# An integral with a delta-function

• I
• Haorong Wu
In summary, the conversation discussed an equation from David Tong's note for QFT that involved a 4-vector and the delta function. The speaker provided a calculation that resulted in two possible outcomes, one being infinite. Another speaker shared the key to this, which involved an identity for the delta function. This identity was then applied to the original equation, resulting in the correct answer.

#### Haorong Wu

TL;DR Summary
How to calculate this integral?
It is from David Tong's note for QFT. The equation states
##\left . \int d^4 p \cdot \delta \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0} =\left . \int \frac {d^3 p} {2 p_0} \right |_{p_0=E_{\vec p}}##
where ##p## is a 4-vector ##p=\left ( p_0, \vec p \right )##.

In my calculation, I get
##\left . \int d^4 p \cdot \delta \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0} =\left . \int d^3 p \int d p_0 \cdot \delta \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0}=\left . \int d^3 p \cdot 1 \right | _{p_0>0}=1## or
##\left . \int d^4 p \cdot \delta \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0} =\left . \int d^3 p \int d p_0 \cdot \delta^{\left ( 3 \right )} \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \delta ^{\left ( 1 \right )}\left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0}=\left . \int d^3 p \cdot \delta^{\left ( 3 \right )} \left ( 0 \right ) \right | _{p_0>0}## which is infinite.

Maybe I get a wrong definition for the delta function in this integral? What is the dimension of the delta function in this integral?

The key to this is the identity
$$\delta \left(f(x)\right)=\frac{\delta(x-x_0)}{\left| f'(x_0)\right|}$$
where ##x_0## is a zero of f(x). In this case, ##f(x)=x^2 - x_0^2##, ##x=p_0##, and ##x_0=\sqrt{\vec{p}^2+m^2}=E_{\vec{p}}##. This gives me the correct answer.

Haorong Wu
Isaac0427 said:
The key to this is the identity
$$\delta \left(f(x)\right)=\frac{\delta(x-x_0)}{\left| f'(x_0)\right|}$$
where ##x_0## is a zero of f(x). In this case, ##f(x)=x^2 - x_0^2##, ##x=p_0##, and ##x_0=\sqrt{\vec{p}^2+m^2}=E_{\vec{p}}##. This gives me the correct answer.
Oh, thanks, @Isaac0427 .