Delta Function

Integrate Dirac Delta in $F_1$

There are two types of dirac delta function we need to handle. One is due to the existence of $\theta\left(p_0-2E_\pi\right)$, the other is due to the existence of $\delta\left(p_0-2E_\pi\right)$.

For the first one, we can see the plot around $p_0=2E_\pi$

using FRGRealTime,Plots
#k>ps/2
plot(p0->FRGRealTime.loopfunppfix(p0,1.0,2.0,1.0,2.0),2*Epi(2.0,1.0)-0.01,2*Epi(2.0,1.0)+0.01)

Considering that our function has the following form:

\[ F(k)=f(k)\theta\Big(g(k)-p_0\Big)\]

and this relation is only true when $p_0$ is in a infinitesimal neighborhood around $g(k)$. $F'(k)$ is:

\[ F'(k)=f(k) g'(k) \delta \Big(g(k)-p_0\Big)+f'(k) \theta \Big(g(k)-p_0\Big)\]

Now we want to integrate back to $F(k)$, any finite part will disappear due to the infinitesimal measure, so:

\[ F'(k)=f(k) g'(k) \delta \Big(g(k)-p_0\Big)=f(k)\partial_k \theta\Big(g(k)-p_0\Big)\]

$F_1$ is $\theta$ function when $k>\frac{p_s}{2}$ and $p_0$ is around $2E_\pi(k,m)$. It has a $drop=\frac{\left(2 k-p_s\right) \coth \left(\frac{\sqrt{k^2+m^2}}{2 T}\right)}{16 \pi ^2 \sqrt{k^2+m^2}}$ drop in $p_0=2E_\pi(k,m)$. $F_1$ can be written as:

\[ F_1=drop\;\theta\Big(p_0-2E_\pi(k,m)\Big)\]

when $p_0=2E_\pi(k,m)$ :

\[ \partial_k F_1=-drop\;\frac{2k}{E_\pi(k,m)}\delta\Big(p_0-2E_\pi(k,m)\Big)\]

alternatively:

\[ \partial_k F_1=-drop\; \delta\!\left(k-\frac{1}{2} \sqrt{p_0^2-4 m^2}\right)\]

Im Part

The only chance that we can get rid of these annoying $\delta$ function in the Im part is doing the $k'$ or $k$ integral:

\[\mathrm{deltaintkqs}\equiv\int_\Lambda^k\!\!dk'\int_0^{qsmax}\!\!dq_s\;q_s^2\int_{-1}^{1}\!\!d\cos\theta\;\tilde{\partial_k}F_1\]

the $q_s$ and $\cos\theta$ integration will act on the $drop$, the $k'$ integration will act on the $\delta$ function.

We also noticed that this delta function only appears in $k>p/2$, so the internel momentum integration can't performed on the whole range. Assuming that the solution of $p_0=2E_\pi$ is $k_0$. So we only do the $k'$ integration around $k_0$, then $p<2k_0$.

Integrate[((2 k - Sqrt[ps^2 + qs^2 + 2*ps*qs*x])*
   Coth[Sqrt[k^2 + m]/(2 T)])/(16 \[Pi]^2 Sqrt[k^2 + m]) qs^2, {x, -1,
   1}, Assumptions -> qsmax > 0 && 1 > x > -1 && ps > 0 && qs > 0]

we get:

\[\begin{aligned} &p_s<k_0,\;\;\; \int_0^{k_0}\\ &p_s>3k_0,\;\;\; 0\\ &p_s>3k_0,\;\;\; 0\\ &p_s>3k_0,\;\;\; 0\\ \end{aligned}\]