Ytukyg

I need to integrate the following expression involving a $delta$-function
$$int_0^1 mathrm{d}x int_0^1 mathrm{d}y int_0^1 mathrm{d}z , delta(x+y+z-1)$$ The textbook I'm using suggests this can be simplified by taking the $z$ integral first, then what remains is
$$int_0^1 mathrm{d}x int_0^{1-x} mathrm{d}y $$

This seems fine to me, although I'm not sure if my understanding is correct: after $z$-integration, the $delta$-function imposes the constraint $x+y=1$, which is just a straight line $y=-x+1$. Graphically, we have the following 2D integration region, which shows that for every allowed $x$, $, 0 le y le 1-x$, which is reflected by the bounds of integration.

I also have to deal with longer versions of the above integral. If we add another variable and integrate over $x_4$, we now have this 3D integration region and so my guess is that:

$$
small int_0^1 mathrm{d}x_1 int_0^1 mathrm{d}x_2 int_0^1 mathrm{d}x_3 int_0^1 mathrm{d}x_4 , delta(x_1+x_2+x_3+x_4-1) =
int_0^1 mathrm{d}x_1 int_0^{1-x_1} mathrm{d}x_2 int_0^{1-x_1-x_2} mathrm{d}x_3
$$

In general, I would like to evaluate this for an arbitrary numer of variables,

$$int_0^1 prod_{i=1}^nmathrm{d}x_i ; delta left(sum_{i=1}^n x_i -1 right)$$

where for $n>4$ such drawings are not possible any more as the integration region becomes 4D.

First of all, I would like to clarify if the two results above for $n=3$ and $n=4$ are correct. Moreover, is there another way of deriving these results, without having to resort to graphical interpretations of the constraint imposed by the $delta$-function?

You are correct.

$ int_0^1 delta(x+y+z-1); dz = 1$ if $0$ is in the interval $(x+y-1, x+y)$, i.e. $0 < x+y < 1$, and $0$ if $x+y > 1$. Since $x+y>0$ is guaranteed when $0 < x < 1$ and $0 < y < 1$, the constraint is $x + y < 1$.

More generally,

$$ int_0^1 delta(x_1 + ldots + x_n - 1); d x_n = 1 text{if} 0 < x_1 + ldots + x_{n-1} < 1$$
so your constraint is $x_1 + ldots + x_{n-1} < 1$, and the integral can be written as

$$int_0^1 dx_1 int_0^{1-x_1} dx_2 ldots int_0^{1-x_1 - ldots - x_{n-2}} dx_{n-1} ; 1$$

A heuristic idea: treat $delta$ as an ordinary function.

Then, the $n$th anti-derivative of $delta(x)$ is
$$underbrace{intcdotsint}_{ntext{ integrals}}delta(x)(dx)^n=frac{x^{n-1}}{(n-1)!}H(x)$$ where $H(x)$ is the Heaviside step function.

(The $+C$ for each indefinite integral is intentionally neglected, as we will be using the result to deal with definite integrals.)

Let $displaystyle{g_n=-1+sum^{n}_{i=1}x_i}$.

$$I_nequivunderbrace{int^1_0cdotsint^1_0}_{ntext{ integrals}}
delta(g_n)left(prod^n_{i=1}dx_iright)
=frac{g_n^{n-1}}{(n-1)!}H(g_n)biggvert^{x_n=1}_{x_n=0}cdotsbiggvert^{x_1=1}_{x_1=0}$$

It can be shown that
$$I_n=frac1{(n-1)!}sum^n_{k=2}(-1)^{n-k}(k-1)^{n-1}P^n_k$$

or equivalently
$$I_{n+1}=(n+1)sum^n_{k=1}(-1)^{n-k}frac{k^n}{(k+1)!}$$