Proving differentiability without using the closed-form integration

(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 $g(t)=\min\big(\tfrac12,\frac2{t^2}\big)$.

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 $$F'(0) = - \int_0^\infty \frac{\sin(u)}{u} du = - \frac{\pi}{2}.$$