Let be $I_n = \int_0^{+\infty} \dfrac{\sin(nx^n)}{n x^{n + 1/2}} \textrm{d}x$ for $n \geq 1$.
I'd like to know the limit of $I_n$. Here's what I've done so far:
Let be $f_n : x \mapsto \dfrac{\sin(nx^n)}{nx^{n + 1/2}}$.
First, $f_n$ are measurable, we can also check that $I_n$ are finite.
We have: $\lvert f_n(x) \rvert \leq \dfrac{1}{n x^{3/2}}$.
Also, when $x \to 0$, $f_n(x) \sim \dfrac{1}{\sqrt{x}}$, and $\lim_{\varepsilon \to 0} \int_{\varepsilon}^1 x^{-1/2} \textrm{d}x = 2$.
Thus, $(I_n)_n$ are all finite.
Then, I tried to look for some domination in order to apply the Lebesgue domination theorem, but I'm not even sure I can compute the simple limit of $(f_n)_n$.
I also tried to go continuous by introducing $J : (x, y) \mapsto \dfrac{\sin(y x^y)}{y x^{y + 1/2}}$, but I'm unsure it'll be useful.
Denote $f_n$ the integrand.
Then $$|f_n(x)| \leq\frac{1}{\sqrt{x}}1_{(0,1)}+\frac{1}{x^{1+\frac{1}{2}}}1_{[1,+\infty)} \in L^1(0,+\infty)$$
1.In $(0,1)$ we have that $f_n \to^{n \to +\infty} \frac{1}{\sqrt{x}}$ since $nx^n \to 0$ if $x \in (0,1)$ and $\frac{\sin{y}}{y} \to^{y \to 0}1$
2.If $x>1$ then $nx^{n+\frac{1}{2}} \to^{n \to +\infty} +\infty$ thus $f_n(x) \to^{n \to +\infty}0$
Now use the Dominated convergence theorem.