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Chapter 9: Problem 47

If \(k>0\), show that \(\int_{e}^{\infty} \frac{k}{x(\ln x)^{k+1}} d x=1 .\)

Short Answer

Expert verified
The integral evaluates to 1.

Step by step solution

01

- Substitution

Let鈥檚 make a substitution to simplify the integral. Set u = ln(x), which means du = (1/x) dx. Note that when x = e, u = ln(e) = 1, and as x approaches infinity, u also approaches infinity.
02

- Rewriting the Integral

Rewrite the integral in terms of u: \[ \int_{e}^{\text{infinity}} \frac{k}{x (\ln x)^{k+1}} dx \] becomes \[ \int_{1}^{\text{infinity}} \frac{k}{u^{k+1}} du \].
03

- Integrating

Now integrate \[ \int_{1}^{\text{infinity}} \frac{k}{u^{k+1}} du \]. Integrate using the formula for the integral of a power function: \[ \int u^{-a} du = \frac{u^{-a+1}}{-a+1} \]. Here, a = k + 1.
04

- Simplify the Integral

Using the formula, the integral becomes: \[ k \left[ \frac{u^{-k}}{-k} \right]_{1}^{\text{infinity}} \]. Simplify this to \[ - \left[ u^{-k} \right]_{1}^{\text{infinity}} \].
05

- Evaluate the Limits

Evaluate the expression at the bounds: \[ - (0 - 1) = 1 \]. Since \(u^{-k}\) approaches 0 as \(u\) approaches infinity and is 1 when \(u = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
In calculus, substitution is a powerful technique to make complex integrals easier to handle. The process involves changing the variable of integration to simplify the function inside the integral.
In our exercise, we substitute \(u = \ln(x)\) and consequently, \(du = \frac{1}{x} dx\). A key part of this method is also adjusting the integral's limits according to the new variable. For instance:
  • When \(x = e\), then \(u = \ln(e) = 1\).
  • As \(x\) approaches infinity, \(u\) also approaches infinity.
This makes the rewritten integral look much simpler and more workable, making the subsequent steps easier to manage.
Integration of Power Functions
Integrating power functions often involves recognizing and applying a specific integral formula. For a general power function \(u^{-a}\), the integration formula is:
\[\int u^{-a} \, du = \frac{u^{-a+1}}{-a+1}\]
In the given problem, after the substitution, the integral becomes:
\[\int_{1}^{\text{infinity}} \frac{k}{u^{k+1}} \, du\]
Here, we recognize \(\frac{k}{u^{k+1}}\) as a power function with \(a = k + 1\). Applying the power function integral formula, we get:
\[\int u^{-(k+1)} \, du = \frac{u^{-k}}{-k}\]
Multiplied by \(k\), the final integrated function is:
  • \[k \cdot \frac{u^{-k}}{-k}\]
  • Which simplifies to: \[- u^{-k}\]
From here, we proceed to evaluate the bounds.
Limit Evaluation
Evaluating improper integrals often requires careful limit evaluation, especially when one of the bounds is infinite. In our example:
\[- \left[ u^{-k} \right]_{1}^{\text{infinity}}\]
We need to consider what happens to the function \(u^{-k}\) at the limits:
  • As \(u\) approaches infinity, \(u^{-k}\) approaches 0 because raising a large number to a negative power yields a very small number.
  • When \(u = 1\), \(u^{-k} = 1^{k} = 1\), since any number to the zero power is 1.
Thus the expression becomes:
\[0 - (-1) = 1\]
This shows that the value of the improper integral is indeed 1, as required by the problem.

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