1 Computation for $\phi^3$ Theory
All of the calculation below is based on Srednicki, and Peskin and Schroeder, so nothing here is original at all.
Here, we will use the metric signature $(-,+,+,+)$. The Lagrangian of $\phi^3$ is given by \[\mathcal{L} = -\frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2\phi^2 + \frac{1}{3!}g\phi^3.\tag{1.1}\] To make the theory satisfying \[\bra{k}\phi(x)\ket{\Omega} = e^{-ik\cdot x},\quad \bra{\Omega}\phi(x)\ket{\Omega} = 0,\tag{1.2}\] we make field redefinitions so that the Lagrangian becomes \[\mathcal{L} = -\frac{1}{2}Z_\phi\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}Z_mm^2\phi^2 + \frac{1}{3!}Z_gg\phi^3 + Y\phi = \mathcal{L}_0 + \mathcal{L}_1,\tag{1.3}\] where \[\mathcal{L}_0 = -\frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2\phi^2,\tag{1.4}\] and \[\mathcal{L}_1 = -\frac{1}{2}(Z_\phi-1)\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}(Z_m-1)m^2\phi^2 + \frac{1}{3!}Z_gg\phi^3 + Y\phi.\tag{1.5}\]
The partition functional can be written as \[Z = \int\mathcal{D}\phi(x) e^{i\int d^dx (\mathcal{L}_0 + J\phi + \mathcal{L}_1)} = e^{i\int d^dx \mathcal{L}_1(-i\delta/\delta J(x))}\int\mathcal{D}\phi(x) e^{i\int d^dx (\mathcal{L}_0 + J\phi)} = e^{i\int d^dx \mathcal{L}_1(-i\delta/\delta J(x))} Z_0,\tag{1.6}\] where \[Z_0 = \int\mathcal{D}\phi(x) e^{i\int d^dx (\mathcal{L}_0 + J\phi)}.\tag{1.7}\] Making the field redefinition \[\chi(x) = \phi(x) - iD_F(x-y) J(y),\tag{1.8}\] where \[(\partial^2 - m^2)D_F(x-y)= i\delta^d(x-y).\tag{1.9}\] Note that since the above is just a shift of field, the measure is unchanged $\mathcal{D}\phi = \mathcal{D}\chi$. Then, the integrand of $Z_0$ becomes \[\begin{aligned} \exp\left[i\int d^dx \left(\frac{1}{2}\chi(x)(\partial^2-m^2)\chi(x) +\frac{i}{2}\int d^dy J(x)D_F(x-y)J(y) \right)\right], \end{aligned}\tag{1.10}\] and hence \[Z_0 = \mathcal{N}\exp\left[-\frac{1}{2}\int d^dx \int d^dy J(x)D_F(x-y)J(y) \right],\tag{1.11}\] where $\mathcal{N}$ is formally infinite, but since it is not a measurable quantity, it does not boarder us. Now the full partition functional becomes \[Z = e^{i\int d^dx \mathcal{L}_1(-i\delta/\delta J(x))}\exp\left[-\frac{1}{2}\int d^dx \int d^dy J(x)D_F(x-y)J(y) \right]\tag{1.12}\] up to some irrelevant constant.
Now, we are interested in finding the exact two-point correlation function to one-loop order. That is \[\bra{\Omega}T\phi(x_1)\phi(x_2)\ket{\Omega} = (-i)^2 \frac{\delta^2}{\delta J(x_1)\delta J(x_2)}\ln{Z[J]}\bigg|_{J=0} = i(-i)^2\frac{\delta^2}{\delta J(x_1)\delta J(x_2)}W[J]\bigg|_{J=0},\tag{1.13}\] where $W[J]$ is the connected diagram generating functional \[Z[J] \equiv e^{iW[J]}.\tag{1.14}\]
In practice, one can simply write down the Feynman rules and compute the correlation functions. However, for our purpose, we shall compute it directly from functional derivatives. We shall first compute $Z[J]$: \[\begin{aligned} Z[J] &= \left[1 + i\int d^dx \frac{1}{2}\left(-i\frac{\delta}{\delta J(x)}\right)(A\partial_x^2 - Bm^2)\left(-i\frac{\delta}{\delta J(x)}\right) + \cdots \right]Z_0\\ &=Z_0 -\frac{i}{2}\int d^dx\left(-i\frac{\delta}{\delta J(x)}\right)(A\partial_x^2 - Bm^2)\left(-\int d^dy J(y)D_F(x-y)\right)Z_0 + \cdots\\ &=Z_0- \frac{i}{2}\int d^dx \left(\int d^dz J(z)D_F(x-z)\right)(A\partial_x^2 - Bm^2)\left(\int d^dy J(y)D_F(x-y)\right)Z_0 + \cdots , \end{aligned}\tag{1.15}\] where we have focused on the connected two-point term since we will eventually take $J=0$ after the functional derivative. However, one should not forget the other term \[\begin{aligned} &\frac{i^2}{(3!)^2}g^2\int d^dx \int d^dy\left(-i\frac{\delta}{\delta J(x)}\right)^3\left(-i\frac{\delta}{\delta J(y)}\right)^3 Z_0 \\ =&-\frac{g^2}{2}\int d^dx \int d^dy J(x)D_F(x-y)^2 J(y) Z_0 + \cdots \end{aligned}\tag{1.16}\] Now, we can compute the two-point correlation function; that is \[\begin{aligned} D(x_1-x_2) =\bra{\Omega}T\phi(x_1)\phi(x_2)\ket{\Omega} &= -\frac{\delta^2}{\delta J(x_1)\delta J(x_2)}Z[J]\bigg|_{J=0} \\ &=D_F(x_1-x_2) +i\int d^dx D_F(x-x_1)(A\partial_x^2 - Bm^2) D_F(x-x_2)\\ &\qquad- \frac{g^2}{2}\int d^dx \int d^dy D_F(x_1-x)D_F(x-y)^2D_F(y-x_2) + \mathcal{O}(g^4), \end{aligned}\tag{1.17}\] where we have been quite sloppy with the overall constant throughout. These are exactly the diagrams one would expect from Feynman rules
In momentum space, we have \[\tilde{D}(k^2) = \tilde{D}(k^2) + \tilde{D}_F(k^2)i\Pi(k^2)\tilde{D}_F(k^2) + \mathcal{O}(g^4) ,\tag{1.18}\] where \[i\Pi(k^2) =-i(Ak^2+Bm^2) - \frac{g^2}{2}\int \frac{d^dl}{(2\pi)^d} \frac{i}{l^2+m^2-i\epsilon}\frac{i}{(k+l)^2+m^2-i\epsilon} + \mathcal{O}(g^4).\tag{1.19}\] Consider the con exact propagator, where $\Pi(k^2)$ is now to all order of $g$, \[\tilde{D}(k^2) = \tilde{D}(k^2)\sum_{n=0}^\infty[i\Pi(k^2)\tilde{D}_F(k^2)]^2 = \frac{i}{k^2+m^2-i\epsilon -\Pi(k^2)}.\tag{1.20}\] For the particle to have mass $m$, we need the exact propagator with residue one at $k^2=-m^2$. Therefore, we require \[\Pi(-m^2) = \Pi'(-m^2) = 0,\tag{1.21}\] which helps us fix $A$ and $B$. Now, we need to compute the integral above.
Now, we shall perform the integral in $d$ dimensions. To do that, we use the Feynman's formula.
Claim: \[\prod_{j=1}^n\frac{1}{A_j^{\alpha_j}} = \frac{\Gamma(\sum_i\alpha_i)}{\prod_i\Gamma(\alpha_i)}\frac{1}{(n-1)!}\int dF_n \frac{\prod_ix_i^{\alpha_i-1}}{(\sum_ix_iA_i)^{\sum_i\alpha_i}},\tag{1.22}\] where \[dF_n = (n-1)!\int_0^1dx_1\cdots\int^1_0 dx_n \delta(1-\sum_ix_i).\tag{1.23}\] Proof: Consider \[\frac{\Gamma(\alpha_i)}{A_i^{\alpha_i}} = \int_0^\infty dt_i\, t_i^{\alpha_i-1}e^{-A_it_i}.\tag{1.24}\] Then, we take \[\prod_{i=1}^n \frac{\Gamma(\alpha_i)}{A_i^{\alpha_i}} = \prod_{i=1}^n\int_0^\infty dt_i\, t_i^{\alpha_i-1}e^{-A_it_i}.\tag{1.25}\] We can multiply the right-hand side by \[1 = \int^\infty_0ds\,\delta(s-\sum_it_i),\tag{1.26}\] which gives \[\begin{aligned} \prod_{i=1}^n \frac{\Gamma(\alpha_i)}{A_i^{\alpha_i}} &= \int^\infty_0ds\,\delta(s-\sum_it_i)\prod_{i=1}^n\int_0^\infty dt_i\, t_i^{\alpha_i-1}e^{-A_it_i}. \end{aligned}\tag{1.27}\] Now, making the substitution $t_i =sx_i$, we find \[\begin{aligned} \prod_{i=1}^n \frac{\Gamma(\alpha_i)}{A_i^{\alpha_i}} &= \int^\infty_0ds\,\delta(s-s\sum_ix_i)\prod_{i=1}^n\int_0^\infty dx_i\, s^{\alpha_i}x_i^{\alpha_i-1}e^{-sA_ix_i}\\ &=\frac{1}{(n-1)!}\int dF_n\left(\prod_{i=1}^nx_i^{\alpha_i-1}\right)\int ^\infty_0 ds\,s^{\sum_i \alpha_i -1}e^{-\sum_isA_ix_i}\\ &=\frac{1}{(n-1)!}\int dF_n\left(\prod_{i=1}^nx_i^{\alpha_i-1}\right)\frac{\Gamma(\sum_i\alpha_i)}{(\sum_iA_ix_i)^{\sum_i\alpha_i}}. \end{aligned}\tag{1.28}\] ∎We use \[\frac{1}{A_1A_2} = \int^1_0dx_1\int^1_0dx_2\,\delta(x_1+x_2-1)\frac{1}{(x_1A_1 + x_2A_2)^2} = \int^1_0 dx\frac{1}{[xA_1 +(1-x)A_2]^2}.\tag{1.29}\] Therefore, we have \[\begin{aligned} \frac{1}{l^2+m^2}\frac{1}{(k+l)^2+m^2} &= \int^1_0dx[(1-x)(l^2+m^2) + x((k+l)^2+m^2)]^{-2}\\ &=\int^1_0 dx [l^2+2xkl+m^2+xk^2]^{-2}\\ &=\int^1_0 dx [(l+xk)^2+m^2 +x(1-x)k^2]^{-2}, \end{aligned}\tag{1.30}\] where we have abbreviated $m^2-i\epsilon$ as $m^2$. Now, we write $q\equiv l+xk$ and $D \equiv m^2 +x(1-x)k^2$. So that the integral is now \[\begin{aligned} - \frac{g^2}{2}\int \frac{d^dl}{(2\pi)^d} \frac{i}{l^2+m^2-i\epsilon}\frac{i}{(k+l)^2+m^2-i\epsilon} &=\frac{g^2}{2}\int^1_0dx\int\frac{d^dq}{(2\pi)^d}\frac{1}{(q^2+D)^2}. \end{aligned}\tag{1.31}\] Then, we perform the Wick rotation so that $q^0 = iq^0_E$, $\mathbf{q} = \mathbf{q}_E$, $q^2 = q^2_E$, and $d^dq = id^dq_E$. Therefore, the above integral becomes \[I(k^2) = i\int^1_0dx\int\frac{d^dq_E}{(2\pi)^d}\frac{1}{(q_E^2+D)^2},\tag{1.32}\] where \[i\Pi(k^2) = \frac{g^2}{2}I(k^2)-i(Ak^2+Bm^2) + \mathcal{O}(g^4).\tag{1.33}\] With a slight abbreviation of notation, we now write the integral measure as $d^dq_E = q_E^{d-1}dq_Ed\Omega_{d-1}$, where $q_E$ is now the length of the vector $q_E$.
Claim: \[\Omega_{d-1} = \frac{2\pi^{d/2}}{\Gamma(d/2)}.\tag{1.34}\]
Proof: Consider the integral \[\int_{\mathbb{R}^d}d^dx\, e^{-x^2} = \int d\Omega_{d-1}\int_0^\infty dr\,r^{d-1}e^{-r^2}.\tag{1.35}\] The left-hand side gives \[\left(\int_\mathbb{R}dx\, e^{-x^2}\right)^d = \pi^{d/2},\tag{1.36}\] and the right-hand side gives \[\Omega_{d-1}\,\frac{1}{2}\int_0^\infty dr \, r^{d/2-1}e^{-r} = \Omega_{d-1}\frac{\Gamma(d/2)}{2}.\tag{1.37}\] ∎We shall also prove the general identity, that would turn out to be useful in many cases when dealing with loop corrections.
Claim: \[\int\frac{d^dq_E}{(2\pi)^d}\frac{(q_E^2)^n}{(q_E^2+D)^{m}} = \frac{\Gamma(m-n-d/2)\Gamma(n+d/2)}{(4\pi)^{d/2}\Gamma(m)\Gamma(d/2)}D^{-(m-n-d/2)}.\tag{1.38}\]
Proof: \[\begin{aligned} \int\frac{d^dq_E}{(2\pi)^d}\frac{(q_E^2)^n}{(q_E^2+D)^{m}} = \frac{2\pi^{d/2}}{\Gamma(d/2)}\frac{1}{(2\pi)^d}\int_0^\infty dq_E\frac{q_E^{2n+d-1}}{(q_E^2+D)^m}. \end{aligned}\tag{1.39}\] Making the substitution $s^2 = q^2/D +1$, we find \[\begin{aligned} \int\frac{d^dq_E}{(2\pi)^d}\frac{(q_E^2)^n}{(q_E^2+D)^{m}} = \frac{2D^{n+d-m-1}}{(4\pi)^{d/2}\Gamma(d/2)}\int^\infty_1ds \frac{(s^2-1)^{n+d/2-1}}{s^{2m-1}}. \end{aligned}\tag{1.40}\] Then, we substitute $t = 1/s^2$, we find \[\begin{aligned} \int\frac{d^dq_E}{(2\pi)^d}\frac{(q_E^2)^n}{(q_E^2+D)^{m}} &= \frac{D^{n+d-m}}{(4\pi)^{d/2}\Gamma(d/2)}\int^1_0 dt\, t^{m-n-d/2-1}(t-1)^{n+d/2-1}\\ &=\frac{D^{n+d-m}}{(4\pi)^{d/2}\Gamma(d/2)}\beta(m-n-d/2,n+d/2-1)\\ &=\frac{D^{n+d-m}}{(4\pi)^{d/2}\Gamma(d/2)}\frac{\Gamma(m-n-d/2)\Gamma(n+d/2)}{\Gamma(m)}. \end{aligned}\tag{1.41}\] ∎For our case, we have $n=0$ and $m=2$, and hence \[I(k^2) = i \int^1_0dx\frac{\Gamma(2-d/2)}{(4\pi)^{d/2}}D^{-2+d/2}.\tag{1.42}\]
Now let us focus on the $\phi^3$ theory in $d=6$, where $[g] =0$ and the theory is renormalizable. For general dimensions, $[g] = 3-d/2\equiv \varepsilon/2$. We shall redefine the coupling constant to make it dimension less, that is \[g \to g\tilde{\mu}^{\varepsilon/2}.\tag{1.43}\] Now, we find \[g^2\tilde{\mu}I(k^2) = i g^2\tilde{\mu}\int^1_0dx\frac{\Gamma(\varepsilon/2-1)}{(4\pi)^{3-\varepsilon/2}}D^{-\varepsilon/2+1} = ig^2\frac{\Gamma(-1+\varepsilon/2)}{(4\pi)^3}\int^1_0 dx\, D \left(\frac{4\pi\tilde{\mu}^2}{D}\right)^{\varepsilon/2}.\tag{1.44}\] Using dimensional regularization, we take $\varepsilon\to 0$ so that \[\Gamma(-1+\varepsilon/2) \sim -\frac{2}{\varepsilon}+\gamma-1+ \cdots,\tag{1.45}\] and \[\left(\frac{4\pi\tilde{\mu}^2}{D}\right)^{\varepsilon/2} \sim 1 + \frac{\varepsilon}{2}\ln\left(\frac{4\pi\tilde{\mu}^2}{D}\right) + \cdots.\tag{1.46}\] Then, \[\Gamma(-1+\varepsilon/2)\left(\frac{4\pi\tilde{\mu}^2}{D}\right)^{\varepsilon/2} \sim -\frac{2}{\varepsilon} + \gamma -1 - \ln\left(\frac{4\pi\tilde{\mu}^2}{D}\right) + \mathcal{O}(\varepsilon).\tag{1.47}\] Defining \[\mu = \sqrt{4\pi}e^{-\gamma/2}\tilde{\mu},\quad \alpha = \frac{g^2}{(4\pi)^3}\tag{1.48}\] we find \[g^2\tilde{\mu}I(k^2) \sim i\alpha \int^1_0 dx\,D\left[-\frac{2}{\varepsilon} -1 + \ln\left(\frac{m^2}{\mu^2}\right)- \ln\left(\frac{m^2}{D}\right) + \mathcal{O}(\varepsilon)\right].\tag{1.49}\] Now, we evaluate \[\int ^1_0 dx D = \int ^1_0 dx (m^2+x(1-x)k^2) = m^2 + k^2/6.\tag{1.50}\] Therefore, we find \[\begin{aligned} \Pi(k^2) &\sim \frac{\alpha}{2}\left[-(2/\varepsilon+1+2\ln(\mu/m))(m^2+k^2/6)+\int^1_0 dx \,D\ln(D/m^2)\right] - (Ak^2 + Bm^2) + \mathcal{O}(\alpha^2)\\ &= \frac{\alpha}{2}\int^1_0 dx \,D\ln(D/m^2) - \left[\frac{1}{6}\alpha(\varepsilon+\ln(\mu/m)+1/2) + A\right]k^2 \\ &\qquad \qquad\qquad\qquad\qquad\qquad\qquad- \left[\alpha(\varepsilon+\ln(\mu/m)+1/2) + A\right]m^2 + \mathcal{O}(\alpha^2). \end{aligned}\tag{1.51}\] To further simplify the expression, we shall let $A$ and $B$ to take the form \[A = -\frac{1}{6}\alpha(\varepsilon+\ln(\mu/m)+1/2) +\alpha \kappa_A - \mathcal{O}(\alpha^2),\tag{1.52}\] \[B = -\alpha(\varepsilon+\ln(\mu/m)+1/2) - \alpha\kappa_B + \mathcal{O}(\alpha^2).\tag{1.53}\] Then, we find \[\Pi(k^2) \sim \frac{\alpha}{2}\int^1_0 dx \,D\ln(D/m^2) - \alpha\left(\frac{1}{6}\kappa_A k^2 + \kappa_B m^2\right) + \mathcal{O}(\alpha^2).\tag{1.54}\] Now, we impose the renormalization condition $\Pi(-m^2) = \Pi'(-m^2) = 0$ to fix $\kappa_A$ and $\kappa_B$. To do that, we note that $\Pi(-k^2)$ must take the form \[\Pi(-k^2) \sim \frac{\alpha}{2}\int^1_0dx\, D\ln(D/D_0) + c(k^2+m^2) + \mathcal{O}(\alpha^2),\tag{1.55}\] where $c\in \mathbb{R}$ and \[D_0 = D\big|_{k^2=-m^2} = [1-x(1-x)]m^2.\tag{1.56}\] Then, we differentiate with respect to $k^2$ to find \[\Pi'(-k^2) \sim \frac{\alpha}{2}\int^1_0dx\,\left[x(1-x)\ln(D/D_0) + D_0x(1-x)\right] +c +\mathcal{O}(\alpha^2).\tag{1.57}\] Therefore, \[\Pi'(-m^2) = 0\implies c = -\frac{\alpha}{12}.\tag{1.58}\] Finally we obtain the 1-loop correction for the two-point correlation function, which is \[\Pi(-k^2) \sim \frac{\alpha}{2}\int^1_0dx\, D\ln(D/D_0) -\frac{\alpha}{12}(k^2+m^2) + \mathcal{O}(\alpha^2).\tag{1.59}\] We might as well right down the asymptotic value for $\kappa_A$ and $\kappa_B$. That is \[\kappa_A = \frac{1}{4},\quad \kappa_B = \frac{1}{24}.\tag{1.60}\]
Note that the integral in our final expression is real if $D>0$. That is $x(1-x)k^2>-m^2$, and since $x(1-x) \in[0,1/4)$ we must have $k^2 >-4m^2$ for real $\Pi(k^2)$. Then, one might be tempting to ask: what happens if $k^2 < -4m^2$, and $\text{Im}\,\Pi(k^2)\neq 0$? To understand what is happening, let us write the one-loop correction in Lehmann-Källén form, \[\tilde{D}(k^2) = \frac{i}{k^2+m^2-i\epsilon} + \int _{4m^2}^\infty ds \,\rho(s)\frac{i}{k^2+s-i\epsilon} = \frac{i}{k^2 +m^2-i\epsilon - \Pi(k^2)}.\tag{1.61}\] Using the identity \[\frac{1}{x-i\epsilon} = P\frac{1}{x} + i\pi\delta(x),\tag{1.62}\] we find \[\text{Im}(-i\tilde{D}(k^2)) = \pi\delta(k^2+m^2) + \pi\rho(-k^2) = \pi\delta(k^2+m^2-\Pi(k^2)),\tag{1.63}\] where $k^2 >-4m^2$. Therefore, for $k^2 >-4m^2$, $\rho(-k^2) = \rho(s)= 0$. However, for $k^2 <-4m^2$, $\text{Im}\, \Pi(k^2) \neq 0$. Hence, one would find \[\pi\rho(s) = \frac{\text{Im}\, \Pi(-s)}{(-s+m^2+\text{Re}\, \Pi(-s))^2 + (\text{Im}\, \Pi(-s))^2}.\tag{1.64}\]
Now, let us compute the vertex correction for the $\phi^3$ theory. In the previous computation we took a slightly more complicated root by evaluating the path integral straight away. Now, let us go back and write down the momentum space vertex Feynman rules of the theory: \[iZ_gg.\tag{1.65}\]
We can define an exact three-point vertex $iV_3(k_1,k_2,k_3)$, where $k_1+k_2+k_3 =0$. To one-loop correction, the exact three-point is given by \[iV_3(k_1, k_2,k_3)/g = iZ_g + i(ig)^2\int\frac{d^dl}{(2\pi)^d}\tilde{D}_F(l^2)\tilde{D}_F((l-k_1)^2) \tilde{D}_F((l+k_2)^2) + \mathcal{O}(g^5),\tag{1.66}\] where we have used the fact that $Z_g = 1 + \mathcal{O}(g^2)$. We shall impose the natural renormalization condition, \[iV_3(0,0,0) = 0,\tag{1.67}\] to fix the value of $Z_g$. Just like before, we use the Feynman integration formula to get \[\begin{aligned} &\tilde{D}_F(l^2)\tilde{D}_F((l-k_1)^2) \tilde{D}_F((l+k_2)^2)\\ =& i^3\int dF_3\frac{1}{[x_1(l^2+m^2) + x_2((l-k_1)^2+m^2) + x_3((l+k_2)^2+m^2)]^3}\\ =&-i\int dF_3\frac{1}{[l^2+m^2-2x_2k_1l + 2x_3k_2l+x_2k_1^2 + x_3k_2^2]^3}\\ =&-i\int dF_3\frac{1}{[(l - x_2k_1 + x_3k_2)^2 + m^2 + x_2k_1^2 + x_3k_2^2]^3}\\ =&-i\int dF_3 \frac{1}{(q^2+D)^3}, \end{aligned}\tag{1.68}\] where $q\equiv l -x_2k_1 + x_3k_2$ and $D \equiv m^2 + x_2k_1^2 + x_3k_2^2$. Then, we have \[\begin{aligned} \int\frac{d^dl}{(2\pi)^d}\tilde{D}_F(l^2)\tilde{D}_F((l-k_1)^2) \tilde{D}_F((l+k_2)^2) & = -i\int dF_3 \int\frac{d^dq}{(2\pi)^d}\frac{1}{(q^2+D)^3}. \end{aligned}\tag{1.69}\] Perform a Wick rotation, we find \[\begin{aligned} -i\int\frac{d^dq}{(2\pi)^d}\frac{1}{(q^2+D)^3} &= \int\frac{d^dq_E}{(2\pi)^d}\frac{1}{(q_E^2+D)^3}\\ &=\frac{\Gamma(3-d/2)}{2(4\pi)^{d/2}}D^{-3+d/2}. \end{aligned}\tag{1.70}\] Now, substitute $\varepsilon = 6-d$ and redefine the coupling constant just as before. We look for the asymptotic series of $\varepsilon\to0$. That is \[\begin{aligned} -i(i^3g^2)\tilde{\mu}^{\varepsilon}\int\frac{d^dq}{(2\pi)^d}\frac{1}{(q^2+D)^3} &=\frac{ig^2}{2(4\pi)^3}\Gamma(\varepsilon/2)\left(\frac{4\pi \tilde{\mu}^2}{D}\right)^{\varepsilon/2}\\ &\sim\frac{ig^2}{2(4\pi)^3}(2/\varepsilon-\gamma+\cdots)\left(1+\frac{\varepsilon}{2}\ln\left(\frac{4\pi \tilde{\mu}^2}{D}\right)+ \cdots\right)\\ &=\frac{ig^2}{2(4\pi)^3}\left(\frac{2}{\varepsilon}-\gamma +\ln\left(\frac{4\pi \tilde{\mu}^2}{D}\right) + \cdots\right)\\ &=\frac{i\alpha}2\left(\frac{2}{\varepsilon}+\ln\left(\frac{\mu^2}{D}\right) + \cdots\right), \end{aligned}\tag{1.71}\] where we have used $\mu^2 = 4\pi e^{-\gamma}\tilde{\mu}^2$ and $\alpha \equiv g^2/(4\pi)^3$. Finally, we find the three-point vertex to be \[iV_3(k_1, k_2,k_3)/g \sim iZ_g + i\frac{\alpha}{2}\left(\frac{2}{\varepsilon} + \int dF_3\ln(\mu^2/D) + \cdots \right) + \mathcal{O}(g^5).\tag{1.72}\] By setting $Z_g = 1 +C$, we find \[V_3(k_1, k_2,k_3)/g \sim 1 + \alpha\left[\frac{1}{\varepsilon}+\ln(\mu/m)+\frac{1}{2}\int dF_3\ln(m^2/D) + \dots\right] +C + \mathcal{O}(\alpha^2).\tag{1.73}\] Again, we let $C$ to take the form \[C = -\alpha\left[\frac{1}{\varepsilon}+\ln(\mu/m) + \kappa_C\right] + \mathcal{O}(\alpha^2),\tag{1.74}\] which gives \[V_3(k_1, k_2,k_3)/g \sim 1 - \frac{\alpha}{2}\int dF_3\ln (D/m^2) -\alpha\kappa_C + \mathcal{O}(\alpha^2).\tag{1.75}\] Now, we impose the renormalization condition, $V_3(0,0,0) = 0$, which gives $D = m^2$, and hence \[\kappa_C= 0.\tag{1.76}\] Therefore, our final expression for the three-point vertex to first order in $\alpha$ is given by \[V_3(k_1, k_2,k_3)/g \sim 1 - \frac{\alpha}{2}\int dF_3\ln (D/m^2)C + \mathcal{O}(\alpha^2).\tag{1.77}\]
Now, we have determined all three of the renormalization factor to one-loop order.