To solve the integral

\[
\int_{0}^{\infty} \frac{1}{x^3} \, dx,
\]

we can rewrite it as:

\[
\int_{0}^{\infty} x^{-3} \, dx.
\]

This is an improper integral due to the lower limit \( x = 0 \) and the upper limit \( x = \infty \). We'll need to consider both limits separately.

### Step 1: Split the Integral
We can rewrite the integral as a limit:

\[
\int_{0}^{\infty} x^{-3} \, dx = \lim_{a \to 0^+} \int_{a}^{1} x^{-3} \, dx + \lim_{b \to \infty} \int_{1}^{b} x^{-3} \, dx.
\]

### Step 2: Evaluate Each Part

1. **First Part: \( \lim_{a \to 0^+} \int_{a}^{1} x^{-3} \, dx \)**

   The antiderivative of \( x^{-3} \) is:

   \[
   \int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}.
   \]

   Applying the limits:

   \[
   \lim_{a \to 0^+} \left[-\frac{1}{2x^2}\right]_{a}^{1} = \lim_{a \to 0^+} \left(-\frac{1}{2} + \frac{1}{2a^2}\right).
   \]

   As \( a \to 0^+ \), \( \frac{1}{2a^2} \to \infty \). Thus, the first part diverges to infinity.

2. **Second Part: \( \lim_{b \to \infty} \int_{1}^{b} x^{-3} \, dx \)**

   Using the same antiderivative:

   \[
   \lim_{b \to \infty} \left[-\frac{1}{2x^2}\right]_{1}^{b} = \lim_{b \to \infty} \left(-\frac{1}{2b^2} + \frac{1}{2}\right).
   \]

   As \( b \to \infty \), \( -\frac{1}{2b^2} \to 0 \), so this part is:

   \[
   \frac{1}{2}.
   \]

### Conclusion

Since the first part diverges to infinity, the overall integral diverges. Therefore, the integral

\[
\int_{0}^{\infty} \frac{1}{x^3} \, dx
\]

**does not converge**.

Would you like to go into more detail or have any specific questions?

Here are some related questions to consider:

1. What does it mean for an integral to diverge, and why is this significant?
2. How does the choice of bounds affect whether an integral converges or diverges?
3. What are some common techniques to determine the convergence of improper integrals?
4. Can we find similar results with other power functions, like \( \int_{0}^{\infty} \frac{1}{x^n} \, dx \)? What values of \( n \) result in convergence or divergence?
5. How can we use comparison tests to evaluate the convergence of improper integrals?

**Tip:** When evaluating improper integrals, always check both the behavior near zero and at infinity separately to understand the overall behavior of the integral.

Learn how to evaluate the improper integral of 1/x^3 from 0 to infinity. Understand why the integral diverges and explore the concept of improper integrals and their limits. Suitable for advanced undergraduate students and anyone interested in calculus.

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