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As a new year's resolution, I'm going through basic exercises and here is one that causes me some trouble. One needs to study the following limit using the dominated convergence theorem:

$$\lim_{n\rightarrow \infty}\ n^2 \int_0^1 (1-x)^n sin(\pi x)\, dx$$

What I can say:

  • By integration by parts, and solving a difference equation, one can certainly obtain the answer but that does not seem to be the goal of the exercise.
  • The integral alone (without the $n^2$ factor) tends to 0: one can simply dominate by the constant function 1, and since the integrand converges to 0. Hence we are with an indetermined limit, something that goes to infinity times something that goes to 0.
  • Doing the change of variable $y=nx$ leads to $$ n^2 \int_0^1 (1-x)^n sin(\pi x)\, dx = n \int_0^n \left(1- \frac{y}{n}\right)^n sin\left(\pi \frac{y}{n}\right)\, dy$$ while $y=n^2 x$ leads to $$ n^2 \int_0^1 (1-x)^n sin(\pi x)\, dx = \int_0^{n^2} \left(1- \frac{y}{n^2}\right)^n sin\left(\pi \frac{y}{n^2}\right)\, dy$$ In the first case the integral alone goes to 0 while n goes to infinity. The last formula seems promising, unfortunately I'm not able to dominate the integrand by an integrable function.
  • I also thought about cutting the integral in two parts but at the moment it leads nowhere.
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  • $\begingroup$ $n^2 \int_0^1 (1-x)^n sin(\pi x)\, dx== \int_0^1 n^2(1-x)^n sin(\pi x)\, dx$ $\endgroup$
    – David C. Ullrich
    Jan 8, 2019 at 17:33
  • $\begingroup$ How do you dominate the integrand by sthg independent of n? $\endgroup$
    – Noix07
    Jan 8, 2019 at 17:52
  • $\begingroup$ I'm not sure - was just pointing out that writing it that way is how you might be able to apply DCT. I have to go - if I were trying to do this I'd start by using calculus to try to find the maximum of $t^2\lambda^t$ for real $t>>0$ (given $0<\lambda<1$). $\endgroup$
    – David C. Ullrich
    Jan 8, 2019 at 18:01
  • $\begingroup$ Indeed by the change of variable $\lambda=1-x$ the integral is equal to $\int_0^1 n^2 \lambda^n sin(\pi \lambda)\, d\lambda$. I actually thought about looking at the maximum at $t$ (playing the role of n) fixed but you are right, one should look at the maximum for $\lambda$ fixed. I find $t=-\frac{2}{ln \lambda}$ and plugging this back to $t^2 \lambda^t sin (\pi \lambda)$ yields $ \frac{4 sin (\pi x)}{(ln x)^2 x^2}$ which is not integrable according to wolfram alpha... $\endgroup$
    – Noix07
    Jan 9, 2019 at 11:00
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    $\begingroup$ $-2/\log(\lambda)$ yes, but pluggin in does not give what you said - you made an error manipulating the exponential. $\endgroup$
    – David C. Ullrich
    Jan 9, 2019 at 14:30

2 Answers 2

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Ok, do this :$$n^2\int_0^1(1-x)^n\sin(\pi x)=\int_0^n\left(1-\frac xn\right)^n n\sin\left(\frac{\pi x}{n}\right).$$Of course here $$\left|n\sin\left(\frac{\pi x}{n}\right)\right|\le\pi x;$$a little calculus shows that $$\log(1-x/n)\le-x/n\quad(0<x<n)$$and there's your DCT. (In fact we have $|f_n|\le g=\lim f_n=\pi xe^{-x}$.)

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  • $\begingroup$ @Noix07 I think I got it - considering the record you should look carefully at the details. $\endgroup$
    – David C. Ullrich
    Jan 10, 2019 at 0:29
  • $\begingroup$ Yeah that's it!! Actually I also thought about another possibility using DCT but after an integration by parts: $\displaystyle n^2 \int_0^1 (1-x)^n \sin (\pi x)\, dx = \left[ \frac{n^2 (1-x)^{n+1}}{-(n+1)} \sin (\pi x)\right]_ 0^1 + \int_0^1 \frac{ \pi n^2 (1-x)^{n+1}}{n+1} \cos (\pi x)$. The first term vanishes and we perform the change of variable $y=nx$ in the second integral. One factorises out a $\pi \frac{n}{n+1}$ factor and dominates the remaining integrand by $e^{-y} \times 1$... $\endgroup$
    – Noix07
    Jan 10, 2019 at 9:42
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First, compute the limit when you replaced $\sin$ by the identity function.

Then, prove that if $x \in [0,1]$, $0 \geq \sin(\pi x)-\pi x\geq -Cx^2$ for some positive constant $C$.

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  • $\begingroup$ Ok it works: $\int_0^1 (1-x)^n \pi x\, dx= \frac{\pi}{(n+1)(n+2)}$ and $\int_0^1 (1-x)^n C x^2\, dx= \frac{2 C}{(n+1)(n+2)(n+3)}$. As for the inequality, I can understand it in two ways, 1) $sin(\pi x)$ is below its tangent at 0 in the interval $[0,1]$ or 2) (only valid for $x\in [0,1/ \pi]$ at least at first glance) regroup the terms two by two in the expression of sin as a series. One then has $\sin(\pi x)= \pi x + $negative terms. I'm still thinking whether it uses the dominated cvg thm $\endgroup$
    – Noix07
    Jan 8, 2019 at 17:50

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