The Integral of a Rational Function

Oct 13, 2025 by pinkandtrash.com 82 views

Introduction

In this article, we will explore the evaluation of a specific integral involving a rational function. The integral in question is given by:

I=\int_0^\infty\frac{(1-t^5)t^4}{1-t^{24}}dt.\tag{1}

This integral appears to be a complex one, and it is not immediately clear how to evaluate it using standard techniques. However, we will show that with a clever change of variable, we can simplify the integral and evaluate it without resorting to harmonic numbers or series.

The Change of Variable

The change of variable $t\to t^{\frac1{24}}$ is a key step in simplifying the integral. This substitution is motivated by the fact that the denominator of the integrand is a power of $t$ , specifically $t^{24}$ . By making this substitution, we can rewrite the integral in a more manageable form.

Let's apply the change of variable to the integral:

I=\int_0^\infty\frac{(1-t^5)t^4}{1-t^{24}}dt

=\int_0^\infty\frac{(1-t^{\frac{5}{24}})t^{\frac{4}{24}}}{1-t}dt

=\frac{1}{24}\int_0^\infty\frac{t^{-\frac{21}{24}}-t^{-\frac{19}{24}}}{1-t}dt

The Integral of a Rational Function

Now that we have simplified the integral, we can focus on evaluating it. The integral we are left with is:

I=\frac{1}{24}\int_0^\infty\frac{t^{-\frac{21}{24}}-t^{-\frac{19}{24}}}{1-t}dt

This is a rational function, and we can use the standard techniques of integration to evaluate it. However, we are not allowed to use harmonic numbers or series, so we need to find an alternative approach.

Using the Logarithmic Derivative

One approach to evaluating the integral is to use the logarithmic derivative. The logarithmic derivative of a function $f(x)$ is given by:

\frac{d}{dx}\log f(x)=\frac{f'(x)}{f(x)}

We can use this formula to evaluate the integral by taking the logarithmic derivative of the integrand.

Let's take the logarithmic derivative of the integrand:

\frac{d}{dt}\log\left(\frac{t^{-\frac{21}{24}}-t^{-\frac{19}{24}}}{1-t}\right)

=\frac{(1-t)\left(-\frac{21}{24}t^{-\frac{25}{24}}+\frac{19}{24}t^{-\frac{23}{24}}\right)-(t^{-\frac{21}{24}}-t^{-\frac{19}{24}})(-1)}{(1-t)^2}

Simplifying the Logarithmic Derivative

Now that we have taken the logarithmic derivative of the integrand, we can simplify the expression. We can start by combining like terms:

=\frac{(1-t)\left(-\frac{21}{24}t^{-\frac{25}{24}}+\frac{19}{24}t^{-\frac{23}{24}}\right)-(t^{-\frac{21}{24}}-t^{-\frac{19}{24}})(-1)}{(1-t)^2}

=\frac{-\frac{21}{24}t^{-\frac{25}{24}}+\frac{19}{24}t^{-\frac{23}{24}}+\frac{21}{24}t^{-\frac{25}{24}}-\frac{19}{24}t^{-\frac{23}{24}}}{(1-t)^2}

=\frac{2}{24}\left(\frac{1}{t^{\frac{25}{24}}}-\frac{1}{t^{\frac{23}{24}}}\right)

Evaluating the Integral

Now that we have simplified the logarithmic derivative, we can evaluate the integral. We can start by integrating the logarithmic derivative:

\int\frac{2}{24}\left(\frac{1}{t^{\frac{25}{24}}}-\frac{1}{t^{\frac{23}{24}}}\right)dt

=\frac{1}{12}\left(-\frac{1}{t^{\frac{25}{24}}}+\frac{1}{t^{\frac{23}{24}}}\right)+C

The Final Answer

Now that we have evaluated the integral, we can find the final answer. We can start by substituting the limits of integration:

\left[\frac{1}{12}\left(-\frac{1}{t^{\frac{25}{24}}}+\frac{1}{t^{\frac{23}{24}}}\right)\right]_0^\infty

=\frac{1}{12}\left(0-0\right)+\frac{1}{12}\left(\frac{1}{0^{\frac{23}{24}}}-\frac{1}{\infty}\right)

=\frac{1}{12}\left(\frac{1}{0^{\frac{23}{24}}}-0\right)

=\frac{1}{12}\left(\frac{1}{0^{\frac{23}{24}}}\right)

Conclusion

Introduction

In our previous article, we evaluated the integral of a rational function without using harmonic numbers or series. We used a change of variable and the logarithmic derivative to simplify the integral and find the final answer. In this article, we will answer some common questions that readers may have about the evaluation of the integral.

Q: What is the change of variable used in the evaluation of the integral?

A: The change of variable used in the evaluation of the integral is $t\to t^{\frac1{24}}$ . This substitution is motivated by the fact that the denominator of the integrand is a power of $t$ , specifically $t^{24}$ .

Q: Why is the logarithmic derivative used in the evaluation of the integral?

A: The logarithmic derivative is used in the evaluation of the integral because it allows us to simplify the integrand and make it easier to integrate. The logarithmic derivative of a function $f(x)$ is given by:

\frac{d}{dx}\log f(x)=\frac{f'(x)}{f(x)}

Q: What is the final answer to the integral?

A: The final answer to the integral is $\boxed{\frac{1}{12}\left(\frac{1}{0^{\frac{23}{24}}}\right)}$ . However, it's worth noting that this answer is not a standard result and may not be valid in all cases.

Q: Can the integral be evaluated using other methods?

A: Yes, the integral can be evaluated using other methods. For example, we could use the method of partial fractions to decompose the integrand into simpler fractions and then integrate each fraction separately. However, this method may not be as straightforward as the method used in this article.

Q: What are some common applications of the integral?

A: The integral has many common applications in mathematics and physics. For example, it can be used to calculate the area under curves, the volume of solids, and the work done by a force. It can also be used to solve problems in physics, such as calculating the energy of a system or the momentum of a particle.

Q: What are some common mistakes to avoid when evaluating the integral?

A: Some common mistakes to avoid when evaluating the integral include:

Not simplifying the integrand before integrating
Not using the correct method of integration
Not checking the limits of integration
Not considering the possibility of singularities in the integrand

Conclusion

In this article, we have answered some common questions that readers may have about the evaluation of the integral. We have discussed the change of variable used in the evaluation of the integral, the logarithmic derivative, and the final answer. We have also discussed some common applications of the integral and some common mistakes to avoid when evaluating the integral.