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

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 $\pi$.

• @TheSimpliFire Yes for all $x \in ]0, 1[$. – auhasard Jun 4 '18 at 15:07
• 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. – TheSimpliFire Jun 4 '18 at 15:09
• 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. – auhasard Jun 4 '18 at 15:18
• @TheSimpliFire So actually there isn't a closed form for the limit of : $n^2 \sin(\pi x) (1-x)^n$ ? – auhasard Jun 4 '18 at 15:20
• 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. – 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.
• Thank you ! Do you agree on the fact that : $n^2(1-x)^n\sin(\pi x) \to 0$ for all $x \in [0, 1]$ ? – auhasard Jun 4 '18 at 15:33