I'm trying to compute the derivative of $k(t) = \int^{\infty}_{-\infty} \frac{\sin{tx^2}}{1+x^4}\,\textrm dx$.
I've already showed that it exists, so here's what I've thought of so far: letting $k_n(t) = \int^n_{-n} \frac{\sin{tx^2}}{1+x^4}\,\textrm dx$, and using the $\int^\infty_{-\infty} \frac{\textrm dx}{1+x^4}$ bound (and symmetry to write $2 \int_0^\infty$), I can show the sequence converges uniformly. Now, letting $h(\delta, t, x) = \frac{\sin{tx^2} - \sin{(t+\delta)x^2}}{\delta} - x^2 \cos{tx^2}$, by considering we are in $[-n, n]$, and using uniform continuity and MVT, we get $h(\delta, t, x) < \epsilon$ for sufficiently small delta for a fixed $t$ with $x \in[-n, n]$. Thus, since we are in a finite interval, we have $k_n^\prime (t) = \int^n_{-n} \frac{x^2 \cos{tx^2}}{1+x^4}\,\textrm dx$. I think I could then finish it using the bound $\int^\infty_{-\infty} \frac{\textrm dx}{x^2}$ to show the derivatives converge uniformly and then apply the uniform convergence result on derivatives.
I would like to ask if my reasoning is correct and I can finish it this way or if there are any mistakes along it.
Any help is appreciated!
What you did is fine, but is involved and subtle. Abstractly, if $k(t)=\int f(x,t)dx$, then $\frac{k(t+\Delta t)-k(t)}{\Delta t}=\int \frac{f(x,t+\Delta t) - f(x, t)}{\Delta t}dx$. By mean value theorem, $\frac{f(x,t+\Delta t) - f(x, t)}{\Delta t} = f_t(x, t')$ for some $t'$, and in this particular case, $|f_t(x, t')|\le \frac{x^2}{1+x^4}$ is integrable, therefore by Lebesgue's dominated convergence theorem, $k'(t)=\int f_t(x, t)dx$.