<s>[INST] Find limit in use of integrals Find limit $\lim_{n \rightarrow \infty} \int_{-1}^{1} x^5 \cdot \arctan{(nx)} dx $
From mean-value-theorem we have
$$ \frac{1}{2}  c^5 \cdot \arctan{(nc)}  \mbox{ for some c } \in (-1,1) $$
$$  \underbrace{\frac{1}{2}  \cdot \arctan{(-n)}}_{\rightarrow - \pi /4} \le \frac{1}{2}  c^5 \cdot \arctan{(nc)} \le \underbrace{\frac{1}{2}  \cdot \arctan{(n)}}_{\rightarrow \pi /4}   $$
so this bounding doesn't help me. Has somebody better idea how to bound that?
 [/INST] We'll simplify with $y=nx$. Integration by parts  gives $$\int y^5\arctan ydy=\frac{y^{6}}{6}\arctan y-\frac{1}{6}\int\frac{y^{6}}{1+y^{2}}dy\\=\frac{y^{6}}{6}\arctan y-\frac{1}{6}\int\left(y^{4}-y^{2}+1-\frac{1}{1+y^{2}}\right)dy\\=\frac{y^6+1}{6}\arctan y-\frac{1}{30}y^5+\frac{1}{18}y^3-\frac16 y+C.$$Hence $$\frac{1}{n^6}\int_{-n}^n y^5\arctan ydy=\frac{\frac{n^6}{3}\arctan n+o(n^6)}{n^6}\stackrel{n\to\infty}{\to}\frac{1}{3}\arctan\infty=\frac{\pi}{6}.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2\lim_{n\to\infty}\int_0^1 x^5\arctan nxdx$ (since the integrand is even), which by dominated convergence is $$\pi\int_0^1 x^5dx=\frac{\pi}{6}.$$
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