Let \(I_{1}=\int_{0}^{1} \frac{\operatorname{cosec} x}{x} d x\) and \(I_{2}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{m} x}{x^{n}} d x\) then :

Explanation:

Evaluation of Integrals \(I_1\) and \(I_2\)

We are given two integrals:

\(I_{1}=\int_{0}^{1} \frac{\operatorname{cosec} x}{x} dx\)

\(I_{2}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{m} x}{x^{n}} dx\)

We need to determine the conditions under which these integrals converge or diverge.

Analysis of \(I_1\):

The integral \(I_1\) is given by:

\(I_{1}=\int_{0}^{1} \frac{\operatorname{cosec} x}{x} dx\)

To analyze the convergence of this integral, we need to consider the behavior of the integrand near the limits of integration, particularly near \(x = 0\).

As \(x \to 0\), \(\operatorname{cosec} x = \frac{1}{\sin x}\) behaves like \(\frac{1}{x}\). Therefore, near \(x = 0\), the integrand \(\frac{\operatorname{cosec} x}{x}\) behaves like \(\frac{1}{x^2}\).

The integral \(\int_{0}^{\epsilon} \frac{1}{x^2} dx\) diverges as \(\epsilon \to 0\). Thus, \(I_1\) diverges.

Analysis of \(I_2\):

The integral \(I_2\) is given by:

\(I_{2}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{m} x}{x^{n}} dx\)

To determine the conditions for convergence of \(I_2\), we consider the behavior of the integrand near \(x = 0\) and \(x = \frac{\pi}{2}\).

Near \(x = \frac{\pi}{2}\), \(\sin x\) is close to 1, so the integrand does not pose any convergence issues. The critical part to analyze is the behavior near \(x = 0\).

As \(x \to 0\), \(\sin x \approx x\). Therefore, near \(x = 0\), the integrand \(\frac{\sin ^{m} x}{x^{n}}\) behaves like \(\frac{x^m}{x^n} = x^{m-n}\).

The integral \(\int_{0}^{\epsilon} x^{m-n} dx\) converges if and only if \(m-n > -1\), which simplifies to \(n < m + 1\).

Therefore, \(I_2\) converges if \(n < m + 1\).

Conclusion:

Based on the above analysis, we can conclude that:

\(I_1\) diverges and \(I_2\) converges if \(n < m + 1\).

Therefore, the correct option is:

Option 4: \(I_1\) diverges and \(I_2\) converges if \(n < m + 1\).

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: \(I_1\) converges and \(I_2\) converges if \(n \geq m + 1\).

This option is incorrect because \(I_1\) diverges as shown in the analysis. \(I_2\) converges when \(n < m + 1\), not \(n \geq m + 1\).

Option 2: \(I_1\) converges and \(I_2\) converges if \(n, m \in \mathbb{N}\).

This option is incorrect because \(I_1\) diverges regardless of the values of \(n\) and \(m\).

Option 3: \(I_1\) diverges and \(I_2\) diverges if \(n < m 1\).

This option is incorrect because \(I_2\) converges if \(n < m + 1\), which is the opposite condition to that stated.

Understanding the convergence properties of these integrals helps in correctly identifying the conditions under which they converge or diverge. This is crucial for solving problems in advanced calculus and mathematical analysis.