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

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.

• You can differentiate $F$ by differentiating under the integral sign. Just check that the conditions are satisfied for it.
– Gary
Aug 14 '21 at 8:13
• 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 $\pi 2$.
– Alan
Aug 14 '21 at 8:16
• @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. Aug 14 '21 at 8:28
• – 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)