Evaluate $\lim_{n \to \infty} \int_{0}^1 x^{n+1} f(x) \ dx$, for integrable $f$

Question

Let $f : [0, 1] \to \mathbb{R}$ be Riemann integrable. Evaluate $$\lim_{n \to \infty} \int_{0}^1 x^{n+1} f(x) \ dx$$ I have tried applying by parts by assuming $$\\ \int_{0}^1 f(x) \ dx = F(1)-F(0)$$ but it doesn't seem to get me anywhere. Do I have to use partitions and if so how do I go about it?

Answer 1

Since $f$ is integrable, say $\int_0^1 f(x)\,dx = a$, it is bounded, say $m\le f(x)\le M$ for $x\in [0,1]$. In particular, $g(x) = f(x)-m \ge 0$. Then we have $$ \int _0^1 x^{n+1} f(x)\,dx = \int _0^1 m x^{n+1}\,dx + \int_0^1 (f(x)-m)x^{n+1}\,dx $$The first integral clearly vanishes in the limit. For the second integral, we can use the First MVT for integrals to deduce $$ \int _0^1 (f(x)-m)x^{n+1}\,dx = c^{n+1} \int _0^1 f(x)-m\,dx = (a-m)c^{n+1}, $$for $c\in(0,1)$; this also vanishes in the limit.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\lim_{n \to \infty}\, \int_{0}^{1}x^{n + 1}\,\on{f}\pars{x}\,\dd x}:\ {\Large ?}}$.

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As $\ds{n \to \infty}$, the main contribution to the integral comes from $\ds{x \lesssim 1}$. So, I'll make the change $\ds{x \mapsto 1 - x}$ such that the main contribution comes from $\ds{x \gtrsim 0}$. Namely, \begin{align} &\bbox[5px,#ffd]{\lim_{n \to \infty}\, \int_{0}^{1}x^{n + 1}\,\on{f}\pars{x}\,\dd x} \\[5mm] = &\ \lim_{n \to \infty}\, \int_{0}^{1}\pars{1 - x}^{n + 1} \,\on{f}\pars{1 - x}\,\dd x \\[5mm] = &\ \lim_{n \to \infty}\, \int_{0}^{1}\expo{\pars{n + 1}\ln\pars{1 - x}} \,\,\,\on{f}\pars{1 - x}\,\dd x \\[5mm] = &\ \lim_{n \to \infty}\, \int_{0}^{\infty}\expo{-\pars{n + 1}x} \,\,\,\on{f}\pars{1 - 0}\,\dd x \end{align}

where I used the Laplace's Method.

\begin{align} &\mbox{Then,} \\ &\ \bbox[5px,#ffd]{\lim_{n \to \infty}\, \int_{0}^{1}x^{n + 1}\,\on{f}\pars{x}\,\dd x} = \lim_{n \to \infty}\,\,{\on{f}\pars{1} \over n + 1} = \bbx{\large 0} \\ & \end{align}