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.

Use Frullani's Integral: https://en.wikipedia.org/wiki/Frullani_integral
$$\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x}dx=(f(\infty)-f(0))\ln\frac{a}{b}$$
$$F(t)=\int_{0}^{\infty} \frac{\tan^{-1}(tx)-\tan^{-1}{x}}{x}dx=(\pi/2-0)\ln t=\frac{\pi}{2}\ln t$$
$$F'(t)=\frac{\pi}{2t}$$
