I would like to find :
lim
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.
Answer
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.
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