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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?

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  • $\begingroup$ Is there a reason to not just do the integration? $\endgroup$ – Clayton May 25 at 15:45
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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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Hint: For the integral $$\int_{-1}^{1}x^5\arctan(nx)dx$$ we get $$\frac{15 \left(n^6+1\right) \tan ^{-1}(n)-3 n^5+5 n^3-15 n}{45 n^6}$$

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  • $\begingroup$ When I am doing integration I stuck there $\int x^6/(n^2\cdot x^2 +1)$ $\endgroup$ – Trebacz112 May 25 at 15:46
  • $\begingroup$ Partial fractions, @Trebacz112 $\endgroup$ – Clayton May 25 at 15:47
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    $\begingroup$ Hint: $$\frac{x^6}{n^2x^2+1}={\frac {{x}^{4}}{{n}^{2}}}-{\frac {{x}^{2}}{{n}^{4}}}+{n}^{-6}-{\frac {1}{{n}^{6} \left( {n}^{2}{x}^{2}+1 \right) }} $$ $\endgroup$ – Dr. Sonnhard Graubner May 25 at 15:49

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