 Solution

A useful differentiation under the integral sign theorem is this: if $f(\cdot ,y)$ is integrable, $f$ is differentiable in $y$ almost everywhere, and on every compact parameter interval $|{\partial }_{y}f(x,y)|$ is bounded by an integrable function of $x$, then $F^{\prime }(y)={\int }_I{\partial }_{y}f(x,y)dx$, under the usual pointwise continuity hypotheses. It follows by applying dominated convergence to difference quotients. The theorem can be iterated when a corresponding bound holds for the second derivative.

On a compact interval $0<a\le y\le b$, 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 arewith integrable bounds $(1+x^4{)}^{-1}$ and $x^2/(1+x^4)$. Thus $F$ is twice differentiable. Subtracting the integrals givesThe final integral is finite: its integrand tends to one at zero and is absolutely bounded by $t^{-2}$ at infinity. Consequentlywhich establishes the required parameter-independent constant without needing an explicit Fresnel integral evaluation.