 Solution

One version of dominated differentiation under the integral sign is the following. Let the parameter range be open. Suppose $f(\cdot ,y)$ is integrable, ${\partial }_{y}f(x,y)$ exists for almost every $x$, and near each parameter value its absolute value is bounded by an integrable function of $x$ independent of $y$. ThenThis follows by applying the dominated convergence theorem to difference quotients, bounded by that same integrable function through the mean value theorem.

On any compact parameter interval $0<y\le B$, the original integrand is bounded by $B/(1+x^4)$, using $|\sin (x^{2}y)|\le B{x}^2$. Its first and second parameter derivatives areTheir absolute values are bounded respectively by $(1+x^4{)}^{-1}$ and $x^2/(1+x^4)$, both integrable on $(0,\infty )$. Applying the theorem twice proves that $F$ is twice differentiable and givesThis last integral is absolutely convergent: its integrand is bounded near zero and bounded by $x^{-2}$ at infinity. Substituting $t=x\sqrt{y}$ yieldsThe expression for $c$ is finite and independent of $y$, which is all that is required.