Math Problem Statement

integrate 1/x^3 with bounds infiniy,0

Solution

To solve the integral

01x3dx,\int_{0}^{\infty} \frac{1}{x^3} \, dx,

we can rewrite it as:

0x3dx.\int_{0}^{\infty} x^{-3} \, dx.

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

Step 1: Split the Integral

We can rewrite the integral as a limit:

0x3dx=lima0+a1x3dx+limb1bx3dx.\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: lima0+a1x3dx\lim_{a \to 0^+} \int_{a}^{1} x^{-3} \, dx

    The antiderivative of x3x^{-3} is:

    x3dx=x22=12x2.\int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}.

    Applying the limits:

    lima0+[12x2]a1=lima0+(12+12a2).\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 a0+a \to 0^+, 12a2\frac{1}{2a^2} \to \infty. Thus, the first part diverges to infinity.

  2. Second Part: limb1bx3dx\lim_{b \to \infty} \int_{1}^{b} x^{-3} \, dx

    Using the same antiderivative:

    limb[12x2]1b=limb(12b2+12).\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 bb \to \infty, 12b20-\frac{1}{2b^2} \to 0, so this part is:

    12.\frac{1}{2}.

Conclusion

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

01x3dx\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 01xndx\int_{0}^{\infty} \frac{1}{x^n} \, dx? What values of nn 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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Improper Integrals
Limit of Integrals

Formulas

Integral of x^{-n}

Theorems

Convergence of Improper Integrals

Suitable Grade Level

Advanced Undergraduate