Asymptotics of $\int^{\pi/2}_0 xf(x)(\cos x)^n\,dx$

Question

The following is an old Analysis qualifying problem I am tried to solve but I have not been able to solve rigorously.

Suppose $f$ is continuous in $[0,\pi/2]$. A simple application of dominated convergence shows that $$I_n=\int^{\pi/2}_0 x f(x) (\cos x)^n\, dx \rightarrow0\quad\text{as}\quad n\rightarrow\infty$$ The problem asks to show that $$I_n\sim \frac{1}{n}f(0)+o(1/n)$$

This integral seems to have similarities with integrals where the Laplace method can be use. The function $\cos$ has maximum at $x=0$ in $[0,1]$, so it is not surprise that the values of $x\in[0,\pi/2]$ that contribute to $I_n$ are those near $0$. On the other hand, $\sin x\sim x$ as $x\rightarrow0$. This suggest that $$I_n\sim \int^{\pi/2}_0f(0)\sin x(\cos x)^n\,dx=\frac{1}{n+1}f(0)$$ I can't find a way to put things into a rigorous argument. Any help is appreciated. Thank you!

Answer 1

We have $$I_n = f(0)\int_0^{\pi/2} x(\cos x)^n\,\mathrm{d}x + \int_0^{\pi/2} x(f(x) - f(0))(\cos x)^n \,\mathrm{d} x.$$

Let $$K_n := \int_0^{\pi/2} x(\cos x)^n\,\mathrm{d}x, \quad J_n := \int_0^{\pi/2} x(f(x) - f(0))(\cos x)^n \,\mathrm{d} x.$$

Using $x \ge \sin x$ for all $x \ge 0$, we have $$K_n \ge \int_0^{\pi/2} \sin x \, (\cos x)^n\,\mathrm{d} x = \frac{1}{n + 1}.$$

Using $x \le \sin x + \sin^3 x$ on $[0, \pi/2]$ (easy to prove using $\sin x \ge x - x^3/6$), we have $$K_n \le \int_0^{\pi/2} \left(\sin x + \sin^3 x\right) (\cos x)^n\,\mathrm{d} x = \frac{1}{n+1} + \frac{2}{(n+1)(n+3)}.$$

Thus, we have $$K_n = \frac1n + o(1/n).$$

On the other hand, we have \begin{align*} |J_n| &\le \int_0^{\pi/2} x\, |f(x) - f(0)|\, (\cos x)^n\,\mathrm{d} x\\ &\le \int_0^{\pi/2} 2\sin x\, |f(x) - f(0)|\, (\cos x)^n\,\mathrm{d} x\\ &= \int_0^{1/\sqrt[4]{n}} 2\sin x\, |f(x) - f(0)|\, (\cos x)^n\,\mathrm{d} x + \int_{1/\sqrt[4]{n}}^{\pi/2} 2\sin x\, |f(x) - f(0)|\, (\cos x)^n\,\mathrm{d} x\\ &\le a_n \int_0^{1/\sqrt[4]{n}} 2\sin x (\cos x)^n\,\mathrm{d} x + M \int_{1/\sqrt[4]{n}}^{\pi/2} 2\sin x(\cos x)^n\,\mathrm{d} x\\ &\le a_n \int_0^{\pi/2} 2\sin x (\cos x)^n\,\mathrm{d} x + M \int_{1/\sqrt[4]{n}}^{\pi/2} 2\sin x(\cos x)^n\,\mathrm{d} x\\ &= \frac{2a_n}{n + 1} + \frac{2M}{n + 1}\left(\cos \frac{1}{\sqrt[4]{n}}\right)^{n+1}\\ &\le \frac{2a_n}{n + 1} + \frac{2M}{n + 1}\left(1 - \frac{1}{4\sqrt n}\right)^{n+1}\\ &= o(1/n) \end{align*} where we use $x \le 2\sin x$ for all $x\in [0, \pi/2]$ (easy to prove), and $$a_n := \max_{x\in [0, 1/\sqrt[4]{n}]} |f(x) - f(0)|, \quad M := \max_{x\in [0, \pi/2]} |f(x) - f(0)|,$$ and we use $\lim_{n\to \infty} a_n = 0$ and $\cos u \le 1 - \frac14 u^2$ for all $u\in [0, \pi/2]$ (easy to prove).

We are done.

Answer 2

I've often found that with the seemingly more "elementary" questions the more powerful measure-theoretic theorems become less useful and what I consider more "old-fashioned" approaches are easier to use. I'd be interested to see a more direct proof via the (generalised) dominated convergence theorem etc.

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I assume only that: (i) $f$ is bounded (ii) $f$ is continuous at zero (iii) $f$ is measurable.

$\newcommand{\d}{\,\mathrm{d}}$By definition of $o(1/n)$, it's the same to show that:

$$\lim_{n\to\infty}nI_n=f(0)$$

Let $g$ be any bounded measurable function on $[0,\pi/2]$, bounded say by $C$. Then notice for $0<\delta<\pi/2$ it is true that: $$\begin{align}\left|\int_\delta^{\pi/2}(n+1)x\cos^n(x)g(x)\d x\right|&\le\frac{\pi\cdot C}{2}\int_\delta^{\pi/2}(n+1)\sin(x)\cos^n(x)\d x\\&=\frac{\pi\cdot C}{2}\cdot\cos^{n+1}(\delta)\\&\to0\end{align}$$

Fix $\epsilon>0$. There exists $0<\delta<\pi/2$ with $|f(x)-f(0)|<\epsilon$ whenever $0\le x<\delta$. For any $n\in\Bbb N$: $$\begin{align}\left|\int_0^{\pi/2}(n+1)x\cos^n(x)[f(x)-f(0)]\d x\right|&\le\frac{\pi\cdot\epsilon}{2}\int_0^{\delta}(n+1)\sin(x)\cos^n(x)\d x+o(1)\\&=\frac{\pi\cdot\epsilon}{2}[1-\cos^{n+1}(\delta)]+o(1)\\&\to\frac{\pi\cdot\epsilon}{2}\end{align}$$

Where the first $o(1)$ term arises from the decay of the integral $\int_\delta^{\pi/2}\cdots$ with $g:x\mapsto f(x)-f(0)$.

Then we have deduced that, for arbitrary $\epsilon>0$:

$$\limsup_{n\to\infty}\left|\int_0^{\pi/2}(n+1)x\cos^n(x)[f(x)-f(0)]\d x\right|\le\frac{\pi\cdot\epsilon}{2}$$Implying that the limit superior equals zero, hence: $$\lim_{n\to\infty}(n+1)\int_0^{\pi/2}x\cos^n(x)[f(x)-f(0)]\d x=0$$

Since $\lim_{n\to\infty}I_n=0$ to show the result it now only suffices to show that: $$\tag{$\ast$}1=\lim_{n\to\infty}(n+1)\int_0^{\pi/2}x\cos^n(x)\d x$$

We use a similar idea. Fix any $0<\delta<\pi/2$. Then for any $n\in\Bbb N$: $$\begin{align}\int_0^{\pi/2}(n+1)x\cos^n(x)\d x&=\int_0^\delta(n+1)x\cos^n(x)\d x+\int_\delta^{\pi/2}(n+1)x\cos^n(x)\d x\\&=\int_0^\delta(n+1)(x-\sin x)\cos^n(x)\d x\\&\quad\quad+\int_0^\delta(n+1)\sin(x)\cos^n(x)\d x+o(1)\\&=\int_0^\delta(n+1)(x-\sin x)\cos^n(x)\d x\\&\quad\quad+[1-\cos^{n+1}(\delta)]+o(1)\\\tag{$\ast\ast$}&=1+o(1)+\int_0^\delta(n+1)(x-\sin x)\cos^n(x)\end{align}$$

And we can estimate for any $n\in\Bbb N$, using the same tricks combined with $x-\sin x\le\frac{x^3}{6}=x^2\cdot\frac{x}{6}$: $$\begin{align}\left|\int_0^\delta(n+1)(x-\sin x)\cos^n(x)\d x\right|&\le\frac{\pi}{12}\int_0^\delta x^2\cdot(n+1)\sin(x)\cos^n(x)\d x\\&\le\frac{\pi\cdot\delta^2}{12}\cdot[1-\cos^{n+1}(\delta)]\end{align}$$

Since $\delta$ is arbitrarily taken, we can conclude using similar arguments to before and equation $(\ast\ast)$ that $(\ast)$ is correct and thus that $I_n=n^{-1}f(0)+o(n^{-1})$ as $n\to\infty$.

Answer 3

Notice that $$\lim_{x\rightarrow0}\Big|\frac{x f(x)}{\sin x}-f(0)\Big|=0$$ Given $\varepsilon>0$, choose $0<\delta<\pi/2$ so that $$|x f(x)-f(0)\sin x|<\varepsilon\sin x\quad\text{whenever}\quad 0<x\leq\delta$$ Then \begin{align} \Big|I_n- f(0)\int^{\pi/2}_0\sin(x)\,\cos^n(x)\,dx\Big|&\leq\int^{\pi/2}_0\big|x f(x)-f(0)\sin x\big|\cos^n(x)\,dx\\ &\leq \varepsilon\int^{\delta}_0\sin(x)\cos^n(x)\,dx +\\ &\qquad \pi\|f\|_\infty \int^{\pi/2}_\delta\cos^n(x)\,dx\\ &\leq \varepsilon\frac{1-\cos^{n+1}(\delta)}{n+1} + \frac{\pi^2}{2}\|f\|_\infty\cos^n(\delta) \end{align} Consequently $$n\big|I_n-f(0)\int^{\pi/2}_0\sin(x)\,\cos^n(x)\,dx\big|\leq\frac{\varepsilon n}{n+1}(1-\cos^{n+1}(\delta))+\frac{\pi^2}{2}n\cos^n(\delta)\xrightarrow{n\rightarrow\infty}\varepsilon$$

In other words $$\limsup_n n\big|I_n-\frac{f(0)}{n+1}\big|\leq\varepsilon$$ for all $\varepsilon>0$. Consequently $$ I_n=\frac{f(0)}{n+1}+o\big(\frac1n\big)=\frac{f(0)}{n}+ o\big(\frac1n\big)$$

Answer 4

If you let $g(x) = f(x){x \over \sin x}$, then $g(x)$ extends to a continuous function on $[0,\pi/2]$ with $g(0) = f(0)$ and $I_n$ can be written as $$I_n = \int^{\pi/2}_0 g(x)\sin x (\cos x)^n\,dx$$ Since $g(0) = f(0)$, it suffices to prove that $$I_n\sim \frac{1}{n}g(0)+o(1/n)$$ Write $g(x) = g(0) + h(x)$, where $h(x)$ is continuous with $h(0) = 0$. Then $$I_n = \int^{\pi/2}_0 g(0)\sin x (\cos x)^n\,dx + \int^{\pi/2}_0 h(x)\sin x (\cos x)^n\,dx$$ The first term is immediately evaluated through a $u$ substitution to ${1 \over n + 1}g(0) = {1 \over n} g(0) + o(1/n)$. Thus it suffices to show that $$\int^{\pi/2}_0 h(x)\sin x (\cos x)^n\,dx = o(1/n) \tag{1}$$ Since $h(x)$ is continuous with $h(0) = 0$, given $\epsilon > 0$ there's a $\delta > 0$ such that $x < \delta$ implies $|h(x)| < \epsilon$. We then have $$\bigg|\int^{\pi/2}_0 h(x)\sin x (\cos x)^n\,dx\bigg| \leq \epsilon \int^{\delta}_0 \sin x (\cos x)^n\,dx + \int^1_{\delta}|h(x)|\sin x (\cos x)^n\,dx $$ Letting $M = \sup_{[0,{\pi \over 2}]} |h(x)|$, the above is bounded by $$\epsilon \int^{\delta}_0 \sin x (\cos x)^n\,dx + M \int^1_{\delta}\sin x (\cos x)^n\,dx $$ $$= \epsilon {1 - (\cos \delta)^{n + 1} \over n + 1} + M {(\cos \delta)^{n + 1} \over n + 1} $$ For large enough $n$ this is at most ${2 \epsilon \over n + 1}$ since $0 < \cos \delta < 1$. Since $\epsilon > 0$ was arbitrary, we have that equation $(1)$ holds as desired.

Answer 5

To the many posted nice solutions it would be interesting to add one more, which exploits the dominated convergence theorem. $$I_n=\int^{\pi/2}_0 x f(x) \cos^n x\, dx\overset{t=nx^2}{=}\frac1{2n}\int_0^{(\frac\pi2)^2n}f\left(\sqrt \frac tn\right)\cos^n\left(\sqrt \frac tn\right)dt$$ Now, $$\cos\left(\sqrt \frac tn\right)=\left(1-\frac t{2n}+\frac{t^2}{24n^2}-...\right)=-\frac t{2n}-\frac t{\sqrt{24}n}+1+\frac t{\sqrt{24}n}-\frac{t^2}{24n^2}+-...$$ $$<-\frac t{2n}-\frac t{\sqrt{24}n}+1+\frac t{\sqrt{24}n}+\frac{t^2}{24n^2}+\frac{t^3}{(\sqrt{24}n)^3}...=-\frac t{2n}-\frac t{\sqrt{24}n}+\frac1{1-\frac t{\sqrt{24}n}}$$ $$=1+\frac{-\frac t{2n}+\frac t{2n}\left(\frac t{12n}+\frac t{\sqrt{24}n}\right)}{1-\frac t{\sqrt{24}n}}$$ Using the fact that at $t\in[0;(\frac\pi2)^2n]\quad\left(\frac1{12}+\frac1{\sqrt{24}}\right)\frac tn-1<0$ $$\cos\left(\sqrt \frac tn\right)<1-\frac t{2n}\left(1-\big(\frac\pi2\big)^2\Big(\frac1{12}+\frac1{\sqrt{24}}\Big)\right)=1-a\cdot\frac t{2n}$$ It is easy to check that $$a=1-\big(\frac\pi2\big)^2\Big(\frac1{12}+\frac1{\sqrt{24}}\Big)=0.29072... \,,\,\,\text{and}\,\,a\frac t{2n}\in[0;1)\,\,\text{for}\,\,t\in[0;(\pi/2)^2n]$$ It follows that $$\cos^n\left(\sqrt \frac tn\right)<\left(1-a\,\frac t{2n}\right)^n=e^{n\ln\left(1-a\,\frac t{2n}\right)}=e^{n\left(-a\,\frac t{2n}-a^2\frac{t^2}{8n^2}-...\right)}<e^{-\frac{at}2}$$ Hence, we see that the set of functions $g_n(t)=f\left(\sqrt \frac tn\right)\cos^n\left(\sqrt \frac tn\right)\bf{1}_{[0;(\frac\pi2)^2n]}$ is dominated by $$G(t)=\max_{x\in[0;1]}f(x)\cdot e^{-\frac{at}2}$$ and we are allowed to take the limit under the integral sign: $$\lim_{n\to\infty}nI_n=\frac12\lim_{n\to\infty}\int_0^\infty f\left(\sqrt \frac tn\right)\cos^n\left(\sqrt \frac tn\right)\bf{1}_{[0;(\frac\pi2)^2n]}dt$$ $$=\frac{f(0)}2\int_0^\infty \lim_{n\to\infty}\cos^n\left(\sqrt \frac tn\right)\bf{1}_{[0;(\frac\pi2)^2n]}dt$$ $$=\frac{f(0)}2\int_0^\infty e^{-\frac t2}dt=f(0)$$