When the equivalence rule for improper integrals fails

(a) Convergence of $\int_1^{+\infty}\frac{\cos t}{t^\alpha}\,dt$ for $\alpha>0$. Integrate by parts on $[1,X]$:
$$\int_1^X\frac{\cos t}{t^\alpha}\,dt=\left[\frac{\sin t}{t^\alpha}\right]_1^X+\alpha\int_1^X\frac{\sin t}{t^{\alpha+1}}\,dt=\frac{\sin X}{X^\alpha}-\sin 1+\alpha\int_1^X\frac{\sin t}{t^{\alpha+1}}\,dt.$$
Since $\left|\frac{\sin t}{t^{\alpha+1}}\right|\le\frac{1}{t^{\alpha+1}}$ and $\alpha+1>1$, the integral on the right converges absolutely. The boundary term $\frac{\sin X}{X^\alpha}\to0$. Hence $\int_1^{+\infty}\frac{\cos t}{t^\alpha}\,dt$ converges.

(b) Divergence of $\int_1^{+\infty}\frac{\cos^2 t}{t}\,dt$. Linearization:
$$\frac{\cos^2 t}{t}=\frac{1}{2t}+\frac{\cos 2t}{2t}.$$
By (a) with $\alpha=1$, $\int_1^{+\infty}\frac{\cos 2t}{2t}\,dt$ converges. But $\int_1^{+\infty}\frac{1}{2t}\,dt$ diverges (Riemann with $\alpha=1$). A sum of a convergent and a divergent integral diverges, so $\int_1^{+\infty}\frac{\cos^2 t}{t}\,dt$ diverges.

(2a) Non-absolute convergence. Since $|\cos t|\ge\cos^2 t$,
$$\frac{|\cos t|}{t}\ge\frac{\cos^2 t}{t}.$$
By comparison with (1b), $\int_1^{+\infty}\frac{|\cos t|}{t}\,dt$ diverges. Hence $\int_1^{+\infty}\frac{\cos t}{t}\,dt$ converges (by (a) with $\alpha=1$) but not absolutely: it is conditionally convergent.

(2b) An equivalence that misleads. Let $f(t)=\frac{\cos t}{\sqrt t}$ and $g(t)=f(t)+\frac{\cos^2 t}{t}$. Then
$$\frac{g(t)}{f(t)}=1+\frac{\cos t}{\sqrt t}\xrightarrow[t\to+\infty]{}1,$$
so $f\sim g$ at $+\infty$. However, by (a) with $\alpha=1/2$, $\int_1^{+\infty}f$ converges. But $\int_1^{+\infty}g=\int_1^{+\infty}f+\int_1^{+\infty}\frac{\cos^2 t}{t}$ diverges (sum of a convergent and a divergent integral by (1b)). So the equivalence of integrands does not preserve convergence.

(2c) Why the equivalence rule fails. The rule "$f\sim g\Rightarrow\int f$ and $\int g$ have the same nature" requires that $f$ and $g$ have a constant sign near $+\infty$. Here, the oscillations of $\cos t$ change sign infinitely often, so the comparison fails: even though $f(t)/g(t)\to1$, the small "correction" $\frac{\cos^2 t}{t}$ accumulates a divergent contribution, since $\cos^2 t\ge0$ is not sign-changing.