Question

Given that: \[ If M = \int_{0}^{\infty} \frac{\log t}{1 + t^3} \, dt, and \quad N = \int_{-\infty}^{\infty} \frac{e^{2t} t}{1 + e^{3t}} \, dt. \] Then, the relation between \( M \) and \( N \) is:

• A. \( N = 2M \)
• B. \( N = M \)
• C. \( N = 3M \)
• D. \( N = -M \)

Hint:

For integrals involving logarithms and rational expressions, substitutions such as \( t = \frac{1}{u} \) or symmetry properties can simplify evaluation.

Answer 1

The Correct Option is D

Step 1: Evaluating \( M \)
Consider the given integral: \[ M = \int_{0}^{\infty} \frac{\log t}{1+t^3} dt. \] We use the transformation \( t = \frac{1}{u} \), so \( dt = -\frac{du}{u^2} \). Under this transformation: \[ M = \int_{0}^{\infty} \frac{\log \left( \frac{1}{u} \right)}{1+\left( \frac{1}{u} \right)^3} \left(-\frac{du}{u^2}\right). \] Rewriting \( \log \frac{1}{u} \) as \( -\log u \), we obtain: \[ M = -\int_{0}^{\infty} \frac{\log u}{1+u^3} du. \] Thus, we conclude: \[ M = -M \Rightarrow M = 0. \] Step 2: Evaluating \( N \)
Similarly, considering the symmetry properties of the integral: \[ N = \int_{-\infty}^{\infty} \frac{e^{2t} t}{1+e^{3t}} dt. \] Using the substitution \( t = -u \), it can be shown that: \[ N = -M. \] Step 3: Conclusion
Thus, we have: \[ \boxed{N = -M}. \]