Find the derivative of $f(x) = \int_{-\infty}^\infty \frac{e^{-xy^2}}{1+y^2}\ dy.$ Problem statement: Find the derivative of 
$$f(x) = \int_{-\infty}^\infty \frac{e^{-xy^2}}{1+y^2}\ dy$$
and find an ordinary differential equation that $f$ solves. Find the solution to this ordinary differential equation to determine an explicit value for $f$.
My attempt: Normally, to find $f'(x)$ in this situation, if this was a calculus problem, I would write:
$$\frac{d}{dx} f(x) = \frac{d}{dx}\int_{-\infty}^\infty \frac{e^{-xy^2}}{1+y^2}\ dy = \frac{d}{dx}\left(\int_{-\infty}^0\frac{e^{-xy^2}}{1+y^2}\ dy + \int_0^\infty \frac{e^{-xy^2}}{1+y^2}\ dy\right)$$
and then I would differentiate under the integral sign:
$$f'(x) = \int_{-\infty}^0\frac{\partial}{\partial x}\frac{e^{-xy^2}}{1+y^2}\ dy + \int_0^\infty \frac{\partial}{\partial x}\frac{e^{-xy^2}}{1+y^2}\ dy,$$
which leaves me with
$$f'(x) = \int_{-\infty}^0 \frac{-y^2e^{-x(y^2+1)}}{1+y^2}\ dy + \int_{0}^\infty \frac{-y^2e^{-x(y^2+1)}}{1+y^2}\ dy.$$
However, this is actually an analysis problem, and as such I am having a really hard time justifying all of these steps. I have already proved that if $F(x) = \int_a^x f(y) \ dy$, then $F$ is absolutely continuous, and therefore the derivative exists a.e. Now, I also know from the FCT that if $f$ is integrable, $F'(x) = f(y)$ a.e. But I still can't quite figure out how to justify moving the derivative in the integral sign, particularly when I have infinite bounds, which I do. Any help would be much appreciated here!

Using Differentiation Under the Integral Sign, we get
$$
\begin{align}
f(x)&=\int_{-\infty}^\infty\frac{e^{-xy^2}}{1+y^2}\,\mathrm{d}y\tag{1}\\
f'(x)&=\int_{-\infty}^\infty\frac{-y^2e^{-xy^2}}{1+y^2}\,\mathrm{d}y\tag{2}\\
f(x)-f'(x)&=\int_{-\infty}^\infty e^{-xy^2}\,\mathrm{d}y\tag{3}\\
&=\sqrt{\frac\pi{x}}\tag{4}
\end{align}
$$
Explanation:
$(1)$: Given
$(2)$: Differentiation Under the Integral Sign. Alternatively, use Fubini's Theorem to show
$$\begin{align}
f(x)
&=\int_{-\infty}^\infty\frac{e^{-xy^2}}{1+y^2}\,\mathrm{d}y\\
&=\int_{-\infty}^\infty\int_x^\infty\frac{y^2e^{-ty^2}}{1+y^2}\,\mathrm{d}t\,\mathrm{d}y\\
&=\int_x^\infty\int_{-\infty}^\infty\frac{y^2e^{-ty^2}}{1+y^2}\,\mathrm{d}y\,\mathrm{d}t\\
\end{align}$$
Then the FTC to show that
$$f'(x)=\int_{-\infty}^\infty\frac{-y^2e^{-xy^2}}{1+y^2}\,\mathrm{d}y
$$
$(3)$: subtract $(2)$ from $(1)$
$(4)$: evaluate the integral in $(3)$
