Can I say that $\left\vert\int_0^\infty \cos x \; dx \right\vert\leq 1$? We know that $\int_0^\infty \cos x \; dx$ is undefined. This seems like a very simple question, but I just want to confirm it is okay to do so:

Can I say that $$\bigg|\int_0^\infty \cos x \; dx \;\bigg  |\leq 1 \,\textrm{?}$$

Since it is equal to 
$$\bigg[\sin x\bigg]^\infty_0 = \lim_{x \rightarrow \infty} \sin x$$
I have genuinely not seen this done before, it seems like we can say it for sure. Since the limit does not exist but it is not "unbounded"?
 A: You are almost correct. We have
$$ \int_{0}^{M}\cos x\,dx = \sin M \tag{1}$$
but $\lim_{M\to +\infty}\sin M$ does not exist. In english terminology every non-convergent sequence is usually called "divergent", and this case brilliantly explains why I prefer the italian/french terminology, where we have convergent, divergent and non-convergent sequences: not every divergent sequence (in the english terminology) is actually unbounded, so it is kind of awkward to call them all "divergent".
There also is a way to extend the concept of convergence and to assign to such integral an actual value. For every $\lambda>0$ the integral $\int_{0}^{+\infty}\cos(x)\,e^{-\lambda x}\,dx$ does exist, and it equals $\frac{\lambda}{\lambda^2+1}$. So:
$$\text{ER}\int_{0}^{+\infty}\cos(x)\,dx = \lim_{\lambda\to 0^+}\int_{0}^{+\infty}\cos(x)e^{-\lambda x}\,dx = 0 \tag{2}$$
where $\text{ER}$ stands for exponential regularization.
A: No, this is not true. The integral $\int_0^{\infty} \cos x \,dx$ does not converge: By definition it is equal to the limit $\lim_{K \to \infty} \int_0^k \cos x \,dx = \lim_{K \to \infty} \sin K$ if that limit exists (it does not) and it is undefined otherwise. So its absolute value does not exist, and hence that expression cannot be compared with $\leq$. The best one can do (without introducing new ideas) is say that
$$\left\vert \int_0^K \cos x \,dx \right\vert \leq 1 \qquad \textrm{for all $K \in \Bbb R$.}$$
A: What you've said is "incorrect" as written.  The closest thing to what you're trying to say is
$$
\limsup_{t \to \infty}\bigg|\int_0^t \cos x \; dx \;\bigg  | = 1
$$
A: $\int_0^\infty \cos x dx$ is undefined, meaning that it has no meaning as a mathematical expression.  So, $\left|\int_0^\infty \cos x dx\right|$ doesn't make sense, since you can't take the absolute value of something that doesn't exist.  On the other hand, you could reason about $$\lim_{t \to \infty} \left|\int_0^t\cos x \ dx\right|$$
