Let $x\in[0,1]$, study the pointwise and uniform convergence of $$f_n(x)=n^2 \int_0^x \frac{\sin(x^n t)}{t} dt$$ My attempt: if $x=0$ it is $f_n(0)=0$ and so its limit is $f(x)=0$ as well.
If $x\in(0,1)$: since $\sin u \leq u$ for all $u \geq 0$, $x\in(0,1) \implies x^n \in (0,1)$ and $t\in(0,x) \land x\in(0,1) \implies t \in (0,1)$, it is $x^n t \in (0,1)$ and so $0 \leq n^2\frac{\sin(x^n t)}{t} \leq n^2\frac{x^n t}{t}=n^2x^n$; so integrating both sides in the interval $(0,x)$ and letting $n\to\infty$ it follows that $$0 \leq \lim_{n\to\infty} n^2 \int_0^x \frac{\sin(x^n t)}{t} dt \leq \lim_{n\to\infty} \int_0^x n^2 x^n dt=\lim_{n \to \infty} n^2 x^{n+1}=0$$ Where the latter limit is $0$ because $x\in(0,1) \implies x^{n+1} \in (0,1)$.
If $x=1$, since $\sin u \geq \frac{2u}{\pi}$ for all $u\in \left[0,\frac{\pi}{2}\right]$ and $[0,1] \subset \left[0,\frac{\pi}{2}\right]$, it is $$f_n(1)=n^2\int_0^1 \frac{\sin t}{t} dt \geq n^2\int_0^1 \frac{2}{\pi}dt \to \infty \ \text{for} \ n \to\infty$$
So $f_n(x)$ converges pointwise to $f(x)=0$ for all $x\in[0,1)$ and diverges to $\infty$ for $x=1$; to study the uniform convergence, let $I_{a}=[0,a)$ with $0 \leq a<1$.
Since, for what said before, it is $f_n(x) \leq n^2 x^{n+1}$ and for the monotonicity of $s^n$ for $s \geq 0$ and $n\in\mathbb{N}$, it is $f_n(x) \leq n^2 x^{n+1} \leq n^2 a^{n+1}$. Since $f_n(x) \geq 0$ for all $x\in[0,1]$ and for all $n\\in\mathbb{N}$, it is $f_n(x)=|f_n(x)|$ and since the right hand side of the inequality $f_n(x) \leq n^2a^{n+1}$ is independent of $x$, the inequality holds for the supremum for $x\in I_a$ as well, hence $\sup_{x \in I_a} |f_n(x)| \leq n^2 a^{n+1}$ and because $a\in[0,1)$ letting $n\to\infty$ in the inequality it is $$\lim_{n \to \infty} \sup_{x\in I_a} |f_n(x)| \leq \lim_{n \to \infty} n^2a^{n+1}=0$$ This shows that $f_n$ is uniformly convergent to $f(x)=0$ for all the compact sets $I_a$, meaning it is uniformly convergent for all $x\in[0,1)$.
Is this correct? If there are mistakes or imprecisions it would be great to have your feedback to improve myself. Any other solutions are welcome in the answers. Thanks.
Consider a sequence $(x_n)$ of numbers in [0,1) that tends to 1, for example the sequence defined by
$\forall n\in \mathbb{N}^*\setminus\{1\}, x_n=1-\frac{1}{n}$
$f_n(x_n)=n^2\int_0^{x_n} \frac{\sin(x_n^n t)}{t}dt$
$\geq n^2\int_0^{x_n} \frac{2(x_n^n t)}{\pi t}dt$
$\geq \frac{2}{\pi} n^2.x_n^{n+1}$
$\geq \frac{2}{\pi} n^2.\left(1-\frac{1}{n}\right)^{n+1}$
So the limit is $+\infty$