- #1

- 83

- 1

[tex] I(p_{1}, p_{2}, p_{3}) = \int \mathrm{d}^{4} q \mathrm{d}^{4}p \, \dfrac{1}{\left[ p_{2} + q \right]^{2} - i0} \dfrac{1}{\left[ p_{1} - q - p \right]^{2} + i0} \Theta(q^{0}) \delta(q^{2}) \Theta(-p_{2}^{0} -p_{3}^{0} - q^{0} -p^{0}) \delta(\left[p_{2} + p_{3} + q + p \right]^{2}) \Theta(p^{0}) \delta(p^{2}) [/tex].

[tex] p_{1}, p_{2}, p_{3} [/tex] are time-like four-vectors, so e.g. [tex] p_{1}^{2} > 0 [/tex]

After some work like exploiting the step- and delta-functions [tex] \Theta(q^{0}) \delta(q^{2}) \Theta(p^{0}) \delta(p^{2}) [/tex] and by choosing a special frame with

[tex] p_{2} + p_{3} = (p_{2}^{0} + p_{3}^{0}, \vec{0}) [/tex] I arrived at:

[tex] I(p_{1}, p_{2}, p_{3}) = \int \dfrac{\mathrm{d}^{3}q \, \mathrm{d}^{3}p}{4 \vert \vec{q} \vert \cdot \vert \vec{p} \vert} \, \dfrac{1}{p_{2}^{2} + p_{2}^{0} \vert \vec{q} \vert + \vert \vec{p}_{2} \vert \cdot \vert \vec{q} \vert \cos \theta_{1} - i0} \, \dfrac{1}{p_{1}^{2} - 2p_{1}^{0} \vert \vec{q} \vert + 2 \vert \vec{p}_{1} \vert \cdot \vert \vec{q} \vert \cos \theta_{2} - 2 p_{1}^{0} \vert \vec{p} \vert + 2 \vert \vec{p}_{1} \vert \cdot \vert \vec{p} \vert \cos \theta_{3} + 2 \vert \vec{q} \vert \cdot \vert \vec{p} \vert - 2 \vert \vec{q} \vert \cdot \vert \vec{p} \vert \cos \eta +i0} [/tex]

[tex] \times \delta((p_{2}^{0}+p_{3}^{0})^{2} + 2 \vert \vec{q} \vert \cdot \vert \vec{p} \vert - 2 \vert \vec{q} \vert \cdot \vert \vec{p} \vert \cos \eta + 2 (p_{2}^{0}+p_{3}^{0}))

\end{split}

[/tex].

So I used spherical coordinates, but I don't know how to integrate that thing. I just know that if [tex] \eta [/tex] is the angle between the vectors [tex] \vec{p} [/tex] and [tex] \vec{q} [/tex] than we must have: [tex] \theta_{3} = \eta - \theta_{2} [/tex].

But how do I continue? I think I must somehow specify the elevation and azimuth angles in a special way, but I don't know how to do that.

Could anyone help me please?