I'm thinking about how to compute $\int_{\frac{3}{2n}}^1 nx^{n-1}(x-\frac{3}{2n})^n dx$ or give limsup of it as n tends to infinity? It seems integrate by part may work, but it is complicated. I try to use dominated convergence theorem, since the function converges to $nx^{2n-1}$, but it seems there is not a uniform function for it. It seems monotone decreasing theorem works.
How to compute or give a limsup of this integral?
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calculus
integration
analysis
2 Answers
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I think you can upper bound the integral using Holder inequality, in fact you have to notice that $nx^{n-1}$ is integrable over $[\frac{3}{2n}, 1]$ and that $\Vert (x-\frac{3}{2n})^n \Vert_{\infty} = (1-\frac{2}{3n})$ (on $[\frac{3}{2n}, 1]$). You get
$\int_{\frac{3}{2n}}^1 nx^{n-1}(x-\frac{3}{2n}))^n dx \le (\int_{\frac{3}{2n}}^{1} nx^{n-1} dx) (1-\frac{3}{2n}))^n = (1 -\frac{3}{2n}^n)(1-\frac{3}{2n})^n \to \frac{1}{e^{3/2}}$
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Here is an evaluation of the integral.
Let $z=\frac{2n}{3}x$
\begin{equation} \int\limits_{3/(2n)}^{1} n x^{n-1} \left(x - \frac{3}{2n} \right)^n dx = n\left(\frac{3}{2n} \right)^{2n} (-1)^n \int\limits_{1}^{2n/3} z^{n-1} (1-z)^n dz \end{equation}
\begin{align} I(n) &= \int\limits_{1}^{2n/3} z^{n-1} (1-z)^n dz \\ &= \int\limits_{0}^{2n/3} z^{n-1} (1-z)^n dz - \int\limits_{0}^{1} z^{n-1} (1-z)^n dz \\ &= \mathrm{B}_{2n/3}(n,n+1) - \mathrm{B}(n,n+1) \\ &= \frac{1}{n} \left(\frac{2n}{3} \right)^{n} {}_{2}\mathrm{F}_{1}\left(n,-n;n+1;\frac{2n}{3} \right) - \mathrm{B}(n,n+1) \end{align}
We have used the incomplete beta function and Gauss's hypergeometric function.