Limits of integration on a delta function of many arguments

Question

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,

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

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?

Answer 1

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

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

Answer 2

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

Then, the $n$th anti-derivative of $\delta(x)$ is $$\underbrace{\int\cdots\int}_{n\text{ integrals}}\delta(x)(dx)^n=\frac{x^{n-1}}{(n-1)!}H(x)$$ 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 $\displaystyle{g_n=-1+\sum^{n}_{i=1}x_i}$.

$$I_n\equiv\underbrace{\int^1_0\cdots\int^1_0}_{n\text{ integrals}} \delta(g_n)\left(\prod^n_{i=1}dx_i\right) =\frac{g_n^{n-1}}{(n-1)!}H(g_n)\bigg\vert^{x_n=1}_{x_n=0}\cdots\bigg\vert^{x_1=1}_{x_1=0}$$

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