Question

Prove that the integral $$ \int_{0}^{1} \frac{1}{x} \sin \frac{1}{x} d x $$ exists as an improper Riemann integral, but not as a Lebesgue integral.

Short answer

The integral exists as an improper Riemann integral due to bounded oscillations, but not as a Lebesgue integral because of divergence.

Step-by-step solution

Understanding the Integral

We need to evaluate if the integral \( \int_{0}^{1} \frac{1}{x} \sin \frac{1}{x} \; dx \) exists as an improper Riemann integral. Integrals of this kind are considered improper because the integrand becomes singular at the lower limit \( x = 0 \).

Converting to Limit Form

Rewrite the integral using limit form to handle the singularity:\[\lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} \sin \frac{1}{x} \; dx.\] Evaluating this limit as \( a \to 0^+ \) will tell us if the integral exists as an improper Riemann integral.

Evaluating the Inner Integral

Evaluate the integral \( \int_{a}^{1} \frac{1}{x} \sin \frac{1}{x} \; dx \). Use substitution: let \( u = \frac{1}{x} \), then \( du = -\frac{1}{x^2} \; dx \), or \( dx = -\frac{1}{u^2} \; du \). Hence, when \( x = a \), \( u = \frac{1}{a} \), and when \( x = 1 \), \( u = 1 \).

Simplified Integral in Terms of u

Substitute and simplify the integral:\[\int_{1/a}^{1} \sin u \; du.\] The negative sign from substitution reverses the integral limits:\[\int_{1}^{1/a} \sin u \; du.\] Integrate to get:\[[-\cos u]_{1}^{1/a} = -\cos(1/a) + \cos(1).\]

Taking the Limit

Now, compute the limit as \( a \to 0^+ \):\[\lim_{a \to 0^+} (-\cos(1/a) + \cos(1)).\] \(-\cos(1/a)\) oscillates between -1 and 1 indefinitely as \( a \to 0^+ \), but the overall contribution to the integral converges to \( \cos(1) \), hence the improper Riemann integral exists.

Justifying the Improper Riemann Integral's Existence

Despite the oscillation, the integral converges because the oscillating part \(-\cos(1/a)\) is bounded and does not affect the overall convergence of the integral as \( a \to 0^+ \).

Evaluating as a Lebesgue Integral

The integral \( \int_{0}^{1} \frac{1}{x} \sin \frac{1}{x} \; dx \) is known to be divergent with respect to Lebesgue integration due to the improper behavior at the origin and the nature of the singularities.

Key concepts

Lebesgue Integral

The Lebesgue integral is a fundamental concept within modern analysis, providing a broader framework than the traditional Riemann integral. One of its key strengths is its ability to integrate a wider class of functions by focusing on measuring the range of the function rather than the domain.For instance, in Lebesgue integration, the integral measures how much of the function's range lies within a specific set. This approach makes it particularly adept at handling functions that may have many points of discontinuity, unlike the Riemann integral which may encounter more limitations.- Powerful when dealing with functions over complicated or infinite intervals- More flexible than Riemann integration, especially for functions with significant oscillationsIn the case of the exercise involving the integral \( \int_{0}^{1} \frac{1}{x} \sin \frac{1}{x} \, dx \), the Lebesgue integral interpretation fails due to the singularity at \( x = 0 \). The nature of the discontinuity created by this singularity makes it unsuitable for Lebesgue integration, as the function diverges in a manner that Lebesgue integrals cannot remedy.Ultimately, functions that can be properly integrated via the Lebesgue method may have a more complex behavior, with the emphasis on "how much" the function exceeds a certain value over intervals rather than looking at infinitesimal area slices exclusively.

Improper Integral

Improper integrals are integrals that require handling limits due to singularities or infinite intervals. They provide a way to evaluate integrals with limits that would otherwise be undefined under normal Riemann integration conditions.When considering an improper integral, we can encounter:- Infinite bounds: Integrals over \([a, \infty)\) or \((-\infty, b]\)- Discontinuous points: such as \( x=0 \) in you integralThe exercise highlights this concept with the integral \( \int_{0}^{1} \frac{1}{x} \sin \frac{1}{x} \, dx \), where a singularity at \( x = 0 \) arises from the \( \frac{1}{x} \) term. By transforming this integral using a limit, we can redefine it as:\[ \lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} \sin \frac{1}{x} \, dx. \]The utility of improper integrals is evident here, as they enable us to consider the limit of a function's behavior over a challenging domain. This reveals insights about the function's convergence over a potentially infinite or singular interval.

Mathematical Convergence

In mathematics, convergence refers to the property of a sequence or series approaching a limit. Specifically, in integration, we are concerned about whether the calculated value approaches a definitive number.Convergence is crucial for determining whether an integral - proper or improper - eventually settles at a finite value. In the context of the exercise,- Despite the oscillation of \(-\cos(1/a)\) as \( a \to 0^+ \),- The oscillation is bounded between -1 and 1.This bounded behavior ensures that the part of the integral incorporating this term does not contribute unbounded values. Instead, it highlights that, overall, the integral evaluates to a finite quantity: \( \cos(1) \).Understanding convergence is essential when evaluating improper integrals. If not for this bounded oscillation, the integral might diverge, confirming convergence as a key aspect of concluding if the improper integral exists.