Math Problem Statement

onsider the following two integrals. I1=∫1−1dxx2I2=∫10cosxx−−√dx

Solution

The integrals given are:

  1. I1=11dxx2I_1 = \int_{-1}^{1} \frac{dx}{x^2}
  2. I2=01cosxxdxI_2 = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \, dx

Let's analyze each integral for convergence or divergence.

Integral I1=11dxx2I_1 = \int_{-1}^{1} \frac{dx}{x^2}

The integrand 1x2\frac{1}{x^2} is undefined at x=0x = 0, making it an improper integral. We can split this integral as follows:

I1=10dxx2+01dxx2I_1 = \int_{-1}^{0} \frac{dx}{x^2} + \int_{0}^{1} \frac{dx}{x^2}

Evaluating 01dxx2\int_{0}^{1} \frac{dx}{x^2}

This part of the integral is:

01dxx2=limϵ0+ϵ1x2dx\int_{0}^{1} \frac{dx}{x^2} = \lim_{\epsilon \to 0^{+}} \int_{\epsilon}^{1} x^{-2} \, dx

Calculating the integral:

=limϵ0+[1x]ϵ1=limϵ0+(1+1ϵ)= \lim_{\epsilon \to 0^{+}} \left[ -\frac{1}{x} \right]_{\epsilon}^{1} = \lim_{\epsilon \to 0^{+}} \left( -1 + \frac{1}{\epsilon} \right)

As ϵ0+\epsilon \to 0^{+}, 1ϵ\frac{1}{\epsilon} \to \infty, so the integral diverges.

Thus, I1I_1 diverges due to the behavior at x=0x = 0.

Integral I2=01cosxxdxI_2 = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \, dx

This integral is also improper because x\sqrt{x} approaches zero as x0x \to 0, creating a singularity.

Rewrite I2I_2 as:

I2=limϵ0+ϵ1cosxxdxI_2 = \lim_{\epsilon \to 0^{+}} \int_{\epsilon}^{1} \frac{\cos x}{\sqrt{x}} \, dx

Since cosx\cos x is bounded and x\sqrt{x} in the denominator implies that 1x\frac{1}{\sqrt{x}} has a singularity at x=0x = 0, we can approximate the behavior near x=0x = 0 by bounding cosx\cos x.

Convergence of 01dxx\int_{0}^{1} \frac{dx}{\sqrt{x}}

Consider:

ϵ1dxx=[2x]ϵ1=2(1ϵ)\int_{\epsilon}^{1} \frac{dx}{\sqrt{x}} = \left[ 2 \sqrt{x} \right]_{\epsilon}^{1} = 2(1 - \sqrt{\epsilon})

As ϵ0\epsilon \to 0, this expression approaches 22, so the integral converges.

Thus, I2I_2 converges.

Summary

  • I1I_1 diverges due to a singularity at x=0x = 0.
  • I2I_2 converges despite the singularity at x=0x = 0.

Would you like a more detailed breakdown of any steps?

  1. How do improper integrals differ in convergence based on singularity types?
  2. Can integrals converge even with infinite discontinuities within the integration range?
  3. What does it mean for an integral to be conditionally convergent?
  4. How does the limit process help in evaluating improper integrals?
  5. Are there specific functions that often lead to divergent integrals?

Tip: When dealing with improper integrals, always identify any points of discontinuity and assess the behavior of the integrand around those points.

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Math Problem Analysis

Mathematical Concepts

Calculus
Improper Integrals
Convergence and Divergence

Formulas

I1 = ∫_{-1}^{1} (1/x^2) dx
I2 = ∫_{0}^{1} (cos(x)/√x) dx

Theorems

Divergence and Convergence of Improper Integrals

Suitable Grade Level

Undergraduate Calculus