Let $F(t)=\int_{0}^{\infty}\frac{\arctan(tx)-\arctan(x)}{x}dx$ Prove $F$ is $C^1(0,\infty)$ and find $F'(t)$ 
Let $F(t)=\int_{0}^{\infty}\frac{\arctan(tx)-\arctan(x)}{x}dx$
Prove $F$ is $C^1(0,\infty)$ and find $F'(t)$

I've thought that maybe I should prove that $f(t,x)=\frac{\arctan(tx)-\arctan(x)}{x}$ is riemann integrable and therefore it's lebesgue integrable on $(0,\infty)$ and then use differentiation under integral (in the real analysis sense) and to show $\frac{\partial f}{\partial t}$ is bounded.
 A: Fix $t>0$, for $\Delta t$ such that $|\Delta t|<\frac{t}{2}$, there is
$\theta\in(0,1)$ such that
$$ \arctan((t+\Delta t)x)-\arctan(tx)=\frac{1}{1+(t+\theta\Delta t)^2x^2}\Delta t x. $$
So
\begin{eqnarray}
&&\frac{F(t+\Delta t)-F(t)}{\Delta t}\\
&=&\int_{0}^{\infty}\bigg[\frac{\arctan((t+\Delta t)x)-\arctan(tx)}{x}dx\\
&=&\int_{0}^{\infty}\frac{1}{1+(t+\theta\Delta t)^2x^2}dx.\\
\end{eqnarray}
Since
$$ |t+\theta\Delta t|\ge t-\theta |\Delta t|>\frac{t}{2} $$
we have
$$ \bigg|\frac{1}{1+(t+\theta\Delta t)^2x^2}\bigg|<\frac4{4+t^2x^2}. $$
Since
$$ \int_{0}^{\infty}\frac4{4+t^2x^2}dx $$
converges, by the DCT,
\begin{eqnarray}
\lim_{\Delta t\to0}\frac{F(t+\Delta t)-F(t)}{\Delta t}=\int_{0}^{\infty}\lim_{\Delta t\to0}\frac{1}{1+(t+\theta\Delta t)^2x^2}dx=\int_{0}^{\infty}\frac{1}{1+t^2x^2}dx=\frac{\pi}{2t}\equiv f(t).
\end{eqnarray}
which impies $F(t)$ is differentiable and $F'(t)=f(t)$. Since $f(t)$ is continuous for $t>0$, we conclude $F(t)\in C^1(0,\infty)$.
A: Use Frullani's Integral: https://en.wikipedia.org/wiki/Frullani_integral
$$\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x}dx=(f(\infty)-f(0))\ln\frac{a}{b}$$
$$F(t)=\int_{0}^{\infty} \frac{\tan^{-1}(tx)-\tan^{-1}{x}}{x}dx=(\pi/2-0)\ln t=\frac{\pi}{2}\ln t$$
$$F'(t)=\frac{\pi}{2t}$$
