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I am going through de bruijn's book on asymptotic methods. In the end of a chapter on Laplace's method for integrals, there is an exercise to show the following asymptotic: $$\int_0^\pi x^n\sin(x)dx\sim \frac{\pi^{n+2}}{n^2}, n\to\infty $$ I couldn't relate this to the examples in the chapter, where he dealt mainly with integrals of the form $\int_I e^{-tx^2}f(x)dx $, where $t>0$ a real number, and where that width of the interval contribuiting the most for the result was small (here it is of constant length to my understanding, $[1,\pi]$). However I manged to show (using the obviouse bound $\sin(x)<x$) a weaker result. I tried to read through the chapter again, but I have no idea how to do better here.

I would very appreciate any hints or sketches of solution.

Fabio Lucchini
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Madarb
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2 Answers2

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For any $x\in(0,\pi)$ we have

$$ \frac{\sin x}{x(\pi -x)} = \frac{1}{\pi}+K x(\pi-x),\qquad K\in\left[\frac{1}{\pi^3},\frac{4(4-\pi)}{\pi^4}\right] \tag{1}$$ hence

$$ \int_{0}^{\pi}x^n\sin(x)\,dx = \frac{\pi^{n+2}}{(n+2)(n+3)}+\pi^n O\left(\frac{1}{n^3}\right)\tag{2}$$ due to $\int_{0}^{\pi}x^\alpha(\pi-x)^{\beta}\,dx = \pi^{\alpha+\beta+1}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)}$.

Jack D'Aurizio
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\pi}x^{n}\sin\pars{x}\,\dd x} = \int_{0}^{\pi}\pars{\pi - x}^{n}\sin\pars{x}\,\dd x \\[5mm] = &\ \pi^{n + 1}\int_{0}^{1}\pars{1 - x}^{n}\sin\pars{\pi x}\,\dd x \\[5mm] = &\ \pi^{n + 1}\int_{0}^{1}\exp\pars{n\ln\pars{1 - x}} \sin\pars{\pi x}\,\dd x \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,& \pi^{n + 1}\int_{0}^{\infty}\expo{-nx}\pars{\pi x}\dd x = \bbx{\pi^{n +2} \over n^{2}} \end{align}

Felix Marin
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