Proof of this definite integral? Saw this sometime in my calculus book, from the Putnam Math Challenges listed:
$$\lim _{ n\rightarrow \infty  }{ \int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \underbrace{\dots}_{n-3 \, times} \int _{ 0 }^{ 1 }{ \cos ^{2} { \left\{ \frac { \pi  }{ 2n } \left({x}_{1}+{ x }_{ 2 }+\dots +{ x }_{ n }\right)\right\}  \quad { dx }_{ 1 }{ dx }_{ 2 }\cdots { dx }_{ n } }  }  }  } = \dfrac{1}{2} } $$ 
Thank you to whoever can help me understand this! : )
 A: Using the transformation $y_1 = 1 - x_1, y_2 = 1 - x_2,\ldots, y_n= 1 - x_n$, your integral becomes
$$\int_0^1\cdots \int_0^1 \cos^2\left\{\frac{\pi}{2n}(n - y_1 - \cdots - y_n)\right\}\, dy_1\ldots dy_n,$$
which is the same as
$$\int_0^1\cdots \int_0^1 \sin^2\left\{\frac{\pi}{2n}(y_1+\cdots + y_n)\right\}\, dy_1\ldots dy_n.$$
So your integral is equal to 
$$\frac{1}{2}\int_0^1\cdots \int_0^1 \left[\cos^2\left\{\frac{\pi}{2n}(x_1 + \cdots + x_n)\right\} + \sin^2\left\{\frac{\pi}{2n}(x_1 + \cdots + x_n)\right\}\right]\, dx_1\cdots dx_n,$$
which reduces to $1/2$. That's why the limit is $1/2$.
A: $$
\cos^2\left(\frac{\pi}{2n}\sum_{i=0}^nx_i\right) = \frac{\cos \left(\frac{\pi}{n}\sum_{i=0}^nx_i\right)+1}{2} = \frac{\cos \left(\frac{\pi}{n}\sum_{i=0}^nx_i\right)}{2} + \frac{1}{2}
$$
thus
$$
\lim_{n\to \infty}\int_0^1..\int_0^1\frac{\cos \left(\frac{\pi}{n}\sum_{i=0}^nx_i\right)}{2} + \frac{1}{2}dx_1..dx_n = \lim_{n\to \infty}\int_{x_1..x_n}\frac{\cos \left(\frac{\pi}{n}\sum_{i=0}^nx_i\right)}{2}dx_1..dx_n + \lim_{n\to \infty}\int_0^1..\int_0^1\frac{1}{2}dx_1..dx_n
$$
then
$$
\cos \left(\frac{\pi}{n}\sum_{i=0}^nx_i\right) = \mathcal{Re}\left(\mathrm{e}^{i\frac{\pi}{n}\sum_{i=0}^nx_i}\right) = \mathcal{Re}\left(\prod_{k=0}^{n}\mathrm{e}^{i\frac{\pi}{n}x_i}\right)
$$
thus the integral
$$
\int_{x_1..x_n}\frac{\cos \left(\frac{\pi}{n}\sum_{i=0}^nx_i\right)}{2}dx_1..dx_n = \int_{x_1..x_n}\mathcal{Re}\left(\prod_{k=0}^{n}\mathrm{e}^{i\frac{\pi}{n}x_i}\right)dx_1..dx_n = \mathcal{Re}\left[\int_0^1\mathrm{e}^{i\frac{\pi}{n}x_1}dx_1..\int_0^1\mathrm{e}^{i\frac{\pi}{n}x_n}dx_n\right]
$$
where we have
$$
\int_0^1\mathrm{e}^{i\frac{\pi}{n}x_n}dx_n = -i\frac{n}{\pi}\left[\mathrm{e}^{i\frac{\pi}{n}}-1\right]\\
 \lim_{n\to\infty}-i\frac{n}{\pi}\left[\mathrm{e}^{i\frac{\pi}{n}}-1\right] = -i\frac{n}{\pi}\left[\mathrm{e}^{0}-1\right] = 0
$$
(note this limit should be proven more rigorously, I have assumed that the exponential tends to one faster than $n\to \infty$)
thus the only nonzero is the integral
$$
\lim_{n\to \infty}\int_0^1..\int_0^1\frac{1}{2}dx_1..dx_n = \lim_{n\to \infty}\frac{1}{2}\int_0^1dx_1..\int_0^1 dx_n = \frac{1}{2}1\cdot1...\cdot 1 = \frac{1}{2}
$$
A: I am not convinced the following is correct. However why not just 
$$s_1 = \frac 1n (x_1 + \dots + x_n), s_2 = x_2, \dots, s_n = x_n$$?
The integral becomes 
$$\int_0^1 \int_0^1 \dots \int_0^1 \cos^2(\pi s_1 / 2)\cdot \frac 1n \ \ \ ds_1 \dots ds_n = \frac 12$$
So it appears to be true always not just as $n \to \infty$
