Math Problem Statement

same goes for this

Consider the following improper integrals: I₁ = ∫(0 to 1) dx/((x - 1/2)^m) + ∫(2 to ∞) dx/(x-1)^(2n+m) and I₂ = ∫(1 to ∞) dx/(x-1)^(2n+1), where m and n are real numbers. Which of the following statements is/are correct?

Solution

We are given two improper integrals, I1I_1 and I2I_2, and need to analyze the convergence based on the real numbers mm and nn. The integrals are:

I1=01dx(x12)m+2dx(x1)2n+mI_1 = \int_0^1 \frac{dx}{\left( x - \frac{1}{2} \right)^m} + \int_2^\infty \frac{dx}{(x-1)^{2n+m}} and I2=1dx(x1)2n+1.I_2 = \int_1^\infty \frac{dx}{(x-1)^{2n+1}}.

Step 1: Analyze I1I_1

The first part of I1I_1 involves the integral 01dx(x12)m.\int_0^1 \frac{dx}{\left( x - \frac{1}{2} \right)^m}. We can see that the function behaves near x=12x = \frac{1}{2}. The nature of the singularity will depend on the value of mm, as for large enough mm, this integral may diverge at x=12x = \frac{1}{2}.

The second part of I1I_1 involves the integral 2dx(x1)2n+m.\int_2^\infty \frac{dx}{(x-1)^{2n+m}}. For this to converge, the exponent 2n+m2n + m must be greater than 1. Thus, we require: 2n+m>1orn>1m2.2n + m > 1 \quad \text{or} \quad n > \frac{1 - m}{2}.

Step 2: Analyze I2I_2

For the second integral I2I_2, we have I2=1dx(x1)2n+1.I_2 = \int_1^\infty \frac{dx}{(x-1)^{2n+1}}. For this to converge at infinity, we require 2n+1>12n+1 > 1, i.e., n>0n > 0.

Step 3: Evaluate the Statements

Let’s now evaluate each of the given statements:

  • (A) I2I_2 is not convergent for any value of nn: This is false since I2I_2 converges for n>0n > 0.

  • (B) I1I_1 is convergent when m>1m > 1 and n>1m2n > \frac{1 - m}{2}: This is true, since for m>1m > 1, the first integral in I1I_1 converges, and for n>1m2n > \frac{1 - m}{2}, the second integral converges.

  • (C) I2I_2 is convergent for n<0n < 0: This is false because for I2I_2 to converge, n>0n > 0 is required.

  • (D) I1I_1 is convergent when m<1m < 1 and n>1m2n > \frac{1 - m}{2}: This is false, because the first part of I1I_1 diverges for m<1m < 1, and the second part converges only under the condition n>1m2n > \frac{1 - m}{2}.

Conclusion:

The correct answer is (B).


Would you like more details on any of these steps?

Here are 5 related questions:

  1. How do you determine the convergence of improper integrals?
  2. What are some methods to handle singularities in integrals?
  3. How does changing the bounds of integration affect convergence?
  4. What is the difference between pointwise convergence and uniform convergence?
  5. Can convergence criteria be extended to complex-valued functions?

Tip: Always check the behavior of the integrand near singularities or infinity to determine convergence conditions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Improper Integrals
Convergence of Integrals
Singularities in Integrals
Exponents in Integration

Formulas

I₁ = ∫(0 to 1) dx/((x - 1/2)^m) + ∫(2 to ∞) dx/(x-1)^(2n+m)
I₂ = ∫(1 to ∞) dx/(x-1)^(2n+1)

Theorems

Convergence criterion for improper integrals
Divergence of improper integrals at singularities

Suitable Grade Level

University Level (Advanced Calculus)