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In M. Srednicki "Quantum field theory", Section 14 -Loop corrections to the propagator-, the exact propagator $\mathbf {\tilde \Delta} (k^2)$ is stated as

$$\frac{1}{i} \mathbf {\tilde \Delta} (k^2) = \frac{1}{i} \tilde \Delta (k^2) + \frac{1}{i} \tilde \Delta (k^2) [i \Pi (k^2)] \frac{1}{i} \tilde \Delta (k^2) + O(g^4)\tag{14.2}$$
where:
$\tilde \Delta (k^2) = \frac{1}{k^2 + m^2 - i \epsilon}$ free-field propagator
$i \Pi (k^2) = \frac{1}{2} (i g)^2 (\frac{1}{i})^2 \int \frac{d^d l}{(2 \pi)^d} \tilde \Delta ((l + k)^2) \tilde \Delta (l^2) - i (A k^2 + B m^2) + O(g^4)$ self-energy

Many pages are then dedicated to the laborious calculation of the self-energy term. I could understand all of the many passages, however almost at the end of the demonstration to compute the expression

$$\Pi (k^2) = \frac{1}{2} \alpha \int_0 ^1 dx D ln (D / D_0) - \frac{1}{12} \alpha (k^2 + m^2) + O(\alpha^2)\tag{14.43}$$

where:

$D = x (1 - x) k^2 + m^2$
$D_0 = D \vert_{k^2 = -m^2}$

the integral over $x$ is solved in closed form as

$$\Pi (k^2) = \frac{1}{12} \alpha [c_1 k^2 + c_2 m^2 + 2 k^2 f(r)] + O(\alpha^2)\tag{14.44}$$

where:

$c_1 = 3 - \pi \sqrt{3}$
$c_2 = 3 - 2 \pi \sqrt{3}$
$f(r) = r^3 tanh^{-1}(1/r)$
$r = (1 + 4 m^2 / k^2)^{1/2}$

My question is: How to move from Eq. (14.43) to Eq. (14.44)? Any hint, or any link where I can find it?

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    $\begingroup$ Have you tried factorizing $D= k^2(x-\alpha)(x-\beta)$? That's how I would start the task. The $\alpha$ abd $\beta$ parameters involve the quantity $r$ so that is a hint that the strategy would be useful. $\endgroup$
    – mike stone
    Jun 13 '19 at 13:12
  • $\begingroup$ @mike stone. Following your hint I performed the calculation of Eq. (14.43), straightforward even if somewhat laborious. The outcome confirms the structure of Eq. (14.44), however the coefficients $c_1$ and $c_2$ do not match. My computation yields $c_1 = -9 + \pi \sqrt{3}$, $c_2 = -9$. If anyone reading this post provides a reference other than Srednicki's, welcome. $\endgroup$ Jun 16 '19 at 15:29
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It is very easy to integrate it with Mathematica. From Eq.(14.14) and Eq.(14.42) we have $D=x(1-x)k^2+m^2$ and $D_0=[1-x(1-x)]m^2$, so $$ \frac\alpha2\int_0^1\text{d}x\ D\ln\frac D{D_0}=\frac\alpha{12}\left[4(k^2+m^2)-\sqrt3(k^2+2m^2)\pi+2\sqrt{\frac{(k^2+4m^2)^3}{k^2}}\arctan\sqrt{\frac{k^2}{k^2+4m^2}}\right]. $$ Therefore, $$ \begin{aligned} \Pi(k^2)=&\ \frac\alpha2\int_0^1\text{d}x\ D\ln\frac D{D_0}-\frac\alpha{12}(k^2+m^2)+\mathcal O(\alpha^2) \\ =&\ \frac\alpha{12}\left[(3-\pi\sqrt3)k^2+(3-2\pi\sqrt3)m^2+2k^2\left(1+\frac{4m^2}{k^2}\right)^\frac32\arctan\frac1{\left(1+\frac{4m^2}{k^2}\right)^\frac12}\right]+\mathcal O(\alpha^2) \\ =&\ \frac\alpha{12}\left[c_1k^2+c_2m^2+2k^2f(r)\right]+\mathcal O(\alpha^2). \end{aligned} $$ That's all.

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  • $\begingroup$ What is Mathematica? $\endgroup$ Aug 13 '20 at 8:42
  • $\begingroup$ A math software that you can use it to calculate many integrals easily. $\endgroup$ Aug 13 '20 at 15:23
  • $\begingroup$ Thanks for the information. $\endgroup$ Aug 14 '20 at 6:38

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