1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.