Discussion
The value of the following expression
\[ \lim_{x \to \infty} \int_{0}^{x} e^{-t} |\sin t| \, dt \]
is _____.
Answer: A
The limit is \(\frac{e^{\pi} + 1}{2(e^{\pi} - 1)}\).
\(|\sin t|\) has period \(\pi\). Splitting \([0, \infty)\) into intervals \([k\pi, (k+1)\pi]\) and putting \(t = k\pi + s\) pulls out a factor \(e^{-k\pi}\) from each piece:
\[ \begin{aligned} I &= \sum_{k=0}^{\infty} e^{-k\pi} \int_{0}^{\pi} e^{-s}\sin s\,ds \\ \int_{0}^{\pi} e^{-s}\sin s\,ds &= \frac{1 + e^{-\pi}}{2} \end{aligned} \]
The geometric series sums to \(\frac{1}{1 - e^{-\pi}}\), so
\[ \begin{aligned} I &= \frac{1 + e^{-\pi}}{2(1 - e^{-\pi})} \\ &= \frac{e^{\pi} + 1}{2(e^{\pi} - 1)} \approx 0.545 \end{aligned} \]
Two bounds rule out the rest without any working. Since \(|\sin t| \leq 1\), \(I \leq \int_{0}^{\infty} e^{-t}\,dt = 1\), which excludes \(\frac{2(e^{\pi} + 1)}{e^{\pi} - 1} \approx 2.18\) and \(\frac{2(e^{\pi} - 1)}{e^{\pi} + 1} \approx 1.83\). Since \(|\sin t| \geq \sin t\), \(I \geq \int_{0}^{\infty} e^{-t}\sin t\,dt = \frac{1}{2}\), which excludes \(\frac{e^{\pi} - 1}{2(e^{\pi} + 1)} \approx 0.46\).
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