Continuity of the function $f(x)=\int_{1}^{\infty}\frac{\cos t}{x^2+t^2}dt.$

Question

How to prove that function $f(x)=\int_{1}^{\infty}\frac{\cos t}{x^2+t^2}dt$ is continuous on $\mathbb{R}?$ By fundamental theorem of calculus I know that indefinite integral $\int_{1}^{x}\frac{\cos t}{x^2+t^2}dt$ is continuous but not having idea for this type of improper integral. Please help. Thanks.

Answer 1

The continuity of $f(x)=\int_{1}^{\infty}\frac{\cos(t)}{x^2+t^2}dt$ follows from the following inequality: for $x,y\in\mathbb{R}$ we have that $$|f(x)-f(y)|= \left|\int_{1}^{\infty}\frac{\cos(t)(y^2-x^2)}{(y^2+t^2)(x^2+t^2)}dt\right|\leq |x^2-y^2|\int_{1}^{\infty}\frac{dt}{t^4}=\frac{|x^2-y^2|}{3}.$$ Therefore, if $x\to y$ then $\frac{|x^2-y^2|}{3}\to 0$ and, by comparison, $f(x)\to f(y)$.

Answer 2

Let $x_n \to x_0$.

Then $$g_n(t)=\frac{\cos{t}}{x_n^2+t^2} \to \frac{\cos{t}}{x_0^2+t^2}$$

Also $|g_n(t)| \leq \frac{1}{t^2}\in L^1([1,+\infty))$

So from Dominated Convergence Theorem we have the desired conclusion.