Prove that
$$\int_0^{\infty} \frac{\cos{x}}{1+x} dx = \int_0^{\infty} \frac{\sin{x}}{(1+x)^2} dx$$
Things that I tried so far: I tried to create integral (0, infinity) cos x/1+x - sin x/(1+x)^2 and prove that it converges to 0 but it did not work. Another thing that I tried is change variables by x=pi/2 -t , and did not lead me anywhere. hints..?
All you need to do is integrate by parts:
$$\int_0^{\infty} dx \frac{\cos{x}}{1+x} = \underbrace{\left [ \frac{\sin{x}}{1+x}\right ]_0^{\infty}}_{\text{this}=0} + \int_0^{\infty} dx \frac{\sin{x}}{(1+x)^2}$$