1
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Does writing $x^n = e^{n\log x} $ help? $\endgroup$
    – Shashi
    Oct 15, 2018 at 10:12
  • $\begingroup$ Just as $\sin x\approx x$ near $x=0$, $\sin x\approx \pi-x$ near $x=\pi$ $\endgroup$
    – Empy2
    Oct 15, 2018 at 10:47
  • $\begingroup$ Thank you both, I see now how to do it $\endgroup$
    – Madarb
    Oct 15, 2018 at 11:30
  • 1
    $\begingroup$ Keeping both the endpoints as references, $\frac{\sin(x)}{x(\pi-x)}$ is bounded between $\frac{1}{\pi}$ and $\frac{4}{\pi^2}$ for any $x\in(0,\pi)$, i.e. $\sin(x)$ has an approximately-parabolic behaviour over the first half-period. $\endgroup$ Oct 15, 2018 at 11:44

2 Answers 2

4
$\begingroup$

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 $\int_{0}^{\pi}x^\alpha(\pi-x)^{\beta}\,dx = \pi^{\alpha+\beta+1}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)}$.

$\endgroup$
1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.