# Calculating $\lim_{n \to +\infty}\int_0^1 (n + 1)x^{n}(1 - x^3)^{1/5}\,dx$

This question is from a bank of past master's exams. I have been asked to evaluate $$\lim_{n \to +\infty}\int_0^1 (n + 1)x^{n}(1 - x^3)^{1/5}\,dx.$$

I did this problem in a hurried manner, but here's what I think. Since $x^n$ is decreasing in $n$ for fixed $x$ in the closed unit interval, it seems like the integrand, which we may denote by $f_n$, converges pointwise to zero. If I can show that the integrand in fact converges uniformly to zero, by showing $$M_n = \sup_{x\in[0,1]}|f_n(x)|\rightarrow 0,$$ then the question is simply a matter of commuting the limit with the integral. Now, $f_n(x)$ is continuous and differentiable on $[0 , 1]$, so it achieves its supremum, which can be found by differentiating and finding the critical points. I found this critical point to be $x=(\frac{5n}{5n+3})^{1/3}$. The denominator exceeds the numerator in this expression for all $n$, so the critical point is between 0 and 1. It also seems clear to me that this is a local maximum. At this point, $f_n$ achieves the value $$M_n = (n+1)(\frac{5n}{5n+3})^{n/3}(1 - \frac{5n}{5n+3})^{1/5}$$ which goes to infinity. So, my intuition failed me at some point. What is the proper solution to this limit? More importantly, is there a better approach to this type of problem?

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First, integrate by parts: $$\small I_n:=\int_0^1(n+1)x^n(1-x^3)^{1/5}dx=[x^{n+1}(1-x^3)^{1/5}]_0^1-\int_0^1x^{n+1}\frac 15(1-x^3)^{-4/5}(-3x^2)dx,$$ hence $$I_n=\frac 35\int_0^1x^{n+3}(1-x^3)^{-4/5}dx.$$ What you did before shows that now we have uniform convergence to $0$ of the integrand.

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Thank you, Davide. Just the kind of solution I needed. –  user36387 Aug 8 '12 at 19:56
Would it not be easier to substitute $y = x^{n+1}$ to get $$\int_0^1 (n + 1)x^{n}(1 - x^3)^{1/5}\,dx = \int_0^1 \left(1 - y^{3/(n+1)}\right)^{1/5}\,dy$$ and invoke monotone convergence?
Change variable $u=x^3$, which reduce the integral to Euler's integral: $$\int_0^1 (n+1) x^n (1-x^3)^{a} \mathrm{d} x = \frac{n+1}{3} \int_0^1 u^{(n-2)/3} (1-u)^{a} \mathrm{d} u = \frac{n+1}{3} B\left( \frac{n+1}{3}, a+1 \right) = \Gamma\left(a+1\right) \frac{\Gamma\left(\frac{n}{3} + \frac{4}{3}\right)}{\Gamma\left(\frac{n}{3} + \frac{4}{3} + a\right)} = \Gamma\left(a+1\right) \left(\frac{3}{n+4}\right)^a \left( 1 + \mathcal{o}\left(\frac{1}{n}\right) \right)$$ Since $a = \frac{1}{5}>0$, the limit is zero.