Answer
$$
\int_{a}^{\infty} \frac{1}{x^{2}+1} d x
$$
$$ a\approx 1000 $$
Work Step by Step
$$
\int_{a}^{\infty} \frac{1}{x^{2}+1} d x
$$
Observe that the given integral is improper
$$
\begin{split}
\int_{a}^{\infty} \frac{1}{x^{2}+1} d x & = \lim _{t \rightarrow \infty} \int_{a}^{t} \frac{1}{x^{2}+1} d x
\\
& =\lim _{t \rightarrow \infty}\left[\tan ^{-1} x\right]_{a}^{t}
\\
& =\lim _{t \rightarrow \infty}\left(\tan ^{-1} t-\tan ^{-1} a\right) \\
&=\frac{\pi}{2}-\tan ^{-1} a
\end{split}
$$
since
$$
\int_{a}^{\infty} \frac{1}{x^{2}+1} d x \lt 0.001
$$
So,
$$
\frac{\pi}{2}-\tan ^{-1} a \lt 0.001
$$
$\Rightarrow $
$$
\tan ^{-1} a\gt \frac{\pi}{2} - 0.001
$$
$\Rightarrow $
$$
a\gt \tan ( \frac{\pi}{2} - 0.001) \approx 1000
$$
