Prove that $$\int_{0}^{\infty} \frac{\cos(x)}{1+x^2} dx = \frac {\pi}{2e}$$ My approach would be $$\lim_{n \to \infty} \int_{0}^{n} \frac{\cos(x)}{1+x^2} dx$$ and evaluate the limits of the sine and cosine integral functions, but I'm pretty sure there is an easier way.

The second integral is $$\int_{0}^{\infty} \frac {\ln(x)}{x^2+b^2} dx, b > 0$$ My approach; Let $f$ be an analytic function $$f(z)=\frac {\ln(z)}{z^2+b^2}$$ then the poles of $f$ would be at $$z=ib, z=-ib$$ Now I don't know what contour to draw and what would be inside it.