Understanding the Integrands

Oct 13, 2025 by sorusgentrification.com 97 views

Introduction

In this article, we will delve into the computation of a definite integral involving the difference of two rational functions. The integral in question is $\int_{0}^{\infty}\frac{1}{x^7-x}-\frac{27}{x^7-27x}dx$ . This problem requires careful analysis and application of various techniques in integration, including partial fraction decomposition and the use of limits.

Understanding the Integrands

Both the integrands have a general form like $\frac{1}{x^7-x}$ and $\frac{27}{x^7-27x}$ . These rational functions can be decomposed into partial fractions, which will facilitate the integration process. To begin, we need to factor the denominators of the integrands.

Factoring the Denominators

The first integrand can be factored as follows:

\frac{1}{x^7-x} = \frac{1}{x(x-1)(x^2+1)(x^2+x+1)}

Similarly, the second integrand can be factored as:

\frac{27}{x^7-27x} = \frac{27}{x(x-3)(x^2+3)(x^2+3x+3)}

Partial Fraction Decomposition

We can now decompose each integrand into partial fractions. For the first integrand, we have:

\frac{1}{x(x-1)(x^2+1)(x^2+x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{Cx+D}{x^2+1} + \frac{Ex+F}{x^2+x+1}

where $A$ , $B$ , $C$ , $D$ , $E$ , and $F$ are constants to be determined.

Similarly, for the second integrand, we have:

\frac{27}{x(x-3)(x^2+3)(x^2+3x+3)} = \frac{G}{x} + \frac{H}{x-3} + \frac{Ix+J}{x^2+3} + \frac{Kx+L}{x^2+3x+3}

where $G$ , $H$ , $I$ , $J$ , $K$ , and $L$ are constants to be determined.

Determining the Constants

To determine the constants, we can use the method of equating coefficients or the method of substitution. In this case, we will use the method of equating coefficients.

After equating coefficients, we find that:

A = \frac{1}{6}, B = -\frac{1}{6}, C = \frac{1}{2}, D = 0, E = -\frac{1}{2}, F = 0

and

G = -\frac{27}{2}, H = \frac{27}{2}, I = -\frac{27}{4}, J = \frac{27}{4}, K = -\frac{27}{2}, L = \frac{27}{2}

Integrating the Partial Fractions

Now that we have the partial fractions, we can integrate each term separately.

For the first integrand, we have:

\int \frac{1}{x(x-1)(x^2+1)(x^2+x+1)} dx = \frac{1}{6} \int \frac{1}{x} dx - \frac{1}{6} \int \frac{1}{x-1} dx + \frac{1}{2} \int \frac{x}{x^2+1} dx - \frac{1}{2} \int \frac{x}{x^2+x+1} dx

Similarly, for the second integrand, we have:

\int \frac{27}{x(x-3)(x^2+3)(x^2+3x+3)} dx = -\frac{27}{2} \int \frac{1}{x} dx + \frac{27}{2} \int \frac{1}{x-3} dx - \frac{27}{4} \int \frac{x}{x^2+3} dx + \frac{27}{4} \int \frac{x}{x^2+3x+3} dx

Evaluating the Integrals

We can now evaluate each integral separately.

For the first integrand, we have:

\int \frac{1}{x(x-1)(x^2+1)(x^2+x+1)} dx = \frac{1}{6} \ln|x| - \frac{1}{6} \ln|x-1| + \frac{1}{2} \arctan x - \frac{1}{2} \arctan \left(\frac{x+1}{2}\right) + C

Similarly, for the second integrand, we have:

\int \frac{27}{x(x-3)(x^2+3)(x^2+3x+3)} dx = -\frac{27}{2} \ln|x| + \frac{27}{2} \ln|x-3| - \frac{27}{4} \arctan \frac{x}{3} + \frac{27}{4} \arctan \left(\frac{x+3}{2}\right) + C

Computing the Definite Integral

Now that we have the antiderivatives, we can compute the definite integral.

We have:

\int_{0}^{\infty} \frac{1}{x^7-x} - \frac{27}{x^7-27x} dx = \left[ \frac{1}{6} \ln|x| - \frac{1}{6} \ln|x-1| + \frac{1}{2} \arctan x - \frac{1}{2} \arctan \left(\frac{x+1}{2}\right) \right]_{0}^{\infty} - \left[ -\frac{27}{2} \ln|x| + \frac{27}{2} \ln|x-3| - \frac{27}{4} \arctan \frac{x}{3} + \frac{27}{4} \arctan \left(\frac{x+3}{2}\right) \right]_{0}^{\infty}

Evaluating the limits, we get:

\int_{0}^{\infty} \frac{1}{x^7-x} - \frac{27}{x^7-27x} dx = \frac{1}{6} \ln \infty - \frac{1}{6} \ln 1 + \frac{1}{2} \arctan \infty - \frac{1}{2} \arctan \left(\frac{\infty+1}{2}\right) + \frac{27}{2} \ln 0 - \frac{27}{2} \ln 3 + \frac{27}{4} \arctan \frac{0}{3} - \frac{27}{4} \arctan \left(\frac{0+3}{2}\right)

Simplifying, we get:

\int_{0}^{\infty} \frac{1}{x^7-x} - \frac{27}{x^7-27x} dx = \frac{1}{6} \ln \infty - \frac{1}{6} \ln 1 + \frac{1}{2} \frac{\pi}{2} - \frac{1}{2} \frac{\pi}{4} + \frac{27}{2} \ln 0 - \frac{27}{2} \ln 3 + \frac{27}{4} 0 - \frac{27}{4} \frac{\pi}{4}

Simplifying further, we get:

\int_{0}^{\infty} \frac{1}{x^7-x} - \frac{27}{x^7-27x} dx = \frac{1}{6} \ln \infty - \frac{1}{6} \ln 1 + \frac{1}{2} \frac{\pi}{2} - \frac{1}{2} \frac{\pi}{4} + \frac{27}{2} (-\infty) - \frac{27}{2} \ln 3 + \frac{27}{4} 0 - \frac{27}{4} \frac{\pi}{4}

Simplifying even further, we get:

\int_{0}^{\infty} \frac{1}{x^7-x} - \frac{27}{x^7-27x} dx = \frac{1}{6} \ln \infty - \frac{1}{6} \ln 1 + \frac{1}{2} \frac{\pi}{2} - \frac{1}{2} \frac{\pi}{4} + \frac{27}{2} (-\infty) - \frac{27}{2} \ln 3 + \frac{27}{4} 0 - \frac{27}{4} \frac{\pi}{4}

Q: What is the main goal of this article?

A: The main goal of this article is to compute the definite integral of a difference of two rational functions, specifically $\int_{0}^{\infty}\frac{1}{x^7-x}-\frac{27}{x^7-27x}dx$ .

Q: What are the key concepts involved in computing this integral?

A: The key concepts involved in computing this integral include partial fraction decomposition, the use of limits, and the evaluation of definite integrals.

Q: How do you decompose the rational functions into partial fractions?

A: To decompose the rational functions into partial fractions, we need to factor the denominators and then express each fraction as a sum of simpler fractions.

Q: What are the steps involved in evaluating the definite integral?

A: The steps involved in evaluating the definite integral include:

Evaluating the antiderivatives of each partial fraction.
Computing the limits of the antiderivatives as $x$ approaches infinity.
Simplifying the resulting expression to obtain the final answer.

Q: What are some common mistakes to avoid when computing this integral?

A: Some common mistakes to avoid when computing this integral include:

Failing to factor the denominators correctly.
Making errors in the partial fraction decomposition.
Failing to evaluate the limits correctly.

Q: What are some real-world applications of this integral?

A: This integral has several real-world applications, including:

Calculating the area under curves in physics and engineering.
Modeling population growth and decay in biology.
Calculating the volume of solids in geometry.

Q: How can I apply the techniques learned in this article to other problems?

A: The techniques learned in this article can be applied to other problems involving rational functions, including:

Computing definite integrals of rational functions.
Evaluating limits of rational functions.
Solving equations involving rational functions.

Q: What are some additional resources for learning more about this topic?

A: Some additional resources for learning more about this topic include:

Textbooks on calculus and analysis.
Online resources such as Khan Academy and MIT OpenCourseWare.
Research papers and articles on the topic of rational functions and definite integrals.

Conclusion

Computing the integral of a difference of two rational functions is a challenging problem that requires careful analysis and application of various techniques in integration. By following the steps outlined in this article, you can learn how to compute this integral and apply the techniques to other problems involving rational functions.