Understanding Scattering Process in QFT Integral

[LCSphysicist]

Homework Statement

All the problem is printed below

Relevant Equations

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I have been studying scattering process in QFT, but i am stuck now because i can't understand how this integral was evaluated:
$$\int dp\space \frac{1}{\sqrt{p^2+c²}}\frac{1}{\sqrt{p^2+k²}}\space p² \space d\Omega \space \delta(E_{cm}-E_{1}-E_{2})$$$

Where Ecm = c + k, E1 is the factor in the denominator involving c and E2 the factor in the denominator involving k.

Now, $$$\delta(\sqrt{p^2+c²} +\sqrt{p^2+k²} - (k+c))$$.

$$\sqrt{p^2+c²} +\sqrt{p^2+k²} - (k+c) = 0$$ has solution for p=0, so shouldn't the integral becomes

$$\int dp\space \frac{1}{\sqrt{p^2+c²}}\frac{1}{\sqrt{p^2+k²}}\space p² \space d\Omega \space \frac{\delta(p-0)}{(p/E_{1}+p/E_{2})_{p=0}}$$

I just applied the fact that $$\delta\big(f(x)\big) = \sum_{i}\frac{\delta(x-a_{i})}{\left|{\frac{df}{dx}(a_{i})}\right|}$$

Why is this wrong? And also, how to evaluate the integral so?

 

[anuttarasammyak]

May we take
\int d\Omega = 4\pi
in your formula ?

f(p)=\sqrt{p^2+c^2}-c-\sqrt{p^2+k^2}+k
How about calculating f'(p) to use your last formula with $p_i$ satisfying
f(p_i)=\sqrt{p_i^2+c^2}-c-\sqrt{p_i^2+k^2}+k=0?

 

[LCSphysicist]

May we take
\int d\Omega = 4\pi
in your formula ?

f(p)=\sqrt{p^2+c^2}-c-\sqrt{p^2+k^2}+k
How about calculating f'(p) to use your last formula with $p_i$ satisfying
f(p_i)=\sqrt{p_i^2+c^2}-c-\sqrt{p_i^2+k^2}+k=0?

"f(p)=\sqrt{p^2+c^2}-c-\sqrt{p^2+k^2}+k
How about calculating f'(p) to use your last formula with $p_i$ satisfying
f(p_i)=\sqrt{p_i^2+c^2}-c-\sqrt{p_i^2+k^2}+k=0?" i did it in the last step. But as you can see, it would makes the denominator zero.
"May we take
\int d\Omega = 4\pi
in your formula ?
"
As far as i know we can do it, so the real main problem is in fact calculating the variables involving the E and the P.

 

[anuttarasammyak]

Let me clear some points. I assume In CM p_1=-p_2=p.
E_{CM}=\sqrt{p^2+k^2}+\sqrt{p^2+c^2}
with
E_1+E_2=\sqrt{p_1^2+k}+\sqrt{p_2^2+c^2}
Is it right?