Proving differentiability without using the closed-form integration 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 $-\pi/2$.
 A: (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 $g(t)=\min\big(\tfrac12,\frac2{t^2}\big)$.
A: For completeness, we show first that $F$ is differentiable on $(0,\infty)$. Note for $y>0$ that the function
$$h_y(u;h) := \frac{\cos((y+h)u) - \cos(uy)}{h},$$
which can be extended continously in $h$ on $h \in [0,\infty)$
satisfies
$$\int_0^x g_y(u;h) d u \ll \frac{1}{y(y+h)} + \frac{x}{y+h}.$$
Imitating the proof of Dirichlet's test for improper integrals shows that $F$ is differentiable on $(0,\infty)$ and that the derivate is given
$$F'(y) = - \int_0^\infty \frac{u}{1+u^2} \cos(uy) \, du.$$
To show that $F$ is right differentiable in $y=0$, we will need to evaluate some integrals at the end, since the previous argument cannot be applied. In particular, using the above-mentioned explicit formula for $F$, seems to be the simplest way to do this. One alternative way for doing this is to note that
$$\frac{\cos(xh)-1}{h} = -\int_0^x \sin(uh) du$$
and now analyze
$$\int_0^m \frac{\cos(xh)-1}{h} \frac{1}{1+x^2} dx = \int_0^m \sin(uh) ( \arctan(u)-\arctan(m)) \, du. $$
Here I have used Fubini's theorem and, to do this, we need to restrict the integral on a compact set. Since $\tan(\alpha) = \cot(\alpha)^{-1}$ we have a Taylor-formula in $\alpha=\infty$ with $$\arctan(u) = \pi/2 - \frac{1}{u} +  g(u)$$
and  $g(u) \ll u^{-2}$. Thus
\begin{align*}
&\int_0^m \sin(uh) ( \arctan(u)-\arctan(m)) du \\
 &= \frac{\cos(mh)-1}{h} \bigg(g(m) - \frac{1}{m} \bigg) - \int_0^m \frac{\sin(uh)}{u} du + \int_0^m \sin(uh) g(u)du.
\end{align*}
Letting $m \rightarrow \infty$, we get
$$\int_0^\infty \frac{\cos(xh)-1}{h} \frac{1}{1+x^2} dx = -\int_0^\infty \frac{\sin(u)}{u} du + \int_0^\infty \sin(uh) g(u) du.$$
Using the well-known formula $\int_0^\infty \sin(u)/u \, du = \pi/2$ and that $g(u) \ll u^{-2}$ together with the dominated congerence theorem, we conclude that right-hand derivative of $F$ in $y=0$ exists and also that it is given by
$$F'(0) = - \int_0^\infty \frac{\sin(u)}{u} du = - \frac{\pi}{2}.$$
