Problem 64 - Calculus Early Transcendentals

Question

For what values of \(p\) does the integral \(\int_{2}^{\infty} \frac{d x}{x \ln ^{p} x}\) exist and what is its value (in terms of \(p\) )?

Step-by-Step Solution

Answer

Answer: The improper integral converges for values of \(p > 1\), and its value is \(\frac{1}{(p-1)(\ln{2})^{p-1}}\).

Step 1: Identify the type of improper integral

The given integral is an improper integral of Type 1 with the variable \(x\) running from a finite value to infinity. It has the general form: $$\int_{2}^{\infty} \frac{1}{x \ln^p x}dx$$

Step 2: Test for Convergence

To test for convergence, we can use the comparison tests. Start with the integral test. $$\lim_{b\to\infty}\int_{2}^{b} \frac{1}{x \ln^p x}dx$$ Comparing this integral to a simple integral expression, \(\frac{1}{x^\alpha}\) where \(\alpha > 1\) will be convergent. $$\frac{1}{x\ln^{p}x} < \frac{1}{x^\alpha}$$ Let's find the integral of \(\frac{1}{x\ln(x)}\). $$\int_{2}^{\infty} \frac{1}{x\ln(x)} dx$$ Using the substitution method: \(u = \ln x\) \(\frac{du}{dx} = \frac{1}{x}\) \(dx = xdu\) The integral will become: $$\int_{\ln{2}}^{\infty} \frac{xdu}{xu}$$ Which simplifies to: $$\int_{\ln{2}}^{\infty} \frac{du}{u}$$ This is a simple integral that will give us the initial value of \(p\) when the integral converges. $$\left[ \ln u \right]_{\ln 2}^{\infty} = \infty - \ln{(\ln 2)}$$ As this is an infinite value, we can only conclude that the integral converges when \(p > 1\).

Step 3: Evaluate the integral

Now, we will evaluate the integral for values of \(p > 1\). The integral expression is: $$\int_{2}^{\infty} \frac{1}{x \ln^p x}dx$$ Using the same substitution method as earlier: \(u = \ln x\) \(\frac{du}{dx} = \frac{1}{x}\) \(dx = xdu\) Then we have: $$\int_{\ln{2}}^{\infty} \frac{xdu}{xu^p} = \int_{\ln{2}}^{\infty} \frac{du}{u^p}$$ Now, integrate this expression with respect to \(u\): $$\left[ -\frac{1}{(p-1)u^{p-1}} \right]_{\ln{2}}^{\infty}$$ Substitute the limits of integration: $$ -\frac{1}{(p-1)\infty^{p-1}} + \frac{1}{(p-1)(\ln{2})^{p-1}}$$ The term involving infinity goes to zero, so the value of the integral is: $$\frac{1}{(p-1)(\ln{2})^{p-1}}$$

Step 4: Final Result

For \(p > 1\), the integral \(\int_{2}^{\infty} \frac{1}{x \ln^p x}dx\) exists and its value is: $$\frac{1}{(p-1)(\ln{2})^{p-1}}$$

Key Concepts

Convergence Test

To determine whether an improper integral converges, we utilize convergence tests. These tests help us understand if the area under a curve, stretching to infinity, can be calculated to a finite number. For the integral \(\int_{2}^{\infty} \frac{1}{x \ln^p x}dx\), determining convergence involves ensuring the denominator grows faster than the numerator for large values of \(x\).
A common method is the **Comparison Test**, which involves comparing the integral in question with a dominantly more straightforward function, often in the form of \(\frac{1}{x^\alpha}\). Here, \(\alpha > 1\) signifies a known convergent form. If \(\frac{1}{x \ln^p x} < \frac{1}{x^\alpha}\) as \(x\) approaches infinity, then the original integral converges. This ensures the tail of the integral's area is diminishing effectively.
Convergence tests such as these are vital for understanding not just if an integral converges, but also the conditions under which it does.

Substitution Method

One of the powerful techniques in calculus is the substitution method, often used to simplify complex integrals. This approach allows for the transformation of the integral into a more manageable form. The substitution \(u = \ln x\), used in our integral, is chosen because it addresses the complexity in the logarithmic term. Here’s the breakdown:

• Start by setting a new variable: \(u = \ln x\)
• Find the differential \(du\): \(\frac{du}{dx} = \frac{1}{x}\)
• Rearrange to express \(dx\) in terms of \(du\): \(dx = x \, du\)

This substitution simplifies the integral \(\int \frac{1}{x \ln^p x}\) into \(\int \frac{du}{u^p}\), which is much simpler to integrate. Transformations like these are crucial because they allow complex-looking integrals to be tackled using basic integral rules and results.

Integral Test

The integral test is a tool used for analyzing the convergence of infinite series by relating them to the convergence of corresponding improper integrals. When you apply the integral test, you essentially check if the integral of a function that represents the series converges.
For the given problem, the function \(\frac{1}{x \ln^p x}\) is comparable to a known convergent form by the integral test method. You take an integral of this function from a specified range (in this case from 2 to infinity) and analyze whether it converges or diverges. If the integral converges, then the related series \(\sum_{n=2}^{\infty} \frac{1}{n \ln^p n}\) would also converge.
The integral serves as a benchmark; if it diverges, then the series diverges too. Essentially, the integral test provides a solid ground for establishing convergence properties when examined over infinity.

Infinite Limits

Infinite limits in integration refer to the bounds of an integral extending to infinity, or the function having infinite discontinuities. These scenarios define improper integrals, which we encounter often in calculus.
The original problem \(\int_{2}^{\infty} \frac{1}{x \ln^p x}dx\) is an example of such a case, where the upper limit is infinite. To evaluate this type of integral, we rely on finding limits as our upper bound approaches infinity. The convergence depends on whether the limit of the resulting integral as \(b \to \infty\) yields a finite value.
Dealing with infinite limits involves examining the behavior of the function as it stretches across an infinite domain, looking at how quickly it approaches zero, and whether this results in a bounded (finite) area.