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This problem focuses on evaluating the **p-integral** and examining its behavior for different values of \( p \). The integral in question is:

\[
\int_1^\infty \frac{1}{x^p} \, dx = 
\begin{cases}
\frac{1}{p-1} & \text{if } p > 1, \\
\infty & \text{if } p \leq 1.
\end{cases}
\]

Let's work through each part:

### Part (a): \( p > 1 \)

The integral to evaluate is:

\[
\int_1^\infty \frac{1}{x^p} \, dx
\]

1. **Find the antiderivative**:
   \[
   \int \frac{1}{x^p} \, dx = \int x^{-p} \, dx = \frac{x^{1-p}}{1-p} \quad \text{(valid for } p \neq 1 \text{)}.
   \]

2. **Apply limits**:
   \[
   \int_1^\infty \frac{1}{x^p} \, dx = \lim_{b \to \infty} \left[ \frac{x^{1-p}}{1-p} \right]_1^b = \lim_{b \to \infty} \left( \frac{b^{1-p}}{1-p} - \frac{1^{1-p}}{1-p} \right).
   \]
   - As \( b \to \infty \), \( b^{1-p} \to 0 \) if \( p > 1 \).
   - So, the integral becomes:
     \[
     \frac{0}{1-p} - \frac{1}{1-p} = \frac{1}{p-1}.
     \]
   - Therefore, the integral **converges** to \( \frac{1}{p-1} \) when \( p > 1 \).

### Part (b): \( p < 1 \)

We now evaluate the same integral, but with \( p < 1 \):

\[
\int_1^\infty \frac{1}{x^p} \, dx = \lim_{b \to \infty} \left( \frac{b^{1-p}}{1-p} - \frac{1^{1-p}}{1-p} \right).
\]

- As \( b \to \infty \), \( b^{1-p} \to \infty \) because \( 1 - p > 0 \) for \( p < 1 \).
- This causes the integral to diverge, as \( \lim_{b \to \infty} b^{1-p} = \infty \).

Thus, the integral **diverges** for \( p < 1 \).

### Part (c): \( p = 1 \)

For this case, the integral becomes:

\[
\int_1^\infty \frac{1}{x} \, dx.
\]

- The antiderivative of \( \frac{1}{x} \) is \( \ln x \), so:
  \[
  \int_1^\infty \frac{1}{x} \, dx = \lim_{b \to \infty} \ln b - \ln 1 = \lim_{b \to \infty} \ln b = \infty.
  \]

Thus, the integral **diverges** for \( p = 1 \).

### Summary:

- For \( p > 1 \), the integral converges to \( \frac{1}{p-1} \).
- For \( p \leq 1 \), the integral diverges.

Would you like any additional details or have further questions? 

Here are some related questions you might explore:
1. What happens if we change the lower limit of the integral to something other than 1?
2. How does the convergence of the p-integral relate to improper integrals?
3. Can you prove the result using comparison tests or other methods for improper integrals?
4. What is the physical or geometrical interpretation of the p-integral?
5. How does this integral behave if we integrate over a finite interval, say from \( 1 \) to some \( a > 1 \)?

**Tip**: Always check whether a definite integral is improper, as it helps in evaluating the convergence or divergence.

In class, we introduced the p-integral which will often appear in this course. In particular, we said that ∫₁^∞ 1/x^p dx = { 1/(p-1) if p > 1 ∞ if p ≤ 1 }. In the following parts, you will prove that this is true by considering three possible cases of the constant p: a) p > 1, b) p < 1, c) p = 1.

This solution covers the evaluation of the p-integral ∫₁^∞ 1/x^p dx for three cases of p (p > 1, p < 1, and p = 1). The solution explains the convergence for p > 1 and divergence for p ≤ 1, using antiderivatives and logarithms. Learn how to test the convergence of improper integrals in undergraduate calculus.

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