Question

Evaluate each improper integral or show that it diverges. \(\int_{-\infty}^{\infty} \frac{x}{\sqrt{x^{2}+9}} d x\)

Short answer

The integral diverges as it does not converge to a finite value.

Step-by-step solution

Split the Integral

The given integral is defined over the entire real line, so we need to split it into two separate integrals to handle the improper nature: \[\int_{-\infty}^{\infty} \frac{x}{\sqrt{x^{2}+9}} dx = \int_{-\infty}^{0} \frac{x}{\sqrt{x^{2}+9}} dx + \int_{0}^{\infty} \frac{x}{\sqrt{x^{2}+9}} dx\]This separation will help us evaluate the two parts individually, considering convergence.

Evaluate the Integral from 0 to ∞

Set up the integral from \(0\) to \(\infty\): \[\int_{0}^{\infty} \frac{x}{\sqrt{x^{2}+9}} dx\]Use the substitution \(x = 3 \tan \theta\), then \(dx = 3 \sec^2 \theta \, d\theta\), and \[\sqrt{x^2 + 9} = 3 \sec \theta.\]The limits change from \(x = 0\) to \(\theta = 0\) and as \(x \to \infty\), \(\theta \to \frac{\pi}{2}\).Thus, \[\int_{0}^{\pi/2} 3 \tan \theta \cdot \frac{3 \sec^2 \theta}{3 \sec \theta} d\theta = 3 \int_{0}^{\pi/2} \tan \theta \, \sec \theta \, d\theta\]This integral diverges because \(\tan \theta\) becomes infinite as \(\theta\) approaches \(\frac{\pi}{2}\).

Evaluate the Integral from -∞ to 0

Similarly, evaluate the integral from \(-\infty\) to \(0\): \[\int_{-\infty}^{0} \frac{x}{\sqrt{x^{2}+9}} dx\]By using the same substitution with limits from \(-\infty\) to \(0\), the integral similarly diverges since the behavior is symmetric around the y-axis.

Conclusion on Convergence

Since both integrals from \(-\infty\) to \(0\) and from \(0\) to \(\infty\) diverge, the whole integral \(\int_{-\infty}^{\infty} \frac{x}{\sqrt{x^{2}+9}} dx\) diverges. Thus, there is no finite value for this integral.

Key concepts

Convergence

In real analysis, the concept of convergence plays a vital role when dealing with integrals, especially improper integrals. An integral converges if, as the limits of integration approach infinity or a particular point of discontinuity, the value of the integral approaches a finite number. Essentially, it is about the integral summing up to a finite area under the curve.

When evaluating an improper integral, such as:

• \( \int_{0}^{\infty} \frac{x}{\sqrt{x^{2}+9}} dx \)
• \( \int_{-\infty}^{0} \frac{x}{\sqrt{x^{2}+9}} dx \)

we split the problem into two separate integrals and analyze each one for convergence. If either part does not converge, the entire integral diverges. Convergence tells us if it is possible to obtain a finite number as the sum, which isn't the case here due to divergence.

Divergence

Divergence is an indication that an integral does not settle to a finite number; in essence, it "goes to infinity." When dealing with indefinite integrals, especially those calculated over infinite limits or discontinuous points on the curve, determining whether an integral diverges is crucial.

In the exercise given, both sub-integrals:

• \( \int_{0}^{\infty} \frac{x}{\sqrt{x^{2}+9}} dx \)
• \( \int_{-\infty}^{0} \frac{x}{\sqrt{x^{2}+9}} dx \)

were evaluated and found to diverge. The symmetry of the integral's function about the y-axis implies that both contribute equally to the divergence. Since neither part of the split integrals converges, the improper integral over the entire real line also diverges. Therefore, there's no finite result for the original integral, signifying total divergence.

Trigonometric Substitution

Trigonometric substitution is a technique used in calculus to simplify complex irrational expressions. In this specific exercise, the integral involves \(\sqrt{x^2 + 9}\), making it suitable for substitution with a trigonometric identity.

The substitution utilized here is:

• Let \(x = 3 \tan \theta\)
• Then \(dx = 3 \sec^2 \theta \, d\theta\)
• And \(\sqrt{x^2 + 9} = 3 \sec \theta\)

These substitutions transform the integral into a trigonometric form, making it more manageable. The results of these transformations set the path to evaluate the behavior of the integral; for instance, \(\tan \theta\) tends toward infinity as \(\theta\) approaches \(\frac{\pi}{2}\), indicating divergence. This technique is powerful for similar integrals involving square roots and helps uncover the properties of complex integrals.

Real Analysis

Real analysis provides the theoretical foundation to explore problems involving limits, integration, and the behavior of functions. This branch of mathematics delves into deep insights about the convergence and divergence of series and integrals.

In this context, real analysis allows us to understand:

• How and why certain improper integrals diverge or converge based on their mathematical properties.
• The importance of breaking down an integral over infinite intervals and examining each sub-interval separately.
• The application of different mathematical tools, like trigonometric substitution, to tackle and evaluate integrals.

Real analysis questions the intuitive understanding of these concepts and provides a building block for further analysis, ensuring that we apply rigorous logic in determining the fate of complex integrals such as the one posed in our problem.