Evaluate $$\lim_{n \to \infty} n \int_0^1 (\cos x - \sin x)^n dx$$

The answer should be $1$.

I tried to solve it similar to how one user solved this integral Limit of integral with cos and sin

but it seems like it doesn't work because the upper bound will become $1 - \pi/4$.

  • $\begingroup$ What upper bound is $1-\pi/4$? Mind explaining what you mean? $\endgroup$ May 4, 2017 at 13:34

3 Answers 3


If you know how to deal with

$$n \int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx,$$

just split the integral at $\frac{\pi}{4}$ and see whether you can find something useful for the other part.

Since $x \mapsto \cos x - \sin x$ is strictly decreasing on $\bigl[0, \frac{\pi}{2}\bigr]$, and $\cos \frac{\pi}{4} = \sin \frac{\pi}{4}$, we have

$$\lvert \cos x - \sin x\rvert \leqslant c := \sin 1 - \cos 1$$

for $\frac{\pi}{4} \leqslant x \leqslant 1$. Hence

$$\Biggl\lvert \int_{\frac{\pi}{4}}^1 (\cos x - \sin x)^n\,dx\Biggr\rvert \leqslant \int_{\frac{\pi}{4}}^1 c^n\,dx < c^n.$$

Since $0 < c < 1$, it follows that

$$n \int_{\frac{\pi}{4}}^1 (\cos x - \sin x)^n\,dx \to 0.$$

That is useful indeed. Thus we need only consider

$$\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx = 2^{n/2} \int_0^{\frac{\pi}{4}} \sin^n \biggl(\frac{\pi}{4} - x\biggr)\,dx = 2^{n/2}\int_0^{\frac{\pi}{4}} \sin^n x\,dx.$$

Substituting $u = \sin x$, then $v = \sqrt{2}\cdot u$, and afterwards integrating by parts, we find (for $n \geqslant 1$)

\begin{align} n\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx &= n 2^{n/2} \int_0^{\frac{\pi}{4}} \sin^n x\,dx \\ &= n 2^{n/2} \int_0^{\frac{1}{\sqrt{2}}} \frac{u^n}{\sqrt{1-u^2}}\,du \\ &= n \int_0^1 \frac{v^n}{\sqrt{2-v^2}}\,dv \\ &= v^n\cdot \frac{v}{\sqrt{2-v^2}}\biggr\rvert_0^1 - \int_0^1 v^n\biggl(\frac{1}{\sqrt{2-v^2}} + \frac{v^2}{(2-v^2)^{3/2}}\biggr)\,dv \\ &= 1 - \int_0^1 \frac{2v^n}{(2-v^2)^{3/2}}\,dv. \end{align}

Since $\frac{1}{\sqrt{2}} < \frac{2}{(2-v^2)^{3/2}} < 2$ for $0 < v < 1$, we thus have

$$1 - \frac{2}{n+1} < n\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx < 1 - \frac{1}{\sqrt{2}\,(n+1)}.$$


$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Laplace Method:

\begin{align} &\lim_{n \to \infty}\braces{% n\int_{0}^{1}\bracks{\cos\pars{x} - \sin\pars{x}}^{\,n}\,\dd x} = \lim_{n \to \infty}\pars{% n\int_{0}^{\infty}\expo{-nx}\,\dd x} = \bbx{1} \end{align}


This answer is base on integration by parts and Lebesgue's dominated convergence theorem.

If $0<x<1$ then $-1 < \cos(x)-\sin(x)<1$ and $\displaystyle \lim_{n\to \infty}(\cos(x)-\sin(x))^n = 0$. Thus

\begin{gather*} n\int_{0}^{1}(\cos(x)-\sin(x))^n\, dx =\\[2ex] n\int_{0}^{1}(\cos(x)-\sin(x))^n(-\sin(x)-\cos(x))\dfrac{-1}{\sin(x)+\cos(x)}\, dx =\\[2ex] \left[\dfrac{n}{n+1}(\cos(x)-\sin(x))^{n+1}\dfrac{-1}{\sin(x)+\cos(x)}\right]_{0}^{1}-\\[2ex]\dfrac{n}{n+1}\int_{0}^{1}(\cos(x)-\sin(x))^{n+2}\dfrac{1}{(\sin(x)+\cos(x))^2}\, dx = \\[2ex] \dfrac{n}{n+1} -\dfrac{n}{n+1}(\cos(1)-\sin(1))^{n+1}\dfrac{1}{\sin(1)+\cos(1)}-\\[2ex]\dfrac{n}{n+1}\int_{0}^{1}(\cos(x)-\sin(x))^{n+2}\dfrac{1}{(\sin(x)+\cos(x))^2}\, dx \to 1-0-0 = 1, \quad n \to \infty . \end{gather*}


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