a) Is $$F:\mathbb{R} \rightarrow \mathbb{R}: x \mapsto \int_0^\infty \frac{\arctan(xt)}{1+t^2}dt$$ differentiable in $x \not= 0$?

b) What is $\lim_{x\rightarrow \infty} F(x)$?

c) What is $\lim_{x\rightarrow 0} \frac{F(x)}{x}$?

I wanted to prove that the function is differentiable by the Fundamental Theorem of the calculus. Thus when $$\frac{\arctan(xt)}{1+t^2}$$ is continuous but this is everywhere? So I do not know if I am allowed to do this.

Further I wanted to calculate $\lim_{x\rightarrow \infty} F(x)$. When we bring the limit in the integral we get $\arctan(\frac{\pi}{2}t)$, but I do not know which theorem I can use to justify that I can bring the limit inside the integral.

The last part of the question I need to calculate $\lim_{x\rightarrow 0} \frac{F(x)}{x}$. I wanted to use the rule of l'hopital but then I get $$\frac{1}{(1+(xt)^2)(1+t^2)}$$ but then I do not know how to go further. I tried to calculate this further but then I could not solve the integral properly. I did think that it may have something to do with the fact that $t\rightarrow \infty$ but that is just a thought.

Can anybody help me? Thanks in advance.

  • $\begingroup$ You can differentiate $F$ by differentiating under the integral sign. Just check that the conditions are satisfied for it. $\endgroup$
    – Gary
    Aug 14 '21 at 8:13
  • $\begingroup$ Note that $\lim_{y\to \infty} \arctan (y)=\frac \pi 2$. So for every fixed value of $t$ other than $0$, as $x\to \infty$, the $\arctan$ goes to $\pi 2$. $\endgroup$
    – Alan
    Aug 14 '21 at 8:16
  • $\begingroup$ @Gary When I differentiate under the integral I do not prove that it is differentiable or did you mean something else? That is why I wanted the use the fundamental theorem of calculus. $\endgroup$ Aug 14 '21 at 8:28
  • $\begingroup$ See brilliant.org/wiki/differentiate-through-the-integral $\endgroup$
    – Gary
    Aug 14 '21 at 8:41

Note that $\lim_{y\to \infty} \arctan (y)=\frac \pi 2$. So for every fixed value of $t$ other than $0$, as $x\to \infty$, the $\arctan$ goes to $\frac \pi 2$. You can then integrate the function inside to get $\frac \pi 2 \arctan t$. then plugging in your limits gets you to $(\frac \pi 2)^2$ (We can ignore the value at $t=0$ because integrals don't care about changing the values on finite points, or countable, or measure 0)

  • $\begingroup$ What do you use to justify bringing the limit inside the integral? $\endgroup$ Aug 14 '21 at 8:32
  • 1
    $\begingroup$ Everything's continuous, so limits interchange. $\endgroup$
    – Alan
    Aug 14 '21 at 8:34

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