Question

Consider the improper integrals
\[I_1 = \int_{1}^{\infty} \frac{t \sin t}{e^t} \, dt \]
and 
\[I_2 = \int_{1}^{\infty} \frac{1}{\sqrt{t}} \ln\left(1 + \frac{1}{t}\right) \, dt.\]
Then:

• A. \( I_1 \) converges, but \( I_2 \) does NOT converge.
• B. \( I_1 \) does NOT converge, but \( I_2 \) converges.
• C. Both \( I_1 \) and \( I_2 \) converge.
• D. Neither \( I_1 \) nor \( I_2 \) converges.

Answer 1

The Correct Option is C

1. Convergence of \( I_1 \): - As \( t \to \infty \), the term \( \frac{\sin t}{e^t} \) decays exponentially. - Multiplication by \( t \) does not change convergence because \( \frac{t}{e^t} \to 0 \) as \( t \to \infty \). - Thus, \( I_1 \) converges. 2. Convergence of \( I_2 \): - Near infinity, \( \ln\left(1 + \frac{1}{t}\right) \sim \frac{1}{t} \), and \( \frac{\ln\left(1 + \frac{1}{t}\right)}{\sqrt{t}} \sim \frac{1}{t^{3/2}} \). - Since \( \int_1^\infty t^{-3/2} \, dt \) converges, \( I_2 \) converges.