The given problem asks to determine whether the integral

$\int_1^\infty 2x^{-3} \, dx$

is convergent or divergent, and to evaluate it if it converges.

Step 1: Check for Convergence or Divergence

This is an improper integral because it has an infinite upper limit. To determine if it converges, we need to evaluate the integral.

Step 2: Set up the integral for evaluation

First, express the integral:

$I = \int_1^\infty 2x^{-3} \, dx$

The integral can be rewritten as:

$I = 2 \int_1^\infty x^{-3} \, dx$

Step 3: Evaluate the integral

To integrate $x^{-3}$, recall that the integral of $x^n$ is:

$\int x^n \, dx = \frac{x^{n+1}}{n+1} \quad \text{for} \quad n \neq -1$

Here, $n = -3$, so we have:

$\int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$

Now, apply the limits of integration from $1$ to $\infty$:

$I = 2 \left[ -\frac{1}{2x^2} \right]_1^\infty$

Step 4: Apply the limits

First, evaluate the expression at $\infty$ and $1$:

• As $x \to \infty$, $\frac{1}{x^2} \to 0$, so:

$-\frac{1}{2x^2} \Big|_{x=\infty} = 0$

• At $x = 1$:

$-\frac{1}{2x^2} \Big|_{x=1} = -\frac{1}{2}$

Thus, the integral becomes:

$I = 2 \left( 0 - \left(-\frac{1}{2}\right) \right) = 2 \times \frac{1}{2} = 1$

Step 5: Conclusion

Since the integral evaluates to a finite number, the integral converges, and its value is:

$\boxed{1}$

Let me know if you'd like more details or have any questions!

Here are 5 related questions for further practice:

1. What is the convergence condition for the integral $\int_1^\infty x^{-p} \, dx$?
2. How do you test for convergence of improper integrals with infinite limits?
3. What happens if the exponent in the integrand were $-2$ instead of $-3$?
4. How do you compute improper integrals with a singularity at a finite point?
5. How does substitution affect the evaluation of improper integrals?

Tip: When dealing with improper integrals, always check for both infinite limits and discontinuities to decide if the integral converges.