EDIT: Thanks for your answers and comments. There is indeed a classical easy formula, given by Pietro Majer (with a simple nice proof) in his answer below.




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Given $x\in\mathbb{R}^n$, $x\_i$ denotes its $i$-th coordinate. My question is:


* What is $Vol(\{x\in[0,1]^n|\sum\_{i=1}^n x\_i\le t\})$ for $t\in\mathbb{R}$ ? Is there some kind of "easily computable" formula for it ?


A colleague asked me this in relation to a problem in applied mathematics, and I thought that the answer was going to be easy to find, but after trying to compute it by hand in dimension $3$ and then generalizing, I just surrendered. I am pretty sure that it should be somehow classical, but I could not track down a formula or a reference in the general case. I should specify that I know next to nothing about section of convex bodies, and I might be missing something easy. With some browsing, I found the following partial answers (which I hope I do not mix-up):


* If $t$ is an integer, then the volume is the sum of the $t$ first Eulerian numbers divided by $n!$.
* Between and integer $t<n$ and the next, the volume function is a polynomial of degree $n$. The polynomials corresponding to two consecutive intervals must have the same first derivative at their common point. For instance, for $n=2$ the volume is given by $\frac{t^2}{2}$ if $t\in[0,1]$, $1 - \frac{(2-t)^2}{2}$ for $t\in[1,2]$ and constant elsewhere.
* There is a Fourier-analysis approach for the problem of finding the $n-1$-volume of the section of a convex body by an hyperplane, but the formulas I found did not seem to me easily computable (my knowledge of Fourier analysis is approximatively equivalent to that of convex bodies, I must say).
* We can use some probability theory to find an approximation: each $x\_i$ is seen as a random variable following a law of mean $\mu$ and variance $\nu$, then $S=\sum x\_i$ follows (approximatively and for large $n$) the normal law with mean $n\mu$ and variance $n^2\nu$, and the probability of $S\le t$ is then easy to compute. I presume that the meaning of "$n$ is large" and "approximate" can be made explicit by looking more closely at the central limit theorem. (This idea comes from Michael Lugo's answer to [this related question](https://mathoverflow.net/questions/22648/area-of-cross-section-at-midpoint-perpendicular-to-longest-diagonal-in-the-uni/22670#22670). Another answer by Andrey Rekalo points to the Fourier-analysis method mentioned above.)


The last point brings a more general question: given a probability distribution on $[0,1]$ (perhaps in some well behaved class), are there some good estimates for $P(\sum\_{i=1}^n x\_i\le t)$ for smaller $n$ ? 



What's the volume of $\{x\in[0,1]^n|\sum x_i\le t\}$ for real $t$?
The volume $V\_n(t)$ is the cdf of the sum of $n$ independent random variables, uniformly distributed on $[0,1]$. So the density $V\_n'(t)$ is the $n$-fold convolution of the characteristic function of the unit interval, $\chi\_{[0,1]}$. (In particular $V\_n'\in C^{n-2,1}(\mathbb{R})$, with support in $[0,n]$, and polynomial of degree less than $ n $ on any interval between consecutive integers). It is convenient to write $\chi\_{[0,1]}(t)=H(t)-H(t-1)$, where $H(t):=\chi\_{\mathbb{R\_+}}(t)$ is the Heaviside function, or $\chi\_{[0,1]}=H-\tau^1 H$
where $\tau^s:f\mapsto f(\cdot-s)$ is the translation semigroup on functions. Recall that convoluting with $H$ is taking a primitive: $(H\*f)(x)=\int\_{-\infty}^xf(t)dt$, so in particular $H^{\*(n+1)}(t)=t\_+^n/n!$, and that $\tau^s(f\*g)=\tau^sf\*g= f\* \tau^s g$.


Thus $$V\_n=H\*V'\_n=H\*(H-\tau^1H)^{\*n}=\sum\_{k=0}^n(-1)^k{n\choose k}H\*H^{\*(n-k)}\*(\tau^1H)^{\*k}$$
$$=\sum\_{k=0}^n(-1)^k{n\choose k} \tau^kH^{\*(n+1)},$$ 
that is $$V\_n(t)=\sum\_{k=0}^n\frac{(-1)^k}{k!(n-k)!} {(t-k)\_+}{^n}\, .$$ 


