Comparison with a Known Integral

Oct 13, 2025 by magicalceylontours.com 119 views

Introduction

In this article, we will determine whether the given integral is convergent or divergent. The integral in question is $${ \int_e^{\infty} \frac{71}{x(\ln(x))^3} , dx }$$. We will use various techniques to evaluate the integral and determine its convergence.

Convergence Tests

To determine whether the integral is convergent or divergent, we can use various convergence tests. One of the most commonly used tests is the Limit Comparison Test.

Limit Comparison Test

The Limit Comparison Test states that if we have two integrals $${ \int_a^{\infty} f(x) , dx }$$ and $${ \int_a^{\infty} g(x) , dx }$$, and if $${ \lim_{x\to\infty} \frac{f(x)}{g(x)} = L }$$, where $L$ is a finite positive number, then either both integrals converge or both integrals diverge.

Comparison with a Known Integral

We can compare the given integral with a known integral to determine its convergence. Let's compare it with the integral $${ \int_e^{\infty} \frac{1}{x(\ln(x))^3} , dx }$$. We can see that the given integral is a constant multiple of this integral.

Limit of the Ratio

To apply the Limit Comparison Test, we need to find the limit of the ratio of the two integrals as $x$ approaches infinity.

{ \lim_{x\to\infty} \frac{\frac{71}{x(\ln(x))^3}}{\frac{1}{x(\ln(x))^3}} = \lim_{x\to\infty} 71 = 71 \}

Since the limit is a finite positive number, we can conclude that the given integral converges or diverges with the integral $${ \int_e^{\infty} \frac{1}{x(\ln(x))^3} , dx }$$.

Convergence of the Comparison Integral

To determine whether the comparison integral converges or diverges, we can use the p-Test.

p-Test

The p-Test states that if we have an integral of the form $${ \int_a^{\infty} \frac{1}{x^p} , dx }$$, where $p$ is a positive real number, then the integral converges if $p>1$ and diverges if $p\leq 1$ .

In this case, we have $p=1-3=-2$ , which is less than 1. Therefore, the comparison integral diverges.

Conclusion

Since the comparison integral diverges, we can conclude that the given integral also diverges.

Conclusion

In conclusion, the given integral $${ \int_e^{\infty} \frac{71}{x(\ln(x))^3} , dx }$$ is divergent.

Discussion

The given integral is a classic example of a divergent integral. The use of the Limit Comparison Test and the p-Test allowed us to determine the convergence of the integral.

References

[1] Calculus by Michael Spivak
[2] Real and Complex Analysis by Walter Rudin

Code

import sympy as sp

# Define the variable
x = sp.symbols('x')

# Define the integral
integral = sp.integrate(71/(x*sp.log(x)**3), (x, sp.exp(1), sp.oo))

# Print the result
print(integral)

Introduction

In our previous article, we determined that the integral $${ \int_e^{\infty} \frac{71}{x(\ln(x))^3} , dx }$$ is divergent. In this article, we will answer some frequently asked questions related to the integral.

Q: What is the main reason for the divergence of the integral?

A: The main reason for the divergence of the integral is that the exponent of the logarithmic term is negative, i.e., $-3$ . This means that the logarithmic term grows slower than the linear term, causing the integral to diverge.

Q: Can we use any other convergence tests to determine the convergence of the integral?

A: Yes, we can use other convergence tests such as the Comparison Test, the Direct Comparison Test, or the Improper Integral Test. However, the Limit Comparison Test is the most straightforward and efficient method to determine the convergence of this integral.

Q: How does the value of the constant 71 affect the convergence of the integral?

A: The value of the constant 71 does not affect the convergence of the integral. The integral would still be divergent even if the constant were 1 or any other positive value.

Q: Can we evaluate the integral numerically using a computer program?

A: Yes, we can evaluate the integral numerically using a computer program such as Sympy or Mathematica. However, the result would not be relevant to the conclusion of the article, as the integral is divergent.

Q: What is the relationship between the given integral and the comparison integral?

A: The given integral is a constant multiple of the comparison integral. The comparison integral is $${ \int_e^{\infty} \frac{1}{x(\ln(x))^3} , dx }$$.

Q: Why is the comparison integral used in the Limit Comparison Test?

A: The comparison integral is used in the Limit Comparison Test because it has the same form as the given integral, but with a simpler expression. This allows us to determine the convergence of the given integral by comparing it with the comparison integral.

Q: Can we use the p-Test to determine the convergence of the comparison integral?

A: Yes, we can use the p-Test to determine the convergence of the comparison integral. The p-Test states that if we have an integral of the form $${ \int_a^{\infty} \frac{1}{x^p} , dx }$$, where $p$ is a positive real number, then the integral converges if $p>1$ and diverges if $p\leq 1$ . In this case, we have $p=1-3=-2$ , which is less than 1, so the comparison integral diverges.

Conclusion

In conclusion, the given integral $${ \int_e^{\infty} \frac{71}{x(\ln(x))^3} , dx }$$ is divergent. We used the Limit Comparison Test and the p-Test to determine the convergence of the integral.

References

[1] Calculus by Michael Spivak
[2] Real and Complex Analysis by Walter Rudin

Code

import sympy as sp

# Define the variable
x = sp.symbols('x')

# Define the integral
integral = sp.integrate(71/(x*sp.log(x)**3), (x, sp.exp(1), sp.oo))

# Print the result
print(integral)

Note: The code is used to evaluate the integral numerically. However, the result is not relevant to the conclusion of the article.