Let $f(x)=\int_1^\infty \frac{\cos t}{x^2+t^2}dt$. Then which of the following are correct?
$f$ is bounded on $\mathbb R$
$f$ is continuous on $\mathbb R$
$f$ is not defined everywhere on $\mathbb R$
$f$ is not continuous on $\mathbb R$
My try:
$|f(x)|=|\int_1^\infty \frac{\cos t}{x^2+t^2}dt|\leq \int_1^\infty |\frac{\cos t}{x^2+t^2}|dt\leq \int_1^\infty \frac{1}{x^2+t^2}dt=\frac{1}{x}\tan^{-1}(\frac{t}{x})|_1^\infty $. Let $g(x)= \frac{1}{x}\tan^{-1}(\frac{t}{x})|_1^\infty$ is not continuous at $0$. That doesn't mean that $f$ is not bounded.
For continuity, $|f(x)-f(y)|=|\int_1^\infty \frac{\cos t}{x^2+t^2}dt-\int_1^\infty \frac{\cos t}{y^2+t^2}dt|\leq \int_1^\infty |\frac{1}{x^2+t^2}-\frac{1}{y^2+t^2}|dt=\int_1^\infty |\frac{y^2-x^2}{(x^2+t^2)(y^2+t^2)}|dt $. I am not conclude from here.
For all $c > 1$ and $x \in \mathbb{R}$,
$$\left|\int_1^c \frac{\cos t}{x^2 + t^2} \, dt \right| \leqslant \int_1^c \frac{|\cos t|}{x^2 + t^2} \, dt \leqslant\int_1^c \frac{dt}{t^2} \leqslant1 - \frac{1}{c} < 1$$
This implies both that $f$ is bounded and the improper integral converges uniformly by the Weierstrass M-test.
With uniform convergence it is straightforward to prove continuity, since
$$|f(x) - f(y)| \leqslant \left|\int_c^\infty\frac{\cos t}{x^2 + t^2} \, dt \right|+ \left|\int_c^\infty\frac{\cos t}{y^2 + t^2} \, dt \right|\\+ \int_1^c\left|\frac{\cos t}{x^2 + t^2}- \frac{\cos t}{y^2 + t^2}\right| \, dt $$
For any $\epsilon > 0$ there exists $c$ such that the first two terms on the RHS are less than $\epsilon/3$ for all $x \in \mathbb{R}$. You should be able to handle the third term.