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I am trying to find the value of this integral:

$\displaystyle{\lim_{n\to\infty}\int_0^\infty\frac{n\cos^2(x/n)}{n+x^4}dx}$.

The integrand tends to 1 as $n$ goes to infinity. So if some convergence theorem holds, the integral would tend to infinity(not Lebesgue dominated convergence theorem, since if there were an integrable function dominating the integrand, there would be a contradiction). So I guess the value of this integral should be infinity, but I am not sure.

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Substituting $x:=n^{1/4} t$ $\>(0<t<\infty)$ gives $$\int_0^\infty{n\cos^2(x/n)\over n+x^4}\>dx=n^{1/4}\int_0^\infty {\cos^2(tn^{-3/4})\over 1+t^4}\ dt\ .\tag{1}$$ The function $t\mapsto \cos^2(tn^{-3/4})$ is nonnegative and $\geq{1\over2}$ for $0\leq t\leq{\pi\over4}n^{3/4}$. It follows that the integral on the right hand side of $(1)$ is $$\geq{1\over2}\int_0^{\pi/4}{dt\over1+t^4}\>dt=:C>0$$ for all $n\geq1$. This proves that the limit in question is $=\infty$.

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By a direct integration, we may obtain more than the value of the desired limit.

We have, as $n$ tends to $+\infty$:

$$ \int_0^{+\infty}n\:\frac{\cos^2(x/n)}{n+x^4}{\rm d}x =\color{#008B8B}{\frac{\pi\sqrt{2}}{4}} \color{#006699}{\:n^{1/4}}+\color{#008B8B}{\frac{\pi\sqrt{2}}{4}}\color{#006699}{\frac{1}{n^{5/4}}}+\color{#008B8B}{\frac{\pi}{3}}\color{#006699}{\frac{1}{n^2}}+\mathcal{O}\left(\color{#006699}{\frac{1}{n^{11/4}}}\right) \tag1 $$

Of course the desired limit is $+\infty$.

One may recall that we have, using inverse Laplace transform, the standard result: $$ \int_0^{+\infty}\frac{\cos^2(ax)}{b^4+x^4}{\rm d}x =\frac{\pi +e^{-\sqrt{2} a b} \pi \left(\cos\left(\sqrt{2} a b\right)+\sin\left(\sqrt{2} a b\right)\right)}{4 \sqrt{2} \:b^3}\quad a>0,\,b>0. $$ Then putting $a:=\dfrac 1n$ and $b:=n^{1/4}$, we get $$ \int_0^{+\infty}\frac{\cos^2(x/n)}{n+x^4}{\rm d}x =\frac{\pi +e^{-\Large \frac{\sqrt{2}}{n^{3/4}}} \pi \left(\cos\left(\Large \frac{\sqrt{2}}{n^{3/4}}\right)+\sin\left(\Large \frac{\sqrt{2}}{n^{3/4}}\right)\right)}{4 \sqrt{2} \:n^{3/4}}. $$ Now, as $x$ is near $0$, we apply the standard Taylor series expansions: $$ \begin{align} e^{-x} & =1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+\mathcal{O}\left(x^5\right) \\ \cos x & =1-\frac{x^2}{2!}+\frac{x^4}{4!}+\mathcal{O}\left(x^5\right) \\ \sin x & =x-\frac{x^3}{3!}+\mathcal{O}\left(x^5\right) \end{align} $$

with $\displaystyle x:=\frac{\sqrt{2}}{n^{3/4}} $ giving $(1)$.

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    $\begingroup$ @ Olivier Oloa It is interesting that the integral with an oscillating integrand $\int_0^{\infty } \frac{n \cos \left(\frac{x}{n}\right)}{n+x^4} \, dx$ still is divergent with the same asymtotics $\frac{\pi }{ \sqrt{2}} n^{\frac{1}{4}}$ $\endgroup$ Sep 18 '19 at 19:29
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I really do not know how much the following could help you; so, please forgive me if I am out of topic.

Consider $$\displaystyle{I=\int\frac{n\cos^2(x/n)}{n+x^4}dx}$$ $$\displaystyle{J=\int\frac{n\sin^2(x/n)}{n+x^4}dx}$$ So $$\displaystyle{I+J=\int\frac{n}{n+x^4}dx}$$ for which the antiderivative can be computed (using partial fraction decomposition). As a consequence $$\displaystyle{\int_0^{\infty}\frac{n}{n+x^4}dx}=\frac{\pi \sqrt[4]{n}}{2 \sqrt{2}}$$ I am not able to demonstrate it, but I have the stange feeling that, using limits, both $I$ and $J$ could have the same behavior.

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The integral

$$i(n) = \int_0^\infty\frac{n\cos^2(x/n)}{n+x^4}dx$$

can be done exactly (Mathematica plus some simplifications) with the result

$$i(n) = \frac{\pi \sqrt[4]{n}}{4 \sqrt{2}} \left(1+e^{-\frac{\sqrt{2}}{n^{3/4}}} \left(\sin \left(\frac{\sqrt{2}}{n^{3/4}}\right)+\cos \left(\frac{\sqrt{2}}{n^{3/4}}\right)\right)\right)$$

The asymptotic behaviour for $n\to\infty$ is easily accessed and is given by

$$i(n) \to \frac{\pi \sqrt[4]{n}}{2 \sqrt{2}}-\frac{\pi n^{-5/4}}{2 \sqrt{2}}$$

This confirmes both that the limit is $= \infty$ and the "feeling" of Claude Leibovici.

EDIT

After finishing the typing I discovered that Olivier Oloa had provided the exact result before.

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