# Putnam Exam Integral: $\lim_{n\to \infty} \int_0^1 \int_0^1…\int_0^1 \cos^2\big(\frac{\pi}{2n}(x_1+x_2+…x_n)\big)dx_1 dx_2…dx_n$

I am trying to evaluate$$\lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n.$$ This is from an old Putnam mathematics competition. Either 1965 or 1987 I forget. Should we re-write the $\cos^2$ term first or how should we approach it? Thanks

• maybe write cosine in the $e^i$ form? – rlartiga Mar 26 '14 at 21:07
• Another idea, you could use this relation: $$\cos^2\Big( \frac{\pi}{2n}(x_1+\ldots + x_n)\Big) = \frac{1}{4}\Big( e^{\frac{i\pi}{n}(x_1 + \ldots + x_n)} + e^{-\frac{i\pi}{n}(x_1 + \ldots + x_n)} \Big) + \frac{1}{2}$$ – Integral Mar 26 '14 at 21:37
• It is a trick question, the expression is constant for all $n:\;\sum x_i\mapsto n-\sum x_i:$ gives $\int_0^1\cdots\int_0^1\cos^{2}\left[\frac{\pi}{2n}\sum x_i\right] \;\mathrm{d}x_1\cdots\mathrm{d}x_n=\int_0^1\cdots\int_0^1\sin^{2}\left[\frac{\pi}{2n}\sum x_i\right] \;\mathrm{d}x_1\cdots\mathrm{d}x_n$. Add. – Julien Godawatta Mar 26 '14 at 22:08
• @JulienGodawatta This is off the Putnam math exam, what do you mean it is a "trick"? This is standard for the exam questions. – Jeff Faraci Mar 27 '14 at 1:43
• Integral's idea works perfectly. Also, if you want some sledgehammer method, just refer to the Strong law of large number. It tells us that $\bar{x}_{n} = \frac{1}{n}(x_{1} + \cdots + x_{n}) \to \frac{1}{2}$ almost surely as $n \to \infty$, where $x_{k}$ are understood as i.i.d. uniform random variables on $[0, 1]$. So by the bounded convergence, the integral converges to $$\Bbb{E}[\cos^{2}(\pi \bar{x}_{n} / 2) ] \to \Bbb{E}[\cos^{2}(\pi/4)] = \frac{1}{2}.$$ – Sangchul Lee Mar 27 '14 at 4:08

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\color{#c00000}{\lim_{n \to \infty}\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1} \cos^{2}\pars{{\pi \over 2n}\,\bracks{x_{1} + x_{2} + \cdots + x_{n}}} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}} \\[3mm]&=\half\bracks{% 1 + \color{#00f}{\lim_{n \to \infty}\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1} \cos\pars{{\pi \over n}\,\bracks{x_{1} + x_{2} + \cdots + x_{n}}} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}}}\tag{1} \end{align}

By replacing $\pars{2}$ in expression $\pars{1}$ we find: $$\color{#00f}{\large\lim_{n \to \infty}\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1} \cos^{2}\pars{{\pi \over 2n}\,\bracks{x_{1} + x_{2} + \cdots + x_{n}}} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n} = \half}$$

• Julien Godawatta notes the addendum in a comment. For some reason, their answer was self-deleted. – robjohn Mar 31 '14 at 9:11
• @robjohn I just deleted the ADDENDUM. I didn't see the JG comment but now I just checked it is alive. Thanks. – Felix Marin Mar 31 '14 at 9:32
• I wasn't suggesting that you delete the addendum. There are many cases where a comment gets promoted to an answer by someone other than the commenter. Good ideas should be in answers, not comments since comments are ephemeral. – robjohn Mar 31 '14 at 9:44
• @FelixMarin Thank you for always answering my integrals:) I am surprised by your solution though! I do not see any $\partial_\mu$ ;] – Jeff Faraci Mar 31 '14 at 16:47
• @Jeff It was not needed. Thanks. – Felix Marin Mar 31 '14 at 17:34

Using $$\cos^2(x)=\frac{1+\cos(2x)}{2}$$ we get that \begin{align} &\int_0^1\int_0^1\cdots\int_0^1\cos^2\left(\frac{a\pi}{2n}(x_1+x_2+\dots+x_n)\right)\,\mathrm{d}x_1\,\mathrm{d}x_2\dots\,\mathrm{d}x_n\\ &=\frac12+\frac12\mathrm{Re}\left(\int_0^1\int_0^1\cdots\int_0^1e^{\frac{ia\pi}{n}(x_1+x_2+\dots+x_n)}\,\mathrm{d}x_1\,\mathrm{d}x_2\dots\,\mathrm{d}x_n\right)\\ &=\frac12+\frac12\mathrm{Re}\left(\left[\int_0^1e^{\frac{ia\pi}{n}x}\,\mathrm{d}x\right]^n\right)\\ &=\frac12+\frac12\mathrm{Re}\left(\left[\frac{n}{ia\pi}\right]^n\left[e^{\frac{ia\pi}{n}}-1\right]^n\right)\\ &=\frac12+\frac12\mathrm{Re}\left(\left[\frac{2n}{a\pi}\sin\left(\frac{a\pi}{2n}\right)\right]^ne^{\frac{ia\pi}{2}}\right)\\ &=\frac12+\frac12\left[\frac{2n}{a\pi}\sin\left(\frac{a\pi}{2n}\right)\right]^n\color{#C00000}{\cos\left(\frac{a\pi}{2}\right)}\\ &\to\frac12+\frac12\cos\left(\frac{a\pi}{2}\right)\\ &=\cos^2\left(\frac{a\pi}{4}\right) \end{align} If $a=1$, $\color{#C00000}{\cos\left(\frac{a\pi}{2}\right)}$ is $0$, so the integral is $\frac12$ for all $n$.

• Very nice solution. Thanks very much!! – Jeff Faraci Mar 31 '14 at 16:46
• Nice solution! (+1) – user 1357113 Mar 31 '14 at 18:56

Hint: $\displaystyle\frac1n\sum_1^nx_k$ is the mean value of $\bar x$, which, for $n\to\infty$, tends to $\dfrac{a+b}2$ , for $x_k\in(a,b)$.

• Ok, can you elaborate? This is more of a comment. Thanks – Jeff Faraci Mar 27 '14 at 2:31
• What are a and b in our case ? Between which values does each $x_k$ vary ? – Lucian Mar 27 '14 at 2:34

$\displaystyle \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n$

$\displaystyle = \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\big(\frac{\pi}{2n}(1-x_1+1-x_2+...+1-x_n)\big)dx_1 dx_2...dx_n$

$\displaystyle = \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \sin^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n$

$\displaystyle = \frac12 \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \sin^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big) + \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...+x_n)\big)dx_1 dx_2...dx_n$

$=\dfrac12$

I have checked the book of Putnam mathematics competition and your question is a little bit different from the original topic.

Since the following relation holds

$$\cos^{2}(π/2n(x_1+\dots+x_n))=\frac14 (e^{iπ/n}(x_1+\dots+x_n)+e^{−iπ/n}(x_1+\dots+x_n))+\frac12$$

it seems that it is not so hard to integrate your function:

$$\lim_{n\to \infty} \int_0^1 \int_0^1\cdots\int_0^1 \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)\,dx_1\, dx_2\dots\,dx_n.$$

you can use the following formulas:

$$\int{dx_{1}} \int{dx_{2}}\dots\int{f(x_{1})f(x_{2})...f(x_{n})}{dx_{n}} = \frac{1}{n!} (\int{f(\tau)}d\tau)^{'}$$

where, $f(x)=e^{i\pi/n}x$.

It seems that this method will be easy to use.