Chapter 8: Problem 45
Evaluate the following integrals or state that they diverge. $$\int_{0}^{1} \frac{x^{3}}{x^{4}-1} d x$$
Short Answer
Expert verified
Answer: Diverge
Step by step solution
01
Check the behavior and singularity of the integrand.
For \(x=0\) and \(x=1\), the integrand behaves well. Nevertheless, for the denominator \(x^4-1\), we can factorize it as \((x^2-1)(x^2+1)\). We have a singularity on \(x^2-1\), which means when \(x=\pm1\). Since the integrand has a singularity at \(x=1\), we must be careful when integrating on the interval \([0,1]\). Next, we will find the antiderivative and then determine if the integral converges by checking the limit at 1.
02
Integration using substitution.
We'll use substitution to integrate the function; let \(u = x^4 - 1\). So \(\frac{du}{dx} = 4x^3 \Rightarrow \frac{1}{4} du = x^3 dx\).
Now rewrite the integral in terms of \(u\):
$$\int_{0}^{1} \frac{x^3}{x^4-1} d x = \int_{-1}^{0} \frac{1}{4}\frac{1}{u}d u$$
03
Evaluate the new integral.
We now have the integral:
$$\frac{1}{4}\int_{-1}^{0} \frac{1}{u} d u$$
The antiderivative of \(\frac{1}{u}\) is \(\ln |u|\). Therefore, we get:
$$\frac{1}{4}\left[\ln|u|\right]_{-1}^{0} = \frac{1}{4}\left[\ln|0+1| - \ln|-1+1|\right]$$
Now, we have a problem: \(\ln(0)\) implies that this integral diverges.
04
State the result.
The integral
$$\int_{0}^{1} \frac{x^3}{x^4-1} d x$$
diverges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Singularity Behavior
In calculus and mathematical analysis, understanding the singularity behavior of a function helps us recognize potential points of divergence in integrals. A singularity occurs at points where a function behaves exceptionally, such as becoming infinitely large or undefined, within the limits of integration. In the original exercise, this happens within the integral \( \int_{0}^{1} \frac{x^{3}}{x^{4}-1} \, dx \).
The denominator \( x^4 - 1 \) can be factored as \((x^2 - 1)(x^2 + 1)\). Singularities appear where the denominator is zero, specifically at \(x^2=1\) or \(x=\pm1\). During integration from 0 to 1, the point \(x=1\) is crucial because it is included in the integration interval and results in division by zero, indicating a singularity. Recognizing this behavior lets us determine the areas of concern when calculating the integral, highlighting the method's role in predicting whether the integral converges (exists) or diverges (does not exist).
Employing appropriate techniques, such as substitution or partial fraction decomposition, helps manage singularities within integrals, minimizing their effect or showing why convergence is impossible. This careful examination ensures accurate solutions and predictions in integral calculus.
The denominator \( x^4 - 1 \) can be factored as \((x^2 - 1)(x^2 + 1)\). Singularities appear where the denominator is zero, specifically at \(x^2=1\) or \(x=\pm1\). During integration from 0 to 1, the point \(x=1\) is crucial because it is included in the integration interval and results in division by zero, indicating a singularity. Recognizing this behavior lets us determine the areas of concern when calculating the integral, highlighting the method's role in predicting whether the integral converges (exists) or diverges (does not exist).
Employing appropriate techniques, such as substitution or partial fraction decomposition, helps manage singularities within integrals, minimizing their effect or showing why convergence is impossible. This careful examination ensures accurate solutions and predictions in integral calculus.
Integration by Substitution
Integration by substitution is a powerful technique in calculus used to simplify integrals by transforming variables. It's analogous to the reverse process of the chain rule in differentiation. By substituting a part of the integral with another variable, we can often solve complex problems more easily.
In the original exercise, we used substitution to address the integral \( \int_{0}^{1} \frac{x^3}{x^4-1} \, dx \) where \( u = x^4 - 1 \). This transformation simplifies the algebra within the integral, converting it into a function of \(u\), specifically \( \int_{-1}^{0} \frac{1}{4}\cdot\frac{1}{u} \, du \).
The derivative \( \frac{du}{dx} = 4x^3 \) allows us to express \( x^3 dx \) as \( \frac{1}{4} \, du \), effectively changing the integration variable from \(x\) to \(u\). As a result, the integration becomes more direct. The key advantage of using integration by substitution is that it can simplify the integrand, making complex or infinite integrals manageable. This technique provides insight into function behavior and is an essential tool for solving integrals involving difficult-to-integrate expressions.
In the original exercise, we used substitution to address the integral \( \int_{0}^{1} \frac{x^3}{x^4-1} \, dx \) where \( u = x^4 - 1 \). This transformation simplifies the algebra within the integral, converting it into a function of \(u\), specifically \( \int_{-1}^{0} \frac{1}{4}\cdot\frac{1}{u} \, du \).
The derivative \( \frac{du}{dx} = 4x^3 \) allows us to express \( x^3 dx \) as \( \frac{1}{4} \, du \), effectively changing the integration variable from \(x\) to \(u\). As a result, the integration becomes more direct. The key advantage of using integration by substitution is that it can simplify the integrand, making complex or infinite integrals manageable. This technique provides insight into function behavior and is an essential tool for solving integrals involving difficult-to-integrate expressions.
Convergence Tests
Convergence tests are methods used to determine whether an integral or series converges or diverges. Understanding when an integral converges allows us to know if it evaluates to a finite number, assuring meaningful results. On the other hand, divergence suggests that the integral does not resolve to a finite limit, indicating potential issues or infinite results.
In the solved exercise, after substituting and simplifying, the resulting integral \( \frac{1}{4}\int_{-1}^{0} \frac{1}{u} \, du \) was evaluated. The antiderivative here is \( \ln|u| \), and substituting the limits \(u = -1\) and \(u = 0\) unveiled a critical point for convergence. At \(u = 0\), the logarithm \( \ln(0) \) implies an undefined expression.
This crucial finding indicates divergence, as the natural logarithm function becomes infinitely negative towards \(-\infty\) as it approaches zero. By examining the behavior around this point, we concluded that the integral does not converge. This conclusion aligns with the singularity detected at \(x=1\), reinforcing how singular points affect convergence conclusions. By applying convergence tests, such as the limit comparison test or direct substitution, one can gain valuable insights into the nature of an integral, efficiently predicting its behavior.
In the solved exercise, after substituting and simplifying, the resulting integral \( \frac{1}{4}\int_{-1}^{0} \frac{1}{u} \, du \) was evaluated. The antiderivative here is \( \ln|u| \), and substituting the limits \(u = -1\) and \(u = 0\) unveiled a critical point for convergence. At \(u = 0\), the logarithm \( \ln(0) \) implies an undefined expression.
This crucial finding indicates divergence, as the natural logarithm function becomes infinitely negative towards \(-\infty\) as it approaches zero. By examining the behavior around this point, we concluded that the integral does not converge. This conclusion aligns with the singularity detected at \(x=1\), reinforcing how singular points affect convergence conclusions. By applying convergence tests, such as the limit comparison test or direct substitution, one can gain valuable insights into the nature of an integral, efficiently predicting its behavior.
