Chapter 7: Problem 16
Evaluate the following integrals or state that they diverge. $$\int_{0}^{\infty} \frac{p}{\sqrt[5]{p^{2}+1}} d p$$
Short Answer
Expert verified
Answer: The integral diverges.
Step by step solution
01
Identify the integral
The given integral is an improper integral with limits from \(0\) to \(\infty\):
$$\int_{0}^{\infty} \frac{p}{\sqrt[5]{p^{2}+1}} d p$$
02
Use substitution method
Let's use substitution to evaluate the integral. Let \(u = p^2 + 1\). Then, \(du = 2p \, dp\). We can update the integral with these substitutions:
$$\frac{1}{2} \int \frac{1}{\sqrt[5]{u}} du$$
Now, we also need to change the limits of integration with respect to \(u\). When \(p=0\), we have \(u=1\). When \(p \to \infty\), \(u \to \infty\). So, the new limits of integration are \(1\) to \(\infty\).
03
Evaluate the integral
Now, we evaluate the integral
$$\frac{1}{2} \int_{1}^{\infty} \frac{1}{\sqrt[5]{u}} du$$
This is an improper integral, and the form is \(\int_{1}^{\infty} u^{r} du\) with \(r=-\frac{1}{5}\). If \(-1<r<0\), the integral converges; otherwise, it diverges. In our case, \(-\frac{1}{5}\) is between \(-1\) and \(0\), so the integral converges. To find the value of the integral, we can apply the power rule for integration. We will get:
$$\frac{1}{2}\left[\frac{5}{4} u^{\frac{4}{5}} \right]_{1}^{\infty}$$
04
Evaluate the limits
Now we evaluate the limits:
$$\frac{1}{2}\left[\frac{5}{4} \cdot (\infty^{\frac{4}{5}} - 1^{\frac{4}{5}})\right]$$
As \(\infty^{\frac{4}{5}}\) approaches infinity, the integral equals:
$$\frac{1}{2}\left[\frac{5}{4} \cdot (\infty - 1)\right] \to \frac{5}{8}\infty$$
Therefore, the given integral converges to \(\frac{5}{8}\infty\), which means it is divergent in the sense that it does not approach a finite number.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a fundamental technique when dealing with integrals, especially improper ones. This method transforms a complex integral into a simpler one by changing variables, making it easier to solve. It's similar to the chain rule in differentiation. By setting a part of the integrand to a new variable, we can simplify calculations.
In the given exercise, we used the substitution \( u = p^2 + 1 \), which resulted in \( du = 2p \, dp \). This substitution helped change the original integral into a more manageable form: \( \frac{1}{2} \int \frac{1}{\sqrt[5]{u}} \, du \).
Keep in mind that with substitution, you also need to adjust the limits of integration. Initially, the limits were from \( p = 0 \) to \( p = \infty \). After substitution, the limits became \( u = 1 \) to \( u = \infty \), reflecting the change in variables. This adjustment ensures that our evaluation aligns with the transformed integral.
In the given exercise, we used the substitution \( u = p^2 + 1 \), which resulted in \( du = 2p \, dp \). This substitution helped change the original integral into a more manageable form: \( \frac{1}{2} \int \frac{1}{\sqrt[5]{u}} \, du \).
Keep in mind that with substitution, you also need to adjust the limits of integration. Initially, the limits were from \( p = 0 \) to \( p = \infty \). After substitution, the limits became \( u = 1 \) to \( u = \infty \), reflecting the change in variables. This adjustment ensures that our evaluation aligns with the transformed integral.
Integral Convergence
Integral convergence is a crucial concept to determine whether an integral has a finite value or not. An improper integral, like the exercise one with infinite limits, tests convergence.
When evaluating an integral like \( \int_{0}^{\infty} \), we need to check whether it converges and has a finite limit, or diverges, meaning it grows without bound.
When evaluating an integral like \( \int_{0}^{\infty} \), we need to check whether it converges and has a finite limit, or diverges, meaning it grows without bound.
- To determine convergence, examine the behavior of the function as it approaches the limits of integration.
- The criteria used involve the exponent \( r \) in the expression \( \int_{a}^{\infty} x^r \, dx \). If \( -1 < r < 0 \), the integral converges.
Power Rule for Integration
The power rule for integration allows us to easily integrate functions of the form \( x^n \). This rule simplifies finding antiderivatives, crucial for evaluating integrals after substitution.
The formula used is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \), where \( C \) is the constant of integration.
Evaluating the antiderivative involved using the formula with the limits substituted for \( u \). The rule immensely simplifies the integral solving process, ensuring our results are exact.
The formula used is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \), where \( C \) is the constant of integration.
- It's used to evaluate parts of an integrand that follow this power pattern.
- The power rule is applied after using substitution to simplify the expression.
Evaluating the antiderivative involved using the formula with the limits substituted for \( u \). The rule immensely simplifies the integral solving process, ensuring our results are exact.