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:
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:
\[I_1 = \int_{1}^{\infty} \frac{t \sin t}{e^t} \, dt \]
and
\[I_2 = \int_{1}^{\infty} \frac{1}{\sqrt{t}} \ln\left(1 + \frac{1}{t}\right) \, dt.\]
Then:
Updated On: Jan 25, 2025
- \( I_1 \) converges, but \( I_2 \) does NOT converge.
- \( I_1 \) does NOT converge, but \( I_2 \) converges.
- Both \( I_1 \) and \( I_2 \) converge.
- Neither \( I_1 \) nor \( I_2 \) converges.
Show Solution
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The Correct Option is C
Solution and Explanation
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.
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