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How to calculate the following limit:$$\lim_{n\to\infty}\int_0^\infty\frac{n^2[\cos(x/n^2)-1]}{1+x^3}dx.$$


I have tried dominated convergence theorem but I cannot find a proper dominated function. I also tried applying the residue theorem, but if we choose the upper-semi circle as the contour then the integrand is not an even function. And I tried to expand the function into power series, but nothing helps. Can someone give me a hint? Thank you.

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Let $f(x) = \int_{x}^{\infty} \frac{\mathrm{d}t}{1+t^3}$. Then $f$ is integrable on $[0, \infty)$. Indeed, $f$ is bounded by $f(0)$ and $f(x) \asymp 1/x^2$ as $x\to\infty$. Now by integration by parts,

\begin{align*} \int_{0}^{\infty} \frac{n^2(\cos(x/n^2)-1)}{1+x^3}\,\mathrm{d}x &= -\int_{0}^{\infty} n^2(\cos(x/n^2)-1)f'(x) \,\mathrm{d}x \\ &= \underbrace{\left[ -n^2(\cos(x/n^2)-1)f(x) \right]_{0}^{\infty}}_{=0} - \int_{0}^{\infty} \sin(x/n^2)f(x) \, \mathrm{d}x. \end{align*}

Now by the dominated convergence theorem,

$$ \lim_{n\to\infty} \int_{0}^{\infty} \frac{n^2(\cos(x/n^2)-1)}{1+x^3}\,\mathrm{d}x = - \int_{0}^{\infty} \lim_{n\to\infty} \sin(x/n^2)f(x) \, \mathrm{d}x = 0. $$


Addendum. A more detailed analysis, with a bit of help from Mathematica 11, shows that

$$ \int_{0}^{\infty} \frac{n^2(\cos(x/n^2)-1)}{1+x^3}\,\mathrm{d}x = -\frac{1}{n^2}\left(\log n + \frac{3}{4} - \frac{\gamma}{2} + o(1) \right) $$

as $n\to\infty$, where $\gamma$ is the Euler-Mascheroni constant.

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    $\begingroup$ Awesome! Thank you. $\endgroup$ – Bach Jun 18 '19 at 7:12
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By letting $t=x/n^2$, we have that $$\begin{align} 0\leq -I_n&:=\int_0^\infty\frac{n^2[1-\cos(x/n^2)]}{1+x^3}dx\\ &= n^4\int_0^\infty\frac{[1-\cos(t)]}{1+n^6 t^3}dt\\ &\leq n^4\int_0^1\frac{t^2/2}{1+n^6t^3}dt +n^4\int_1^\infty\frac{2}{n^6 t^3}dt\\ &=\frac{[\ln(1+n^6t^3)]_0^1}{6n^2} +\frac{1}{n^2}\left[-\frac{1}{t^2}\right]_1^\infty\\ &=\frac{\ln(1+n^6)}{6n^2}+\frac{1}{n^2} \end{align}$$ where we used the fact that $0\leq 1-\cos(t)\leq \min(t^2/2,2)$.

Now it should be easy to see that the RHS goes to zero as $n\to +\infty$. As a matter of fact, the above inequality implies that $I_n\in O(\ln(n)/n^2)$ and therefore even $n^aI_n$ goes to zero for $a<2$.

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$$L=\lim_{n\to\infty}\int_0^\infty\frac{n^2[\cos(x/n^2)-1]}{1+x^3}dx$$ for ease I will let: $$f(x,n)=\frac{\cos(x/n^2)-1}{1+x^3}$$ and so we can say: $$L=\lim_{n\to\infty}\frac{\int_0^\infty f(x,n)dx}{\frac{1}{n^2}}$$ now since this is a $\frac 00$ situation we can use L'Hopitals rule and say: $$L=\lim_{n\to\infty}\left[-\int_0^\infty\frac{x\sin(x/n^2)}{1+x^3}dx\right]$$ where we can observe that for $n\to\infty,\sin(x/n^2)\to0$ and so $L\to0$

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