# approximation of integral of $|\cos x|^p$

Let $$p\in [1,2)$$. Let $$\beta = \frac{1}{2\pi}\int_0^{2\pi} |\cos x|^p\, dx = \frac{\Gamma(\frac{p+1}{2})}{\sqrt{\pi}\Gamma(1+\frac{p}{2})}.$$ Consider the following approximation to the integral definition of $$\beta$$: $$S_n = \frac{1}{2n}\sum_{k=0}^{2n-1} \left| \cos\left(2\pi \cdot \frac{k}{2n}\right)\right|^p.$$ We are interested in the asymptotics of the approximation error $$|S_n - \beta|.$$ I have found empirically that $$|S_n - \beta| \leq \frac{c_p}{n^{p+1}}$$ where $$c_p$$ is a constant that depends only on $$p$$ and $$c_p\to 0$$ as $$p\to 2$$. But I have no idea at all how to prove this. (This bound is actually tight, that is, the correct order should be $$1/n^{p+1}$$.)

I tried to split the integral from $$0$$ to $$2\pi$$ into $$2n$$ pieces and sum up the approximation errors in each piece, which only gives an error of $$O(1/n^2)$$ instead of $$O(1/n^{p+1})$$. Can anyone shed some light on how to deal with the $$p$$-th power or how $$p$$ enters the exponent in the approximation error?

Edit: The following seems to be a proof, up to tiny boundary sloppiness that can be easily fixed.

Assume that $$n$$ is even and so we replace $$n$$ with $$2n$$. Note that $$\cos x$$ is decreasing and positive on $$[0,\pi/2]$$. By symmetry, it suffices to bound \begin{align*} E &:= \frac{1}{2n}\sum_{k=0}^{2n-1} \left|\cos\frac{\pi k}{2n}\right|^p - \frac{1}{\pi}\int_0^\pi |\cos x|^p dx\\ &= \frac{1}{2n}\sum_{k=0}^{n-1} \left(\cos^p\left(\frac{\pi k}{2n}\right) + \cos^p\left(\frac{\pi (k+1)}{2n}\right)\right) - \frac{1}{\pi}\sum_{k=0}^{n-1} 2\int_{k\pi/(2n)}^{(k+1)\pi/(2n)} \cos^p xdx \\ &= \frac{1}{n}\sum_{k=0}^{n-1}\left[ \frac{1}{2}\left(\cos^p\left(\frac{\pi k}{2n}\right) + \cos^p\left(\frac{\pi (k+1)}{2n}\right)\right) - \frac{1}{\pi/(2n)}\int_{k\pi/(2n)}^{(k+1)\pi/(2n)} \cos^p xdx\right]\\ &=: \frac{1}{n}\sum_{k=0}^{n-1} E_k \end{align*} Recall the following mean-value result for the error of trapezoidal rule: $$\frac{1}{b-a}\int_a^b f(x)dx - \frac{f(a)+f(b)}{2} = -\frac{(b-a)^2}{12}f''(\xi)$$

Let $$f(x) = \cos^p x$$, then $$f'(\pi/2-\delta)\asymp \delta^{p-1}$$ and $$f''(\pi/2-\delta)\asymp 1/\delta^{2-p}$$.

Let $$\epsilon \geq 2/n$$ to be determined let $$K = [n\epsilon]$$ so $$K \geq 2$$. Applying the mean-value result to the intervals corresponding to $$k=0,\dots,n-K-1$$, we have $$E = \frac{1}{n}\sum_{k=0}^{n-K-1} \frac{1}{12}\left(\frac{\pi}{2n}\right)^2 f''(\xi_k) + \frac{1}{n}\sum_{k=n-K}^{n-1} E_k =: A + B.$$ where $$\xi_k \in [k\pi/(2n), (k+1)\pi/(2n)]$$.

Write $$A$$ as $$A = \frac{\pi}{24n^2} \sum_{k=0}^{n-K-1} f''(\xi_k)\frac{\pi}{2n} =: \frac{\pi}{24n^2} A'$$ Note that $$A'$$ is the Riemann sum of $$\int_0^{\pi/2-\epsilon} f''(x)dx$$. Also note that $$f''(x)$$ has a unique positive root $$x_0$$ in $$(0,\pi/2)$$ and $$f''(x)$$ is positive and increasing when $$x\geq x_0$$. We can upper bound \begin{aligned} A' &\leq \int_0^{\pi/2-\epsilon} f''(x)dx + \frac{C_1}{n}\max_{x\in [0,x_0]}|f'''(x)| + \frac{C_2}{n}f''\left(\frac{\pi}{2}-\epsilon\right) \\ &\leq f'\left(\frac{\pi}{2}-\epsilon\right) + \frac{C_3}{n} + \frac{C_4}{n\epsilon^{2-p}} \\ &\lesssim \epsilon^{p-1} + \frac{1}{n\epsilon^{2-p}}. \end{aligned}

Next, we deal with the last term $$B$$. Invoke the integral estimation error from Theorem 3 of this paper, which states that, if $$f'$$ is absolutely continuous on $$[a,b]$$ and $$f''\in L^\alpha(a,b)$$ for some $$\alpha\geq 1$$, then $$\left|\frac{1}{b-a}\int_a^b f(x)dx - \frac{f(a)+f(b)}{2}\right| \leq C_\alpha (b-a)^{2-\frac{1}{\alpha}} \|f''\|_\alpha.$$ Since $$f''(\pi/2-\epsilon)\asymp 1/\epsilon^{2-p}$$, $$f''$$ is $$L^1$$ integrable. It is clear that $$E_k > 0$$ when $$k\geq n-K$$, as $$f''(\xi_k) > 0$$ in this case. Applying the error bound above to $$E_k$$ ($$k\geq n-K$$), we obtain that $$B \lesssim \frac{1}{n^2}\int_{\pi/2-\epsilon}^{\pi/2} f''(x)dx \lesssim\frac{1}{n^2}\int_0^{\epsilon} \frac{1}{x^{2-p}}dx \lesssim \frac{\epsilon^{p-1}}{n^2}.$$ Therefore we conclude that $$E \lesssim \frac{\epsilon^{p-1}}{n^2} + \frac{1}{n^3\epsilon^{2-p}}.$$ Taking $$\epsilon = \Theta(1/n)$$ gives the desired result.

• @ClaudeLeibovici I was only asking for $p\in [1,2)$. For even $p$, I think you just get 0 additive error, that is, $c_p = 0$ in this case. Apr 28, 2022 at 5:41
• I am sorry : I misread it. Thanks. Apr 28, 2022 at 5:42
• @user58955 This looks like the trapezoidal rule... Look for the Euler-MacLaurin formula. Apr 28, 2022 at 7:01
• $S_n$ should be the following instead?$$S_n:=\frac\pi{n}\sum_{k=0}^{2n-1}\cos^p\left(\frac{k\pi}{n}\right)$$ Apr 28, 2022 at 7:47
• @Tianlalu Sorry, I should have normalized $\beta$ by $1/(2\pi)$. It's now fixed. Apr 28, 2022 at 8:58

A "Fourier-analytic" approach, with the analysis of $$n^{p+1}|S_n-\beta|$$ as $$n\to\infty$$. Let $$T_N(f):=\frac1N\left(\frac{f(0)+f(2\pi)}{2}+\sum_{k=1}^{N-1}f\Big(\frac{2k\pi}{N}\Big)\right),\\\Delta_N(f):=T_N(f)-\frac1{2\pi}\int_0^{2\pi}f(x)\,dx.$$ Suppose that $$f(x)=\sum_{n\in\mathbb{Z}}c_n e^{inx}$$ with $$\sum_{n\in\mathbb{Z}}|c_n|<\infty$$, then $$T_N(f)=\frac1N\sum_{k=0}^{N-1}\sum_{n\in\mathbb{Z}}c_n e^{2\pi ikn/N}=\sum_{d\in\mathbb{Z}}c_{Nd},$$ because $$\sum_{k=0}^{N-1}e^{2\pi ikn/N}=0$$ if $$n$$ is not a multiple of $$N$$; thus $$\color{blue}{\Delta_N(f)=\sum_{d\neq 0}c_{Nd}}$$. In particular, if $$c_n=O(|n|^{-\alpha})$$ as $$n\to\pm\infty$$ with $$\alpha>1$$, then $$\Delta_N(f)=O(N^{-\alpha})$$ as $$N\to\infty$$.
Take $$f(x)=|\cos x|^p$$ with $$\color{blue}{p>0}$$, then $$c_n=0$$ if $$n$$ is odd, and (see e.g. {1} or {2}) $$c_{2n}=\frac2\pi\int_0^{\pi/2}\cos^p x\cos 2nx\,dx=\frac{\Gamma(p+1)}{2^p\Gamma(p/2+1+n)\Gamma(p/2+1-n)},$$ so that, using the reflection formula for $$\Gamma$$ and that $$\Gamma(x+a)/x^a\Gamma(x)\to1$$ as $$x\to\infty$$, $$\lim_{n\to\infty}(-1)^{n-1}n^{p+1}c_{2n}=\lambda_p:=\frac1{2^p\pi}\Gamma(p+1)\sin\frac{p\pi}{2}.$$
Now, denoting $$a_n:=n^{p+1}(S_n-\beta)=n^{p+1}\Delta_{2n}(f)$$, we get easily $$\lim_{n\to\infty}a_{2n}=-2\lambda_p\zeta(p+1)=2\pi^p\zeta(-p);\\\lim_{n\to\infty}a_{2n+1}=-(1-2^{-p})\lim_{n\to\infty}a_{2n}.$$