My book has this problem:
$$\int_0^{10\pi} (\lfloor \sec^{-1}x\rfloor + \lfloor \cot^{-1}x\rfloor) \;dx = \;?$$
Now, the domain of $\sec^{-1}x$ is $({-\infty}, -1) \bigcup (1, \infty)$. How can you integrate $\sec^{-1}x$ in $(0, 10\pi)$ when $(0, 1)$ lies outside it's domain?
I checked Wolfram|Alpha. It takes $$\lfloor \sec^{-1}x\rfloor = 0 \;\;\forall\; x\in (0, 1) $$
Why? When $\sec^{-1}x$ is not defined for $(0, 1) $, why is it's Greatest Integer defined?
Wolfram Alpha gives $$10\pi - \sec(1) + \cot(1)$$ as the answer, while my book gives
$$\int_0^{10\pi} (\lfloor \sec^{-1}x\rfloor + \lfloor cot^{-1}x\rfloor) \;dx $$
$$ = \int_0^{\sec(1)} (\lfloor \sec^{-1}x\rfloor + \lfloor \cot^{-1}x\rfloor) \;dx + \int_{\sec(1)}^{10\pi} (\lfloor \sec^{-1}x\rfloor + \lfloor \cot^{-1}x\rfloor) \;dx$$
$$ = \int_0^{\sec(1)} (0 + 0) \;dx + \int_{\sec(1)}^{10\pi} (1+0) \;dx$$
$$ = 0 + [x]_{\sec(1)}^{10\pi}$$
$$ = 10\pi - \sec(1)$$