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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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up vote 6 down vote accepted

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?

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Hey, that works, too. Thanks. – user36387 Aug 9 '12 at 4:43

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.

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Thanks for the quick response. I'm pretty sure, however, that I'd have no chance of coming up with that on an exam. At the risk of seeming demanding, is there another more elementary if less elegant way of approaching this problem? – user36387 Aug 8 '12 at 19:40

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