An integral with a delta-function

[Haorong Wu]

TL;DR

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?

 

[Isaac0427]

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]

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 .