# ## \int ~ dx dy dz ~ f(x,y,z)~ \delta (x+y+z-1)##

• I
• Safinaz
In summary: I am not going to write these out for you, but you should be able to do it yourself, given that you have derived yourself the relation of a delta function and an integral, because delta function is not the same "thing" as any other function. It's not a function of the variables of integration in the same way. It's a functional. So f(x) is fine but f(x,y,z) may not be fine, and you have to see what breaks the rule, and therefore the integral.
Safinaz
Hi all ,

I see this integral too much in QFT books when making loop calculations :

## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) = \int_{0}^{1} dz \int_{0}^{1-z} dy ##

Can anyone explain how did we get this ? I mean it's apparent that ##\int_{0}^{1}~ dx \delta (x+y+z-1) ## have been evaluated and equals one, but when applying ##\delta## when integrating ##dy## why it equals ## 1-z## , not ##1-z-x ## ?

Also suppose now I have the following function:

## \int_{0}^{1}~ dx~ dy~ dz~ \log ~\frac{m_1(x+y) + z m_2}{m_1(x+y)+ z(m_2+ m_3)} ##

how can this evaluated ? or even a simpler function:

## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~ (x+y +z) ##

If ##y+z > 1## then ##\int_0^1 \delta(x+y+z-1)dx = 0##

And if ##y+z < 1## then ##\int_0^1 \delta(x+y+z-1)dx = 1##

Last edited:
Safinaz
The delta function constrains x + y + z = 1, and x,y,z in [0,1] so we have an equation for a plane in the positive quadrant intersecting x,y,z axes at (1,0,0), (0,1,0) and (0,0,1) respectively. Draw it out and you'll see we must have 0 < y+z < 1, i.e y < 1-z.

Safinaz
PeroK said:
And if ## y+z<1y+z < 1## then ##\int^1_0 δ(x+y+z−1)=1##

I think you mean ## \int^1_0 ~ dx ~ δ(x+y+z−1)=1 ##

What about if I have :

## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~ (x+y +z) ##

Safinaz said:
I think you mean ## \int^1_0 ~ dx ~ δ(x+y+z−1)=1 ##

What about if I have :

## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~ (x+y +z) ##

Yes, I forgot the ##dx##. Corrected now.

The second one you need to post in the homework section and give it your best attempt.

Indeed It's not a problem I have, I just confused, now I know that
## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) = \int_{0}^{1} dz \int_{0}^{1-z} dy
## because ##\int_{0}^{1}~ dx \delta (x+y+z-1) = 1, ## at ## y+z < 1##, but what if I integrate a function of x, i.e .,
## \int_{0}^{1}~ dx \delta (x+y+z-1) ~ x ##? , it seems naive i know !

Safinaz said:
Indeed It's not a problem I have, I just confused, now I know that
## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) = \int_{0}^{1} dz \int_{0}^{1-z} dy
## because ##\int_{0}^{1}~ dx \delta (x+y+z-1) = 1, ## at ## y+z < 1##, but what if I integrate a function of x, i.e .,
## \int_{0}^{1}~ dx \delta (x+y+z-1) ~ x ##? , it seems naive i know !

Why don't you try yourself? The delta function simply picks out the function value.

Safinaz
Homework Statement

## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~\frac{1}{m_1(x+y)+m_2 z } ##

The attempt at a solution

## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~\frac{1}{m_1(x+y)+m_2 z } ##
## = \int_0^1 ~dz ~\int_0^{1-z}~ dy~ \frac{1}{m_1(y)+m_2 z } , ##

where I considered ## \int_{0}^{1}~ dx~ \delta (x+y+z-1) ~f(x) = \int_{0}^{1}~ dx~ \delta (x+y+z-1) ~\frac{1}{x m_1} = 1 ##, I don't if this is right ? actually this is my problem .

Then the integration run normally over z and y .

Last edited:
The problem requires some clarification in the limits of the integrals. Is it a triple integral with each limit from 0 to 1? I don't have a solution to it yet, but it is interesting. ## \\ ## editing: It looks like you just let ## z=1-(x+y) ## from the dz and delta function integral with the requirement that ## 0<(x+y)<1 ## because ## 0<z<1 ##, (assuming the limits of the z integral are 0 and 1). Outside of the ## 0<(x+y)<1 ## region, the delta function will be zero. The limits of x and y integrals are unclear from the OP. Once you do the dz integration, you would then perform a dx and dy integration.

Last edited:
Safinaz said:
Homework Statement

## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~\frac{1}{m_1(x+y)+m_2 z } ##

The attempt at a solution

## \int_{0}^{1}~ dx~ dy~ dz~ \delta (x+y+z-1) ~\frac{1}{m_1(x+y)+m_2 z } ##
## = \int_0^1 ~dz ~\int_0^{1-z}~ dy~ \frac{1}{m_1(y)+m_2 z } , ##

where I considered ## \int_{0}^{1}~ dx~ \delta (x+y+z-1) ~f(x) = \int_{0}^{1}~ dx~ \delta (x+y+z-1) ~\frac{1}{x m_1} = 1 ##, I don't if this is right ? actually this is my problem .

Then the integration run normally over z and y .

You have:

##\int_0^1 \delta(x+y+z-1)f(x, y, z) dx = f(1-y-z, y, z)## (if ##0 < 1-y-z < 1##)

And the integral is ##0## otherwise.

Note: The previous 3 posts have been merged into this thread from another.

Safinaz said:
where I considered ∫10 dx δ(x+y+z−1) f(x)=∫10 dx δ(x+y+z−1) 1xm1=1 \int_{0}^{1}~ dx~ \delta (x+y+z-1) ~f(x) = \int_{0}^{1}~ dx~ \delta (x+y+z-1) ~\frac{1}{x m_1} = 1 , I don't if this is right ? actually this is my problem .

fact 1 you are taking the case where the following is true:
$\int \int \int dx dy dz G(x,y,z) = \int dz \Big( \int dy \big( \int dx G(x,y,z) \big) \Big)$

I guess you were tired when you came up with this question? because it looks like you don't follow basic delta "integration" rules... let me clarify the rule that you missed or got confused with:
$\int_a^b f(x) \delta(x-x_0) = \begin{cases} 0 & \text{if } x_0 \not \in [a,b] \\ f(x_0) & \text{else} \end{cases}$
Now the thing is that your $x_0$ is not a constant but a function of $y,z,~~x_0(y,z) \equiv 1-y-z$.
Then again as given in the integration rule of the delta function, that $x_0$ can't be just anything, it has to be constrained to $[0,1]$ (integration of x variable), otherwise the integral results to 0. In formula:
$0 \le 1 - y -z \le 1 ~~\Rightarrow ~~0 \le 1 - (y+z) \le 1$
or
$0 \le y+z \le 1$
Now for $y \in [0,1]$ this can give lower/upper limits for $z$ given by it being 0 to $1-y$ for any value of y...

Safinaz and PeroK

## 1. What does the integral ∫ f(x,y,z) δ(x+y+z-1) represent?

The integral ∫ f(x,y,z) δ(x+y+z-1) represents the weighted average of the function f(x,y,z) over the region defined by the constraint x+y+z=1.

## 2. How is the Dirac delta function δ(x) related to this integral?

The Dirac delta function δ(x) is a mathematical tool used to represent a point mass or concentration of mass at a specific point. In this integral, the Dirac delta function is used to constrain the region of integration to a specific surface defined by x+y+z=1.

## 3. Can this integral be solved analytically?

It depends on the function f(x,y,z) and the region defined by the constraint x+y+z=1. In some cases, this integral can be solved analytically using standard integration techniques. However, in more complex cases, numerical methods may be necessary.

## 4. What are some real-world applications of this type of integral?

This type of integral is commonly used in physics and engineering to calculate moments of inertia, center of mass, and other physical properties of objects with complex shapes. It is also used in probability and statistics to calculate joint probability distributions.

## 5. Can this integral be extended to higher dimensions?

Yes, this type of integral can be extended to higher dimensions by adding additional delta functions to the constraint. For example, in four dimensions, the integral would be ∫∫∫∫ f(x,y,z,w) δ(x+y+z+w-1) dx dy dz dw.

Replies
4
Views
1K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
1
Views
2K
Replies
4
Views
1K