<s>[INST] Evaluate $\int_{0}^{1000} \frac{e^{-10x}\sin x}{x} \text{d}x$ to within $\pm 10^{-5}.$ Evaluate $$\displaystyle \int_{0}^{1000} \frac{e^{-10x}\sin x}{x}  \text{d}x$$ to within  $\pm 10^{-5}$. 
 [/INST] Thinking more about this, one don't need to approximate the integral by sending the upper limit to $\infty$ nor know how to evaluate the integral over $[0,\infty)$.
For $x > 0$, $|\frac{\sin x}{x}| < 1$ and $e^{-x}$ drops off to $0$ very quickly.
If we cutoff the integral at $1$ instead of $1000$. The error:
$$\left|\int_{1}^{1000} e^{-10x} \frac{\sin x}{x} dx\right| < \int_{1}^{1000} e^{-10x} dx < \frac{e^{-10}}{10} \sim 4.54 \times 10^{-6}$$
itself is small enough. For $x \in [0,1]$, one can Taylor expand $\frac{\sin x}{x}$ as:
$$\frac{\sin x}{x} = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k+1)!}$$
If one chop off the term for $k \ge 4$, we have:
$$\left|\frac{\sin x}{x} - (1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!})\right| < 1/9! + 1/11! + \cdots \sim 2.781\times 10^{-6}$$
This implies 
$$\begin{align}&\left|\int_{0}^{1} e^{-10x}\frac{\sin x}{x} dx - \int_{0}^{1}e^{-10x}(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!}) dx \right|\\ <& 2.781\times 10^{-6} \int_{0}^{1} e^{-10x} dx\\ \sim & 2.781 \times 10^{-7}\end{align}$$
Within an accuracy of $10^{-5}$, we have:
$$\int_0^{1000} e^{-10x} \frac{\sin x}{x} dx \sim \int_0^1 e^{-10x} (1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!}) dx \\= \frac{20930417}{210000000}-\frac{50976061}{630000000} e^{-10} \sim 0.099665$$ 
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