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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.

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  • $\begingroup$ Hint: Use the uniform convergence of the integral. $\endgroup$
    – Feng
    Sep 15 '19 at 15:05
  • $\begingroup$ 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. $\endgroup$
    – neelkanth
    Sep 15 '19 at 15:07
  • $\begingroup$ @PeterForeman dt........................ $\endgroup$
    – neelkanth
    Sep 15 '19 at 15:07
  • $\begingroup$ if possible please give solution i details . Thanks... $\endgroup$
    – neelkanth
    Sep 15 '19 at 15:09
  • $\begingroup$ @PeterForeman i edited the question... $\endgroup$
    – neelkanth
    Sep 15 '19 at 15:12
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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, $f(x)\to f(y)$.

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

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