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

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.

• Hint: Use the uniform convergence of the integral.
– Feng
Sep 15 '19 at 15:05
• i know the result that if a sequence of continuous function converges uniformly then limit function is also continuous. But did not got the exact point. Sep 15 '19 at 15:07
• @PeterForeman dt........................ Sep 15 '19 at 15:07
• if possible please give solution i details . Thanks... Sep 15 '19 at 15:09
• @PeterForeman i edited the question... Sep 15 '19 at 15:12

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, $$f(x)\to f(y)$$.

• I edited my answer. Sep 15 '19 at 15:19
• Yes ...I think this one is nice... Sep 15 '19 at 15:22
• now will i compare it to $x^2$ in last step? Sep 15 '19 at 15:25
• Just note that when $x\to y$ then $\frac{|x^2-y^2|}{3}\to 0$ and therefore, by comparison, $f(x)\to f(y)$. Sep 15 '19 at 15:27

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.