I would like to find :

$$\lim_{n \to \infty} n^2 \int_0^1 (1-x)^n \sin(\pi x) \mathrm{d}x $$

We have : $$n^2 \int_0^1 (1-x)^n \sin(\pi x) \mathrm{d}x = \int_0^1 n^2(1-x)^n \sin(\pi x) \mathrm{d}x$$

Moreover we have $\forall x \in [0, 1]$ :

$$\lim_{n \to \infty} n^2(1-x)^n \sin(\pi x) = 0$$

So by the dominated convergence theorem we can deduce that :

$$ \lim_{n \to \infty} \int_0^1 n^2(1-x)^n \sin(\pi x) \mathrm{d}x = 0$$

Yet, here my book say the answer is actually $\pi$, and I don't understand why what I've done is wrong, and how I can actually find that the value is $\pi$.

  • $\begingroup$ @TheSimpliFire Yes for all $x \in ]0, 1[$. $\endgroup$ – auhasard Jun 4 '18 at 15:07
  • $\begingroup$ Wait a minute... there seems to be a peak just after $x=0$, say $x_0$ that gets larger and larger as $n$ increases, suggesting that as $n\to\infty$, $x_0\to\infty$? That would contradict your statement. $\endgroup$ – TheSimpliFire Jun 4 '18 at 15:09
  • $\begingroup$ That's actually weird... But it seems to be correct because when I take $x = 1/n$ then we have $n^2(n-1)^n/n^n \to \infty$. But to me this actually very strange. $\endgroup$ – auhasard Jun 4 '18 at 15:18
  • $\begingroup$ @TheSimpliFire So actually there isn't a closed form for the limit of : $n^2 \sin(\pi x) (1-x)^n$ ? $\endgroup$ – auhasard Jun 4 '18 at 15:20
  • $\begingroup$ For the indefinite integral, probably not, seeing as there as so many components to it. I do not know of methods advanced enough to solve your problem. $\endgroup$ – TheSimpliFire Jun 4 '18 at 15:23

You cannot apply the DCT because there is not an integrable function $g$ (independent of $n$) such that $n^2(1-x)^n\sin(\pi\,x)\le g(x)$.

Integrating by parts twice we get $$ \int_0^1(1-x)^n\sin(\pi\,x)\,dx=\frac{\pi}{(n+1)(n+2)}+\frac{\pi^2}{(n+1)(n+2)}\int_0^1(1-x)^{n+2}\sin(\pi\,x)\,dx. $$ Multiply by $n^2$, let $n\to\infty$ and observe that the DCT can be applied to the integral on the left hand side because $n^2/(n+1)/(n+2)$ is bounded.

  • $\begingroup$ Thank you ! Do you agree on the fact that : $n^2(1-x)^n\sin(\pi x) \to 0$ for all $x \in [0, 1]$ ? $\endgroup$ – auhasard Jun 4 '18 at 15:33
  • 2
    $\begingroup$ Yes, but the convergence is pointwise, not uniform. $\endgroup$ – Julián Aguirre Jun 4 '18 at 15:48

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