# A Integration using delta function and step function

1. Oct 7, 2016

### spaghetti3451

I would like to evaluate the following integral:

$\displaystyle{\int_{-\infty}^{\infty} dp^{0}\ \delta(p^{2}-m^{2})\ \theta(p^{0})}$

$\displaystyle{= \int_{-\infty}^{\infty} dp^{0}\ \delta[(p^{0})^{2}-\omega^{2}]\ \theta(p^{0})}$

$\displaystyle{= \int_{-\infty}^{\infty} dp^{0}\ \delta[(p^{0}+\omega)(p^{0}-\omega)]\ \theta(p^{0})}$

$\displaystyle{= \frac{1}{2\omega}\int_{-\infty}^{\infty} dp^{0}\ \delta[(p^{0}+\omega)+(p^{0}-\omega)]\ \theta(p^{0})}$

$\displaystyle{= \frac{1}{2\omega}\int_{-\infty}^{\infty} dp^{0}\ \delta(2p^{0})\theta(p^{0})}$

Have I made a mistake somewhere?

The answer is supposed to be $\frac{1}{\omega}$ but I'm not sure how to proceed next.

2. Oct 8, 2016

### andrewkirk

I don't follow some of your steps so far. But if you want to be able to get rid of the integral, try a change of variable, making the variable over which we are integrating $u=2p^0$.

The integral can then be evaluated quite simply. However I then get the answer $1/4\omega$, which is why I am wondering about your earlier steps.

3. Oct 8, 2016

### spaghetti3451

Would you recommend me to post this on the Relativity forum?

This question is better suited for that forum.

4. Oct 8, 2016

### Fightfish

$$\delta(x^2 - \alpha^2) = \frac{1}{2|\alpha|} \left[\delta(x-\alpha) + \delta(x+\alpha) \right] \neq \frac{1}{2|\alpha|} \left[\delta( (x-\alpha) + (x+\alpha)) \right]$$ which is what you seem to have done.
Are you sure the answer is supposed to be $1/\omega$ and not $1/2\omega$ though?

Last edited: Oct 8, 2016