Evaluating Improper Integrals: A Comprehensive Guide

Improper integrals are a key concept in calculus, especially when dealing with functions that extend to infinity or have singularities. This article provides a detailed method for evaluating a specific class of improper integrals. We will explore an example integral, apply substitution techniques, and use the gamma function to find its value.

Introduction to Improper Integrals

An improper integral is an integral that has one or both limits at infinity, or an integrand that approaches infinity at one or more points within the interval of integration. These integrals often require careful evaluation to determine their convergence or divergence. In this discussion, we focus on a specific improper integral:

Definition of the Integral

Consider the integral:

I int_1^{infty} left(frac{log{x}}{x}right)^{2011} dx

The goal is to determine the value of this integral.

Step-by-Step Solution

Step 1: Substitution

We start by making a substitution. Let:

u log{x}

This implies:

x e^u quad text{and} quad dx e^u , du

When x 1, we have:

u log{1} 0

As x to infty, we have:

u to infty

Therefore, the integral transforms to:

I int_0^{infty} left(frac{u}{e^u}right)^{2011} e^u , du int_0^{infty} frac{u^{2011}}{e^{2011u}} , du

Step 2: Simplify the Integral

The integral can now be simplified as:

I int_0^{infty} u^{2011} e^{-2011u} , du

Step 3: Recognize the Gamma Function

The given integral is in the form of a gamma function:

int_0^{infty} u^n e^{-beta u} , du frac{n!}{beta^{n-1}}

Here, n 2011 and beta 2011. Thus, we have:

I frac{2011!}{2011^{2012}}

Generalization and Conclusion

General Case: Integral In

Consider a more general integral:

text{Define } I_n int_1^{infty} frac{log^n x}{x^k} , dx quad text{where } k > 1

Using the substitution t log{x} or x e^t, the integral transforms to:

I_n -frac{1}{k-1} int_1^{infty} log^n x cdot dleft(frac{1}{x^{k-1}}right)

This further simplifies to:

I_n -frac{1}{k-1} left[frac{log^n x}{x^{k-1}}right]_1^{infty} frac{n}{k-1} int_1^{infty} frac{log^{n-1} x}{x^k} , dx frac{n}{k-1} I_{n-1}

Applying the recursive formula:

I_n frac{n!}{(k-1)^{n-1}} cdot frac{1}{k-1} frac{n!}{(k-1)^{n-1 1}}

Substituting n k 2011, we get:

int_1^{infty} left(frac{log x}{x}right)^{2011} , dx frac{2011!}{2010^{2012}}

Conclusion

The value of the integral is:

boxed{int_1^{infty} left(frac{log x}{x}right)^{2011} , dx frac{2011!}{2010^{2012}}}

This example showcases the power of substitution and the gamma function in evaluating complex improper integrals. For more generalized cases, the method provides a clear and structured approach to finding the solution.