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: Oct 1, 2024
- \( I_1 \) 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.
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The Correct Option is C
Solution and Explanation
The correct option is (C): Both \( I_1 \) and \( I_2 \) converge
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