.

Exploring their properties, the integral of the sine function can be computed over finite bounds using the process of anti-differentiation. The integral \(\int \sin x \, dx\) leads to \(-\cos x + C\), where \(C\) is an integration constant. However, when computed over symmetric bounds like \([-a, a]\), a special phenomenon can be observed. By using the fundamental theorem of calculus, the result is zero because \(-\cos(a) + \cos(-a)\) equals zero since \(\cos(-a) = \cos(a)\).

Understanding these properties is key to discerning the behavior of the integral, specifically, how they allow certain solutions to be derived for finite symmetric intervals, as opposed to handling them over infinite intervals.

Oscillatory Functions

Oscillatory functions are those that exhibit a repeating pattern, usually characterized by regular fluctuation. The sine function, \(\sin x\), is a classic example of an oscillatory function, as it consistently oscillates between two bounds, -1 and 1. This repetition makes it challenging to evaluate certain integrals, especially over infinite intervals.

When evaluating integrals involving oscillatory functions, one must consider how these constant shifts affect convergence. An \