Chapter 8: Problem 20
Evaluate each improper integral or show that it diverges. \(\int_{-\infty}^{\infty} \frac{x}{e^{2|x|}} d x\)
Short Answer
Expert verified
The integral converges to 0.
Step by step solution
01
Rewriting the Integral as a Limit
Before we evaluate the improper integral, we rewrite it as the limit of two definite integrals. Since the function has symmetry about the y-axis and is discontinuous at infinity, the limits will be useful in evaluating the integral. The given integral becomes:\[\lim_{a \to -\infty, b \to \infty} \int_{a}^{b} \frac{x}{e^{2|x|}} \, dx\]
02
Splitting the Integral at Discontinuity
We'll split the integral into two parts to handle the symmetry and piecewise nature of the function with respect to \(|x|\):\[\int_{- ext{infinity}}^{ ext{infinity}} \frac{x}{e^{2|x|}} \, dx = \lim_{a \to -\infty} \int_{a}^{0} \frac{x}{e^{-2x}} \, dx + \lim_{b \to \infty} \int_{0}^{b} \frac{x}{e^{2x}} \, dx\]
03
Evaluating the Left Integral
To solve \(\lim_{a \to -\infty} \int_{a}^{0} \frac{x}{e^{-2x}} \, dx\), we perform a substitution: let \(u = -2x\), then \(du = -2 \, dx\). The integral becomes:\[-\frac{1}{2} \lim_{a \to -\infty} \int_{-2a}^{0} u e^u \, du\]After integrating by parts and evaluating the limits, we find that it simplifies to zero.
04
Evaluating the Right Integral
For \(\lim_{b \to \infty} \int_{0}^{b} \frac{x}{e^{2x}} \, dx\), let \(v = 2x\), then \(dv = 2 \, dx\). The integral becomes:\[\frac{1}{2} \lim_{b \to \infty} \int_{0}^{2b} \frac{v}{e^v} \, dv\]Applying integration by parts again and evaluating the limits, this integral also converges to zero.
05
Conclusion on Convergence of the Integral
Adding the results of both the left and right integrals, we get:\[0 + 0 = 0\]Thus, the improper integral \(\int_{- ext{infinity}}^{ ext{infinity}} \frac{x}{e^{2|x|}} \, dx\) converges and equals zero.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Convergence of Integrals
Improper integrals go beyond the realm of standard definite integrals by including infinite limits or unbounded functions. A crucial first step when dealing with improper integrals is determining their convergence or divergence. To assess convergence, we often replace the infinites with limits, allowing us to reimagine the integral as a limit of definite integrals.
This is where convergence gets interesting. We analyze how the integral behaves as the bounds extend toward infinity. If the limit exists and results in a finite number, the integral converges; if not, it diverges.
For example, when evaluating our given integral, \(\int_{-\infty}^{\infty} \frac{x}{e^{2|x|}} \ dx\), it is transformed into two separate integrals ranging from negative to zero, and zero to positive infinity. By examining the limits of each portion, it is possible to conclude on whether the overall integral converges, as was done in the solution where the integral converges to zero.
This is where convergence gets interesting. We analyze how the integral behaves as the bounds extend toward infinity. If the limit exists and results in a finite number, the integral converges; if not, it diverges.
For example, when evaluating our given integral, \(\int_{-\infty}^{\infty} \frac{x}{e^{2|x|}} \ dx\), it is transformed into two separate integrals ranging from negative to zero, and zero to positive infinity. By examining the limits of each portion, it is possible to conclude on whether the overall integral converges, as was done in the solution where the integral converges to zero.
Symmetry in Calculus
When dealing with functions, recognizing symmetry can be a major simplification strategy. Specifically, for functions symmetric about the y-axis, this often simplifies the computation of integrals. In our scenario, the function \(\frac{x}{e^{2|x|}}\) exhibits this exact symmetry.
Symmetry implies that if you split the integral at the y-axis (x=0), the computations can be handled separately over two symmetric ranges. The left part from -∞ to 0 and the right part from 0 to ∞.
The value of the integral over these intervals could well result in terms that cancel each other out. For symmetric functions, integrating over a symmetric range can either highlight this mutual cancellation or reinforce the need for further evaluation. Here, the symmetry was pivotal in showcasing convergence and played a role in determining that our integral simplifies as shown in the given solution.
Symmetry implies that if you split the integral at the y-axis (x=0), the computations can be handled separately over two symmetric ranges. The left part from -∞ to 0 and the right part from 0 to ∞.
The value of the integral over these intervals could well result in terms that cancel each other out. For symmetric functions, integrating over a symmetric range can either highlight this mutual cancellation or reinforce the need for further evaluation. Here, the symmetry was pivotal in showcasing convergence and played a role in determining that our integral simplifies as shown in the given solution.
Integration by Parts
Integration by parts is a powerful technique for simplifying integrals of products of functions. It's derived from the product rule for differentiation and follows the formula: \(\int u \, dv = uv - \int v \, du\). This process often transforms a complex integral into a simpler one.
In the context of this particular problem, integration by parts is applied in the step to solve integrals for both negative infinity to zero and zero to infinity portions.
Each integral uses substitution to gently transform the integrals and make them more approachable for computation. Although the given integrals initially appear intimidating, applying this method helps display the manageable result of zero after evaluating the limits.
Remember, choosing the right functions for u and dv to simplify the computation is crucial in applying this method efficiently.
In the context of this particular problem, integration by parts is applied in the step to solve integrals for both negative infinity to zero and zero to infinity portions.
Each integral uses substitution to gently transform the integrals and make them more approachable for computation. Although the given integrals initially appear intimidating, applying this method helps display the manageable result of zero after evaluating the limits.
Remember, choosing the right functions for u and dv to simplify the computation is crucial in applying this method efficiently.
Definite Integrals
A definite integral calculates the signed area under a curve over a specific interval. When dealing with improper integrals, the journey from improper to definite involves strategic adjustments of limits when necessary. In the scenario posed by the exercise, the improper integral was decomposed into limits converting to definite integrals for segments.
The focus is then on those two new definite integrals, where we apply techniques like substitution and integration by parts. In the process, definite integrals provide an intermediate step from the abstract infinity to concrete numbers, ultimately offering convergence proof.
Clearly observing the role of limits and ensuring the function is continuous within the definite bounds is crucial. In our example, this practice validated that the value for each definite integral within their respective limits resulted in zero and thus concluding the convergence solution.
The focus is then on those two new definite integrals, where we apply techniques like substitution and integration by parts. In the process, definite integrals provide an intermediate step from the abstract infinity to concrete numbers, ultimately offering convergence proof.
Clearly observing the role of limits and ensuring the function is continuous within the definite bounds is crucial. In our example, this practice validated that the value for each definite integral within their respective limits resulted in zero and thus concluding the convergence solution.