To evaluate the improper integral $\int_0^{15} \frac{1}{x^2} \, dx$, we first need to recognize that the function $\frac{1}{x^2}$ has a singularity at $x = 0$. Therefore, we will treat the integral as an improper integral by approaching the lower limit $0$ using a limit.

Step 1: Set up the improper integral as a limit

We express the improper integral as: $\int_0^{15} \frac{1}{x^2} \, dx = \lim_{\epsilon \to 0^+} \int_\epsilon^{15} \frac{1}{x^2} \, dx$

Step 2: Integrate $\frac{1}{x^2}$

The integral of $\frac{1}{x^2}$ is: $\int \frac{1}{x^2} \, dx = -\frac{1}{x}$

Step 3: Apply the limits of integration

We now apply the limits to the integral: $\int_\epsilon^{15} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_\epsilon^{15} = -\frac{1}{15} + \frac{1}{\epsilon}$

Step 4: Take the limit as $\epsilon \to 0^+$

As $\epsilon \to 0^+$, the term $\frac{1}{\epsilon}$ grows without bound, meaning the integral diverges.

Conclusion

The integral $\int_0^{15} \frac{1}{x^2} \, dx$ diverges because it involves a singularity at $x = 0$.

Let me know if you need further clarification or additional details! Here are some related questions you might find interesting:

1. How does the concept of improper integrals work for different types of singularities?
2. What would happen if the lower limit of integration were a nonzero constant?
3. How do we handle improper integrals involving infinity as a limit of integration?
4. How would the result change if the integrand were $\frac{1}{x^p}$ for some $p$?
5. How do we evaluate improper integrals using comparison tests?

Tip: When dealing with improper integrals, always check the behavior of the integrand at the boundaries of the integration domain to determine if the integral converges or diverges.