# Asymptotics of the integral $\int_0^\pi x^n \sin(x)dx$

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)) 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.

• Does writing $x^n = e^{n\log x}$ help? Oct 15, 2018 at 10:12
• Just as $\sin x\approx x$ near $x=0$, $\sin x\approx \pi-x$ near $x=\pi$ Oct 15, 2018 at 10:47
• Thank you both, I see now how to do it Oct 15, 2018 at 11:30
• 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. Oct 15, 2018 at 11:44

## 2 Answers

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

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