The problem asks for the values of $p$ for which the integral

$\int_2^{\infty} \frac{1}{x (\ln x)^p} \, dx$

converges.

To determine this, we will analyze the integral at its potential points of divergence, specifically at $x = \infty$. The behavior near $x = 2$ is not problematic, as the integrand does not blow up. Therefore, the main issue is with $x \to \infty$.

Step-by-Step Approach:

1. Substitution to Simplify: Let’s use the substitution: $u = \ln x \quad \text{so that} \quad du = \frac{1}{x} dx.$ This changes the bounds of integration:

   \quad \text{When } x \to \infty, \ u \to \infty.$$ The integral becomes: $$\int_{\ln 2}^{\infty} \frac{du}{u^p}.$$

2. Analyzing Convergence: The behavior of the integral depends on the value of $p$.

   • For $p = 1$, the integral becomes $\int_{\ln 2}^{\infty} \frac{du}{u}$, which is a divergent logarithmic integral.
   • For $p > 1$, the integral becomes $\int_{\ln 2}^{\infty} u^{-p} \, du$, which converges because the integral of $u^{-p}$ for $p > 1$ yields a finite result.
   • For $p < 1$, the integral behaves like $u^{1-p}$, which diverges as $u \to \infty$ since the exponent $1-p > 0$.

Conclusion:

The integral converges for $p > 1$.

Would you like further explanation or clarification on any step?

Here are 5 related questions to deepen your understanding:

1. How does the substitution $u = \ln x$ simplify the analysis of the integral?
2. What happens to the integral when $p = 1$, and why does it diverge?
3. Could you explore what happens if the lower bound of the integral changes, for example, to 1 instead of 2?
4. What is the general criterion for convergence for integrals of the form $\int_1^\infty \frac{1}{x^a} dx$?
5. How can you relate this problem to the comparison test for improper integrals?

Tip: Whenever solving improper integrals, always examine the behavior near points of infinity or where the function could become unbounded.