This question has an answer that relates differentiation under the integral to the OP.

Again, here's the original integral: $$\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x$$

...and we let $$ F(y) = \int\nolimits_{0}^{\infty} \frac{\sin xy}{x(1+x^2)} \ dx \ \ \text{for} \quad\quad y > 0$$

The first part of interest is in showing that $\displaystyle F''(y) - F(y) + \pi/2 = 0$. Is it necessary to integrate $F(y)$ to show this? What about the possibility of taking $\lim_{y \to 0+}$ beforehand? I'm wondering if someone can help explain this step in much greater detail. I'm a little hazy with the $y>0$ portion of it, and whether or not integration has to occur here. I'm trying to make sure I thoroughly understand this post so that I can apply it later to different problems.