On the continuity of the function $f(x)=\int_1^\infty \frac{\cos t}{x^2+t^2}dt$.

Let $F(x)=\int_1^\infty \frac{\cos t}{x^2+t^2}dt$. Then which of the following are correct?

1. $f$ is bounded on $\mathbb R$

2. $f$ is continuous on $\mathbb R$

3. $f$ is not defined everywhere on $\mathbb R$

4. $f$ is not continuous on $\mathbb R$

Here is Let $f(x)=\int_1^\infty \frac{\cos t}{x^2+t^2}dt$. Then which of the following are correct? one of the solutions. I am unable to get the continuity of $F$ in the linked solution. I tried as follows.

We are given the function:$$F(x) = \int_1^\infty \frac{\cos(t)}{x^2 + t^2} \, dt,$$and we need to show that $F(x)$ is continuous on $ \mathbb{R} $.

Step 1: Pointwise Convergence

Let $x_0 \in \mathbb{R}$. We need to show that:$$\lim_{x \to x_0} F(x) = F(x_0).$$This means we aim to demonstrate:$$\lim_{x \to x_0} \int_1^\infty \frac{\cos(t)}{x^2 + t^2} \, dt = \int_1^\infty \frac{\cos(t)}{x_0^2 + t^2} \, dt.$$

Step 2: Dominating Function

Next, we seek a dominating function that does not depend on $x $. Observe that:$$\left| \frac{\cos(t)}{x^2 + t^2} \right| \leq \frac{1}{t^2}.$$This holds for all $ x \in \mathbb{R} $ and $ t \geq 1 $. The function $\frac{1}{t^2} $ is integrable over $ [1, \infty) $, since:$$\int_1^\infty \frac{1}{t^2} \, dt = 1.$$Thus, we can take $g(t) = \frac{1}{t^2} $ as the dominating function.

Step 3: Apply the Dominated Convergence Theorem

By the Dominated Convergence Theorem (DCT), we can interchange the limit and the integral. Hence, we have:$$\lim_{x \to x_0} F(x) = \lim_{x \to x_0} \int_1^\infty \frac{\cos(t)}{x^2 + t^2} \, dt = \int_1^\infty \lim_{x \to x_0} \frac{\cos(t)}{x^2 + t^2} \, dt = F(x_0).$$

Thus $F$ is continuous.Please verify that am I correct? Thank you.