# Limit of a Lebesgue integral

What is the value of:

$$\lim_{n\to\infty}\sqrt{n}\int_0^{1}(1-t^2)^ndt$$

I think I have to use the Theorem of dominated convergence

• What have you done so far? People are much more likely to help if they see your dedication to the question. Oct 16 '13 at 16:51
• Yes, but I wouldn't ask here If I knew how to start... I'm totally lost. Oct 16 '13 at 16:55
• Have you tried computing the integral? At least for the first few $n$ and maybe even for general $n$? Oct 16 '13 at 17:01
• A little partial integration gave me ($A_n$ denoting the integral part) $$A_n = A_{n-1} + \frac{2}{n} \int_0^1 t (1-t^2)^n dt = A_{n-1} + \frac{2}{n(n+1)} A_{n+1}$$ So we have a recurrence relation of $$A_{n+1} = \frac{n(n+1)}{2} (A_n - A_{n-1})$$ Oct 16 '13 at 17:04

You can find the value of the limit by squeezing, since: $$\sqrt{n}\int_{0}^{1}(1-t^2)^n dt\leq \sqrt{n}\int_{0}^{1}e^{-nt^2}dt = \int_{0}^{\sqrt{n}}e^{-t^2}dt<\int_{0}^{+\infty}e^{-t^2}dt=\frac{\sqrt{\pi}}{2},$$ and the differences between the first and the second term, the third and the fourth, are $O\left(\frac{1}{\sqrt{n}}\right).$ Through the $t=\cos\theta$ substitution you can also recognize the Wallis product in the LHS.
• Perhaps more quickly, you could substitute $t' = t/\sqrt{n}$. Oct 16 '13 at 17:19
From my comment: \begin{align*} A_n & = \int_0^1 (1-t^2)^n dt = \int_0^1 (1-t^2)^{n-1} dt - \int_0^1 t^2 (1-t^2)^{n-1} dt \\ & = A_{n-1} - \left( \underbrace{\frac{1}{n} \left[ t^2 (1-t^2)^n \right]_0^1}_{=0} - \frac{2}{n} \int_0^1 t (1-t^2)^n dt \right) \\ & = A_{n-1} + \frac{2}{n} \left( \underbrace{\frac 1 {n+1} \left[ t(1-t^2)^{n+1} \right]_0^1}_{=0} - \frac{1}{n+1} \int_0^1 (1-t^2)^{n+1} \right) \\ & = A_{n-1} - \frac 2 {n(n+1)} A_{n+1} \end{align*} So $$A_{n+1} = \frac{n(n+1)} 2 (A_{n-1} - A_n)$$ With $A_0 = 1, A_1 = \frac{2}{3}$.