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We are given two improper integrals, \( I_1 \) and \( I_2 \), and need to analyze the convergence based on the real numbers \( m \) and \( n \). The integrals are:

\[
I_1 = \int_0^1 \frac{dx}{\left( x - \frac{1}{2} \right)^m} + \int_2^\infty \frac{dx}{(x-1)^{2n+m}}
\]
and
\[
I_2 = \int_1^\infty \frac{dx}{(x-1)^{2n+1}}.
\]

### Step 1: Analyze \( I_1 \)
The first part of \( I_1 \) involves the integral
\[
\int_0^1 \frac{dx}{\left( x - \frac{1}{2} \right)^m}.
\]
We can see that the function behaves near \( x = \frac{1}{2} \). The nature of the singularity will depend on the value of \( m \), as for large enough \( m \), this integral may diverge at \( x = \frac{1}{2} \).

The second part of \( I_1 \) involves the integral
\[
\int_2^\infty \frac{dx}{(x-1)^{2n+m}}.
\]
For this to converge, the exponent \( 2n + m \) must be greater than 1. Thus, we require:
\[
2n + m > 1 \quad \text{or} \quad n > \frac{1 - m}{2}.
\]

### Step 2: Analyze \( I_2 \)
For the second integral \( I_2 \), we have
\[
I_2 = \int_1^\infty \frac{dx}{(x-1)^{2n+1}}.
\]
For this to converge at infinity, we require \( 2n+1 > 1 \), i.e., \( n > 0 \).

### Step 3: Evaluate the Statements
Let’s now evaluate each of the given statements:

- **(A) \( I_2 \) is not convergent for any value of \( n \):** This is **false** since \( I_2 \) converges for \( n > 0 \).
  
- **(B) \( I_1 \) is convergent when \( m > 1 \) and \( n > \frac{1 - m}{2} \):** This is **true**, since for \( m > 1 \), the first integral in \( I_1 \) converges, and for \( n > \frac{1 - m}{2} \), the second integral converges.

- **(C) \( I_2 \) is convergent for \( n < 0 \):** This is **false** because for \( I_2 \) to converge, \( n > 0 \) is required.

- **(D) \( I_1 \) is convergent when \( m < 1 \) and \( n > \frac{1 - m}{2} \):** This is **false**, because the first part of \( I_1 \) diverges for \( m < 1 \), and the second part converges only under the condition \( n > \frac{1 - m}{2} \).

### Conclusion:
The correct answer is **(B)**.

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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.

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?

Analyze the convergence of the improper integrals I₁ and I₂, where I₁ involves two terms and I₂ involves one. Explore how the real numbers m and n affect convergence, including conditions for the integrals to diverge or converge. Ideal for advanced calculus students.

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