Problem 17
Question
Evaluate each improper integral or show that it diverges. \(\int_{-\infty}^{\infty} \frac{x}{\sqrt{x^{2}+9}} d x\)
Step-by-Step Solution
Verified Answer
The integral diverges as it does not converge to a finite value.
1Step 1: 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.
2Step 2: 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 \(\frac{\pi}{2}\).
3Step 3: 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.
4Step 4: 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
ConvergenceDivergenceTrigonometric SubstitutionReal Analysis
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:
When evaluating an improper integral, such as:
- \( \int_{0}^{\infty} \frac{x}{\sqrt{x^{2}+9}} dx \)
- \( \int_{-\infty}^{0} \frac{x}{\sqrt{x^{2}+9}} dx \)
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:
In the exercise given, both sub-integrals:
- \( \int_{0}^{\infty} \frac{x}{\sqrt{x^{2}+9}} dx \)
- \( \int_{-\infty}^{0} \frac{x}{\sqrt{x^{2}+9}} dx \)
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:
The substitution utilized here is:
- Let \(x = 3 \tan \theta\)
- Then \(dx = 3 \sec^2 \theta \, d\theta\)
- And \(\sqrt{x^2 + 9} = 3 \sec \theta\)
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:
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.
Other exercises in this chapter
Problem 16
Evaluate each improper integral or show that it diverges. \(\int_{4}^{\infty} \frac{d x}{(\pi-x)^{2 / 3}}\)
View solution Problem 16
Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} \frac{\sin x-\tan x}{x^{2}
View solution Problem 17
Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0^{+}} \frac{x^{2}}{\sin x-x}
View solution Problem 18
Evaluate each improper integral or show that it diverges. \(\int_{-\infty}^{\infty} \frac{d x}{\left(x^{2}+16\right)^{2}}\)
View solution