When evaluating whether an improper integral is convergent or divergent, I'm sometimes unsure whether I simplified enough to be sure. For example, given
$$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sec \theta \, \textrm{d}\theta = \lim_{b \to \frac{\pi}{2}^-} [\ln \left|\frac{1 + \sin b}{\cos b}\right|] + \ln |1 + \sqrt{2}| $$
Is this enough to claim the integral is divergent?
Just to be sure, I tried to simplify the limit further by raising $e$ to the power of the LHS and RHS, which I've never seen before, so that I can apply L'Hôpital's rule
$$ e^y = \lim_{b \to \frac{\pi}{2}^-} e^{\ln \left|\frac{1 + \sin b}{\cos b}\right|} = \lim_{b \to \frac{\pi}{2}^-} \left|\frac{1 + \sin b}{\cos b}\right| = \lim_{b \to \frac{\pi}{2}^-} \left|\frac{\cos b}{\sin b}\right| = 0 \\ y = \ln 0 $$
Is that overkill?
You don't need to evaluate an improper integral to figure out whether it converges or not. If the integrand behaves as $1/(x-a)^b$ at a singularity $x=a$, where $b \ge 1$, then you have a nonintegrable singularity which causes the integral to diverge. In your case, $\sec{\theta} = 1/\cos{\theta}$ which diverges at $\theta=\pi/2$. However, you should further know that, in the neighborhood of $\theta = \pi/2$, $1/\cos{\theta} \sim 1/(\pi/2 - \theta)$. Thus, the integral is indeed divergent. Further, because the integrand diverges as $1/(\pi/2 - \theta)$, the integral will diverge as $\log{(\pi/2 - \theta)}$ as you have seen.