Q:

Hypervolume of the biggest cross-section of a hypercube

Dealing with this question I first gave an erroneous answer, that, however, was quite interesting, so I am willing to share its content in a separate question.
Let $H$ be the hypercube in $\mathbb{R}^{n+1}$ having vertices in $(\pm 1,\pm 1,\ldots,\pm 1)$.
Let $S$ be the cross section given by $H\cap\left\{(x_1,\ldots,x_{n+1})\in\mathbb{R}^{n+1}:\sum_{j=1}^{n+1} x_j = 0\right\}.$
Compute the hypervolume of $S$ as a $n$-dimensional polytope and show that, in such a context, the Central Limit Theorem gives a really simple approximation for a quite ugly sign-alternating sum.

A:

We can write the hypervolume of $S$ in the following way:
$$ V_n = \sqrt{n+1}\int_{x_1,\ldots,x_n\in I}\mathbb{1}_I(x_1+\ldots+x_n)\,d\mu, $$
where $I=(-1,1)$. The integral is just $2^n$ times the probability that the sum of $n$ indipendent random variables, uniformly distributed over $I$, lies in $I$. 
By assuming that $X_i$ is a uniformly distributed random variable over $I$, its characteristic function is given by $\frac{\sin t}{t}$. This gives that the characteristic function of $X_1+\ldots+X_n$ is given by $\left(\frac{\sin t}{t}\right)^n$, so:
$$\begin{eqnarray*}\mathbb{P}[X_1+\ldots+X_n\in I]&=&\frac{1}{2\pi}\int_{-1}^{1}\int_{-\infty}^{+\infty}\left(\frac{\sin t}{t}\right)^n \cos(xt)\,dt\,dx\\&=&\frac{1}{\pi}\int_{-\infty}^{+\infty}\left(\frac{\sin t}{t}\right)^{n+1}dt\end{eqnarray*}$$
and the last integral can be computed through standard complex analytic tools like the residue theorem, or by exploiting integration by parts:
$$\int_{-\infty}^{+\infty}\left(\frac{\sin t}{t}\right)^{n+1}dt=\frac{1}{n}\int_{-\infty}^{+\infty}\frac{\frac{d}{dt}(\sin t)^{n+1}}{t^n}\,dt = \frac{1}{n!}\int_{-\infty}^{+\infty}\frac{\frac{d^n}{dt^n}(\sin t)^{n+1}}{t}\,dt.$$
Since for any $m\in\mathbb{N}^*$ we have $\int_{\mathbb{R}}\frac{\sin(mx)}{x}=\pi$ and:
$$(\sin t)^{n+1}=\frac{1}{(2i)^{n+1}}\sum_{j=0}^{n+1}\binom{n+1}{j}(-1)^j e^{(n+1-2j)it},$$
$$\frac{d^n}{dt^n}(\sin t)^{n+1}=\frac{1}{2^n}\sum_{j=0}^{\lfloor(n+1)/2\rfloor}\binom{n+1}{j}(-1)^j(n+1-2j)^n \sin((n+1-2j)t)$$
it follows that:

$$ V_n = \frac{\sqrt{n+1}}{n!}\sum_{j=0}^{\lfloor(n+1)/2\rfloor}\binom{n+1}{j}(-1)^j (n+1-2j)^n.$$

By approximating $\frac{\sin t}{t}\approx e^{-t^2/6}$, i.e. by using the Central Limit Theorem, we get the really nice approximation:

$$V_n\approx 2^n\sqrt{\frac{6}{\pi}}.$$

