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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?

StubbornAtom
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Algebra
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1 Answers1

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$|\sin x | \leq |x|$ 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 $\int_1^{\infty} \frac 1 {x^{n}} \, dx =\frac 1 {n-1}$.

Kavi Rama Murthy
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  • Thanks! I still have some questions: So your answer would be that the limit exist and is equal to 1? I have difficulties to uderstand the solution: Is DCT the Dominated Convergence Theorem? Can you explain in more detail how it applies? I do understand the second part. You need $n\geq2$ there I guess, but this is not a problem as $n\rightarrow\infty$. – Algebra Jul 25 '18 at 08:36
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    If $x<1$ then $x^{n} \to 0$ as $n \to \infty $. Since $\frac {\sin t } t \to 1$ as $t \to 0$ it follows that the integrand tends to $1$ at every point. Since it is bounded by the constant $1$ which is integrable on $(0,1)$ it follows by Dominated Convergence Theorem that the integral tends to $\int_0^{1}1, dx=1$. – Kavi Rama Murthy Jul 25 '18 at 08:39
  • More elementarily (without the DCT...) – Anne Bauval Dec 15 '23 at 18:16