How to prove this equality with closed form?

An exercise I'm trying to solve gives us

$$F(y) := \int_{0}^{\infty} \frac{sin(xy)}{x(1+x^2)}dx$$ and first asks us to prove that F(y) satisfies the ODE $F''(y)-F(y)+\frac{\pi}{2}=0$, then find the closed form of the ODE and lastly prove that:

$$\int_{0}^{\infty}\frac{cos(yx)}{1+x^2}dx=\frac{\pi e^{-y}}{2}$$

using the closed form of the ODE.

I've gone through the first two and for the closed form I got $F(y)=c_1e^{-y}+c_2e^y+\frac{\pi}{2}$.

But I'm having trouble with solving the last part of the exercise. With complex analysis, I can solve it, but I don't know what to do with the closed form. I'm sharing my original thoughts, but I get stuck:

First, I noticed that the last integral is the derivative of F(y):

$$F'(y)= \int_{0}^{\infty} \frac{d}{dy} \Big(\frac{sin(xy)}{x(1+x^2)} \Big)dx= \int_{0}^{\infty}\frac{cos(yx)}{1+x^2}dx $$

Also: $F'(y)=-c_1e^{-y}+c_2e^y$. Then:

$$ \int_{0}^{\infty}\frac{cos(yx)}{1+x^2}dx = -c_1e^{-y}+c_2e^y $$

Then I'd need to show $ -c_1e^{-y}+c_2e^y = \frac{\pi e^{-y}}{2}$ and I get stuck here.

Am I doing something wrong? Is this even solvable this way? Should I approach it differently?

If you proven the first parts already, then all you need is a limiting condition.

$$F'(0) = \int_0^\infty \frac{\cos(0)}{(1+x^2)}dx = \frac{\pi}{2}$$

and consider the change of variables $z=xy$

$$F(y) = \int_0^\infty\frac{\sin z}{z\left(1+\frac{z^2}{y^2}\right)}dz \implies \lim_{y\to\infty}F(y) = \int_0^\infty\frac{\sin z}{z}\:dz = \frac{\pi}{2}$$

The exponentially increasing term must be $0$ since the limit exists at all, and then we have

$$F'(y) = \frac{\pi e^{-y}}{2}$$