find the values of p for which the following integral converges 1/(x(ln(x))^p) from e to infinity and evaluate the integral at all values of p it converges

We are tasked with finding the values of \( p \) for which the integral

\[
I(p) = \int_e^\infty \frac{1}{x (\ln(x))^p} \, dx
\]

converges, and then evaluating it for the values of \( p \) where it converges.

### Step 1: Substitution
To handle this integral, we start by making the substitution \( u = \ln(x) \). Then, we have:

\[
du = \frac{1}{x} dx \quad \text{or} \quad dx = x du = e^u du.
\]

Under this substitution:
- When \( x = e \), \( u = 1 \).
- When \( x \to \infty \), \( u \to \infty \).

Thus, the integral becomes:

\[
I(p) = \int_1^\infty \frac{1}{(\ln(e^u))^p} \cdot e^u \cdot du = \int_1^\infty \frac{1}{u^p} \, du.
\]

### Step 2: Convergence of the Integral
Now, we need to analyze the convergence of the integral:

\[
\int_1^\infty \frac{1}{u^p} \, du.
\]

- For \( p > 1 \), the integral converges because for large \( u \), \( \frac{1}{u^p} \) decreases rapidly enough to ensure convergence.
- For \( p \leq 1 \), the integral diverges. Specifically:
  - When \( p = 1 \), the integral behaves as \( \int_1^\infty \frac{1}{u} \, du \), which is the harmonic integral and diverges logarithmically.
  - For \( p < 1 \), the term \( \frac{1}{u^p} \) grows too slowly to counteract the large \( u \)-behavior, leading to divergence.

Thus, the integral converges only for \( p > 1 \).

### Step 3: Evaluating the Integral
For \( p > 1 \), the integral is a standard improper integral:

\[
I(p) = \int_1^\infty \frac{1}{u^p} \, du.
\]

This can be evaluated as:

\[
I(p) = \left[ \frac{u^{1-p}}{1-p} \right]_1^\infty = \lim_{u \to \infty} \frac{u^{1-p}}{1-p} - \frac{1^{1-p}}{1-p}.
\]

For \( p > 1 \), as \( u \to \infty \), \( u^{1-p} \to 0 \). Thus, we are left with:

\[
I(p) = \frac{1}{p-1}.
\]

### Conclusion
- The integral converges for \( p > 1 \).
- The value of the integral for \( p > 1 \) is:

\[
I(p) = \frac{1}{p-1}.
\]

Would you like any further details or clarification?

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Here are 5 related questions for further exploration:
1. How would the convergence change if the lower limit of integration were changed to 1 instead of \( e \)?
2. What happens to the integral if we replace \( (\ln(x))^p \) with \( (\ln(x))^q \) for some \( q \neq p \)?
3. How does the convergence behavior change if the integrand is \( \frac{1}{x^a (\ln(x))^p} \) instead?
4. Can you generalize the result for an integral of the form \( \int_e^\infty \frac{1}{x (\ln(x))^{p+k}} \, dx \)?
5. What is the behavior of the integral if \( x \) approaches a different asymptote, say \( 1 \) instead of \( e \)?

Tip: Substitution techniques, especially logarithmic substitutions, are powerful for integrals involving logarithms or powers of logarithms!

Explore the convergence of the integral 1/(x(ln(x))^p) from e to infinity. Learn for which values of p the integral converges and find the evaluated result when p > 1. This problem involves improper integrals, logarithmic substitution, and calculus concepts, suitable for undergraduate students.

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