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
 A: $\newcommand{\+}{^{\dagger}}
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\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}

\begin{align}
&\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}}
\\[3mm]&=\lim_{n \to \infty}\Re\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1}
\int_{-\infty}^{\infty}\expo{\ic\pi x/n}\delta\pars{x - \sum_{k = 1}^{n}x_{k}}\,\dd x
\,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}
\\[3mm]&=\lim_{n \to \infty}\Re\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1}
\int_{-\infty}^{\infty}\expo{\ic\pi x/n}
\int_{-\infty}^{\infty}\exp\pars{\ic q\bracks{x - \sum_{k = 1}^{n}x_{k}}}
\,{\dd q \over 2\pi}\,\dd x\,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}
\\[3mm]&=\lim_{n \to \infty}\Re\int_{-\infty}^{\infty}\dd q\
\overbrace{\int_{-\infty}^{\infty}{\dd x \over 2\pi}\,
\exp\pars{\ic\bracks{q + {\pi \over n}}x}}^{\ds{=\ \delta\pars{q + {\pi \over n}}}}\
\pars{\int_{0}^{1}\expo{-\ic q\xi}\,\dd\xi}^{n}
\\[3mm]&=\lim_{n \to \infty}
\Re\bracks{\pars{\int_{0}^{1}\expo{\ic\pi\xi/n}\,\dd\xi}^{n}}
=\lim_{n \to \infty}
\Re\bracks{\pars{\expo{\ic\pi/n}  - 1 \over \ic\pi/n}^{n}}
\\[3mm]&=\lim_{n \to \infty}
\Re\bracks{%
\expo{-\ic\pi/2}\pars{\expo{\ic\pi/2n}  - \expo{-\ic\pi/2n} \over \ic\pi/n}^{n}}
=\lim_{n \to \infty}
\Re\braces{\expo{-\ic\pi/2}\bracks{\sin\pars{\pi/2n} \over \pi/2n}^{n}}
\\[3mm]&=\color{#00f}{\large 0}
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\qquad\qquad\pars{2}
\end{align}

By replacing $\pars{2}$ in expression $\pars{1}$ we find:
$$
\color{#00f}{\large\lim_{n \to \infty}\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1}
\cos^{2}\pars{{\pi \over 2n}\,\bracks{x_{1} + x_{2} + \cdots + x_{n}}}
\,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n} = \half}
$$
A: $\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 $
$\displaystyle = \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \sin^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n$
$\displaystyle = \frac12 \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \sin^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big) + \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...+x_n)\big)dx_1 dx_2...dx_n$
$=\dfrac12$
A: 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 $x_k\in(a,b)$.
A: 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 $n$.
A: 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.
