TITLE: Proving differentiability without using the closed-form integration
QUESTION [5 upvotes]: It is well known that $F(y) = \int_0^\infty \frac{\cos(xy)}{1+x^2}dx = \frac{\pi e^{-y}}{2}$ for $y \geqslant 0$. Without knowing this a priori I want to show that $F(y)$ is differentiable on $(0,\infty)$ and that the right-hand derivative $F'_R(0)$ also exists.
I already showed that $F'(y)$ exists for $y > 0$ because differentiation under the integral is justified by uniform convergence on intervals excluding $y = 0$. So  for $y > 0$
$$F'(y) = \int_0^\infty \frac{-x\sin(xy)}{1+x^2} dx$$
How can I show that there exists 
$$F_R'(0) = \lim_{y \to 0+} \frac{F(y) - F(0)}{y-0} = \lim_{y \to 0+}\int_0^\infty \frac{\cos(xy) - 1}{y} \frac{1}{1+x^2}dx $$
Perhaps using dominated convergence theorem?  I already know the value must be $-\pi/2$.

REPLY [4 votes]: (Let me take the absolute values instead.)
Substutite $t=xy$:
$$ 
\int_{x=0}^\infty \frac{1-\cos(xy)}{y} \cdot \frac{1}{1+x^2} \mathrm{d}x =
\int_{t=0}^\infty \frac{1-\cos t}{y^2+t^2} \mathrm{d}t.
$$
Pointwise we have $0\le \frac{1-\cos t}{y^2+t^2} \nearrow \frac{1-\cos t}{t^2}$, so the monotone convergence theorem works.
Alternatively, a suitable dominant function is $g(t)=\min\big(\tfrac12,\frac2{t^2}\big)$.