I want to prove that the function defined as the Lebesgue Integral $$F(y)=\int_0^\infty e^{-x^2}\cos(2xy)dx$$ satisfies $F'(y)+2yF(y)=0$, and after that, that $F(y)=\frac{1}{2}\sqrt{\pi}e^{-y^2}$. I tried this:
First, we have that $F'(y)=\int_0^\infty -2xe^{-x^2}\sin (2xy)dx$, and that
$$F'(y)+2yF(y)=\int_0^\infty e^{-x^2}(2y \cos(2xy)-2x\sin(2xy))dx.$$
So I want to prove that this last integral is 0. I suppose I should use in some step that $$\int_0^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{2}.$$
You'll kick yourself: the integrand of $F'(y)+2yF(y)$ is the derivative with respect to $x$ of $$ e^{-x^2}\sin{2xy}, $$ which vanishes at the endpoints. Hence the integral is zero.
HINT:
Integrate by parts the integral $\int_0^\infty e^{-x^2}2y\cos(2xy)\,dx$ with $u=e^{-x^2}$ and $v=\sin(2xy)$.
As Chappers said: \begin{aligned} F^{\prime}(y)+2 y F(y) & =\int_0^{\infty} e^{-x^2}(2 y \cos (2 x y)-2 x \sin (x x y) d) \\ & =\left[e^{-x^2} \sin (2 x y)\right]_0^{\infty}\\&=0 \end{aligned} which is a separable linear differential equation. $$ \begin{aligned} F^{\prime}(y) & =-2 y F(y) \\ \frac{d F(y)}{F(y)} & =-2 y d y \\ \int \frac{d F(y)}{F(y)} & =-\int 2 y d y \\ \ln (F(y)) & =-y^2+C \\ F(y) & =F(0) e^{-y^2} \\ & =\frac{\sqrt{\pi}}{2} e^{-y^2} \end{aligned} $$
