Continuity of parameter dependent integral

Question

Let $F: (0,\infty)\rightarrow \mathbb{R}$ be defined by \begin{equation*}F(t)=\int_0^1\frac{\sin (xt^2)}{(x^2+t^2)^2}\, dx\end{equation*}

I want to show that $F$ is continuous.

Can we apply directly the following proposition?

If $f: I \times J\rightarrow \mathbb{R}$ is continuous, then the function $F:J\rightarrow \mathbb{R}$ is continuous, i.e. the parameter dependent integral is continuous in respect to the parameter.

If yes, the function $f(x,t)=\frac{\sin (xt^2)}{(x^2+t^2)^2}$ is continuous at $(x,t)\neq (0,0)$ as a composition of continuous functions, or not?

It is left to check the continuity at the point $(0,0)$, i.e. we have to check if $\lim_{(x,t)\rightarrow (0,0)}f(x,t)=f(0,0)$.

Do we have to use here sequences? And which is the value of $f(0,0)$ ?

Answer 1

$f$ can't be extended by continuity at $(0,0)$: note that if $x=\sqrt{t}$ then $$\lim_{t\to 0^+}\frac{\sin (xt^2)}{(x^2+t^2)^2}=\lim_{t\to 0^+}\frac{ t^{5/2}}{t^2}=0,$$ whereas if $x=t$ then $$\lim_{t\to 0^+}\frac{\sin (xt^2)}{(x^2+t^2)^2}=\lim_{t\to 0^+}\frac{ t^3}{4t^4}=+\infty.$$ However, for any $t_0>0$, $f$ is continuous in $[0,1]\times [t_0/2,t_0+1]$ which implies that $F$ is continuous at $t_0$.

Answer 2

Since continuity is a local property, you can consider any rectangular neighborhood of $(x, t)$ for any given $t>0$. (For example, choose $I=[0,1]$ and $J=[t/2, 2t]$. Then the proposition shows that the function $F$ is continuous on $J$.