How to find: $\lim\limits _{n \rightarrow \infty} n^{3} \int\limits_{-\pi}^{\pi} \frac{\cos (2 n x)}{1+x^{8}} d x$ Problem:
What is the following limit?
$\lim _{n \rightarrow \infty} n^{3} \int_{-\pi}^{\pi} \frac{\cos (2 n x)}{1+x^{8}} d x$
Attempt:
Denote $ f(x) = \frac{1}{1+x^8} $ for $ x \in [ -\pi , \pi] $ ( Note that this function is infinitely differentiable. )
$ I_n = n^{3} \int_{-\pi}^{\pi} \frac{\cos (2 n x)}{1+x^{8}} d x = \frac{n^{3}}{2} \int_{-\pi}^{\pi} \frac{ e^{2inx} + e^{-2inx} }{1+x^{8}} d x =  \frac{2\pi \cdot n^{3}}{2} ( \hat{f}(2n) + \hat{f}(-2n) ) $.
Now we use  $ \hat{f^{(k)}}(n) = (in)^{k}f(n) $ ( since $ f$ is indeed differentiable on its interval ), thus
$ I_n = \frac{2\pi \cdot n^{3}}{2} ( \hat{f}(2n) + \hat{f}(-2n) ) = \pi \cdot n^3 ( \frac{\hat{f^{(3)}}(2n)}{-2^3n^3} 
 + \frac{\hat{f^{(3)}}(-2n)}{-2^3n^3} ) = \frac{-\pi}{8} ( \hat{f^{(3)}}(2n) + \hat{f^{(3)}}(-2n)  )   $
And by Riemann-Lebesgue we get $ I_n \to 0  $.
I'm not really sure whether the limit I calculated is justified or the proof's specious ( Which, I think it is but I don't know why, probably where I said that $ \hat{f^{(k)}}(n) = (in)^{k}f(n) $ ). then I had doubts whether the limit is $ 0, \infty, -\infty $ or none of them, I think it's none of them but don't have a gist of justification as to why. Can you please help as to how to find the limit? thanks very much for the help in advance!

Too long for a comment
Using the recommendation of @Gribouillis, integrating by part several times, using the fact that the integrand is even and denoting $\,\Big[\frac{1}{1+x^8}\Big]^{(k)}=\frac{d^k}{dx^k}\frac{1}{1+x^8}$
$$I(n)=2\int_0^\pi \frac{\cos (2 n x)}{1+x^{8}} d x=\frac{1}{n}\frac{\sin 2nx}{1+x^8}\bigg|_{x=0}^{x=\pi}+\frac{1}{n}\int_0^\pi \sin (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(1)}d x$$
$$=-\frac{1}{2n^2}\cos (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(1)}\bigg|_{x=0}^{x=\pi}+\frac{1}{2n^2}\int_0^\pi \cos (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(2)}d x$$
$$=-\frac{4}{n^2}\frac{\pi^7}{(1+\pi^8)^2}+\frac{1}{4n^3}\sin (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(2)}\bigg|_{x=0}^{x=\pi}-\frac{1}{4n^3}\int_0^\pi \sin (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(3)}d x$$
$$=-\frac{4}{n^2}\frac{\pi^7}{(1+\pi^8)^2}+\frac{1}{8n^4}\cos (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(3)}\bigg|_{x=0}^{x=\pi}-\frac{1}{8n^4}\int_0^\pi \cos (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(4)}d x$$
Evaluating the third term:
$$\frac{1}{8n^4}\bigg|\int_0^\pi \cos (2 n x)\Big[\frac{1}{1+x^8}\Big]^{(4)}d x\bigg|<\frac{1}{8n^4}\bigg|\int_0^\pi \Big[\frac{1}{1+x^8}\Big]^{(4)}d x\bigg|<\frac{\operatorname{Const}}{8n^4}$$
Therefore,
$$I(n)=-\frac{4}{n^2}\frac{\pi^7}{(1+\pi^8)^2}+O\Big(\frac{1}{n^4}\Big)$$
and
$$\lim_{n\to\infty}n^3I(n)=\lim_{n\to\infty}\Big(-n\frac{4\pi^7}{(1+\pi^8)^2}\Big)=-\infty$$
Numeric check at WolframAlpha confirms the result. At $n=10 \,000$
$$10^{12}\int_{-\pi}^\pi\frac{\cos(20\,000\, x)}{1+x^8}dx=-1.341589316...$$
while
$$-\frac{4\pi^7}{(1+\pi^8)^2}\cdot 10\,000=-1.341589346...$$
