I haven't done Analysis for a long time and now I am looking at some exercises and could need some help:
Does the limit $$\lim_{n\rightarrow\infty}\int_0^\infty \frac{\sin(x^n)}{x^n}\,dx$$ exist and if yes what is the value?
I think I first need to check if there is a function $f$ such that $\dfrac{\sin(x^n)}{x^n}$ converges uniformely to $f$ and if that's the case I can change the order of the limit and the integration and compute $\int f$ ? But I have difficulties to do that. I was thinking about writing $$\dfrac{\sin(x^n)}{x^n} =\dfrac{1}{2x^n}i(e^{-ix^n}-e^{ix^n})$$ but then I am stucked. If $x>1$ the first term is going to $0$ but what else can I say?
$|\sin x | \leq |x|$ for all $x$. So $\int_0^{1} \frac {\sin (x^{n})} {x^{n}} \, dx \to 1$ as $n \to \infty $ by DCT. Clearly, $\int_1^{\infty } \frac {\sin (x^{n})} {x^{n}} \, dx \to 0$ since $|\sin (x^{n})|\leq 1$ and $\int_1^{\infty} \frac 1 {x^{n}} \, dx =\frac 1 {n-1}$.