# Evaluate $1$-loop contribution to the $4$-point Green's function

I am trying to evaluate the following integral $$I = \int \frac{d^d p_\text{E}}{(2 \pi)^d} \frac{1}{(p_\text{E}^2+m^2)((q_\text{E}-p_\text{E})^2 + m^2)} \tag{1}$$ where $p_\text{E}$ is the momentum in Euclidean space (i.e. after having performed a Wick rotation). Using Feynman's parametrization trick, and some effort, equation $(1)$ can be written as: $$I = \int\limits^1_0 \mathrm{d} x \int \frac{d^d p'_\text{E}}{(2 \pi)^d} \frac{1}{\left[p_\text{E}'^2 + q_\text{E}^2 x(1-x) + m^2 \right]^2} \tag{2}$$ where $p'_\text{E} = p_\text{E} - q(1-x)$. I know that equation $(2)$ is correct as I have verified by checking 2 different sources. Now, according to the two aforementioned sources, equation $(2)$ can be written evaluated as: $$I = \frac{\pi^{d/2}}{(2 \pi)^d} \Gamma(2-d/2) \int\limits^1_0 \mathrm{d} x \; (q_\text{E}^2 x(1-x) + m^2)^{d/2-2} \tag{3}$$ where $\Gamma(2-d/2)$ is the gamma function. However, I don't know where this comes from.

So far, the closest I have come is by writing equation $(2)$ as: $$I = \int\limits^1_0 \mathrm{d} x \int \frac{d^d p'_\text{E}}{(2 \pi)^d} \int\limits^\infty_0 \mathrm{d}u \int\limits^\infty_0 \mathrm{d}u \; e^{-u(p_\text{E}'^2 + q_\text{E}^2 x(1-x) + m^2 )} \tag{4}$$ We can perform the integration over $p'_\text{E}$ by using Gaussian integration: $$\int \mathrm{d}^d p_{\text{E}} \; e^{-up'^2_{\text{E}}} = \left(\frac{\pi}{u}\right)^{d/2}$$ Thus equation $(4)$ becomes: $$I = \frac{\pi^{d/2}}{(2 \pi)^d} \int\limits^1_0 \mathrm{d} x \int\limits^\infty_0 \mathrm{d}u \int\limits^\infty_0 \mathrm{d}u \; u^{-d/2}e^{-u(q_\text{E}^2 x(1-x) + m^2 )} \tag{5}$$ Now, letting: $$w = \left(q_\text{E}^2 x(1-x) + m^2 \right) u \Rightarrow \mathrm{d} u = \left(q_\text{E}^2 x(1-x) + m^2 \right)^{-1} \; \mathrm{d} w$$ and substituting the above into equation $(5)$ yields: \begin{aligned} I & = \frac{\pi^{d/2}}{(2 \pi)^d} \int\limits^1_0 \mathrm{d} x \int\limits^\infty_0 \mathrm{d}w \int\limits^\infty_0 \mathrm{d}w \; w^{-d/2}e^{-w} \left(q_\text{E}^2 x(1-x) + m^2 \right)^{d/2-2} \\& = \frac{\pi^{d/2}}{(2 \pi)^d} \int\limits^1_0 \mathrm{d} x \int\limits^\infty_0 \mathrm{d}w \; \Gamma(1-d/2) \left(q_\text{E}^2 x(1-x) + m^2 \right)^{d/2-2} \tag{6} \end{aligned} At this point I'm stuck. I have no idea if the steps taken to get to equation $(6)$ are correct, but equation $(6)$ does kind of look like equation $(3)$ so I hope they are. Could anybody help me and tell me how I can derive equation $(3)$?

• The expression you mentioned is known as "the master formula" when doing loop calculations. QFT books probably have a derivation of it. I am pretty sure Peskin and Schroeder do, when they introduce these techniques. – JeffDror Feb 12 '14 at 12:17
• @JeffDror I have looked at P&S, but couldn't find anything. Maybe I have missed it; I will check again. – Hunter Feb 12 '14 at 12:20
• @JeffDror what expression in my original post is known as "the master formula". I can't find anything in P&S which could resolve my problem. – Hunter Feb 12 '14 at 12:58
• I was referring to equation 3. This equation and its modifications are derived on pg. 249-251. The set of master formula in Minkowski space (i.e., you avoid the hassle of Wick rotation) are just quoted in the appendix, p.g. 807. – JeffDror Feb 12 '14 at 13:01
• Also, I think you have a typo on your equation 6. I don't think it needs the extra $w$ integration. – JeffDror Feb 12 '14 at 13:05

You can prove a general formula $$\int\frac{d^dp_{\mathrm{E}}}{(2\pi)^d} \frac{(p_{\mathrm{E}}^2)^m}{(p_{\mathrm{E}}^2+\Delta)^n}= \frac{1}{(4\pi)^{d/2}}\frac{\Gamma(m+d/2)\Gamma(n-m-d/2)}{\Gamma(d/2)\Gamma(n)} \left(\frac{1}{\Delta}\right)^{n-m-d/2},\quad n>m+d/2$$ by using Gaussian integral and Euler integral of the first kind.
• Thanks, however I am not sure how to apply it to the above problem. According to your identity, $n=2$ and $m=0$, thus: $$I = \frac{1}{(4 \pi)^{d/2}} \frac{\Gamma(d/2)\Gamma(2-d/2)}{\Gamma(d/2)\Gamma(2)} (\Delta)^{2-d/2}$$ which is not equal to equation $(3)$. – Hunter Feb 12 '14 at 13:14
• And $\Delta=q_{\mathrm{E}}^2x(1-x)+m^2$, it is independent of $p_{\mathrm{E}}$. – soliton Feb 12 '14 at 13:17
• Ahhh ok, so it does look very similar. What has happened to the $\int\limits^1_0 \mathrm{d} x$ part that should go in front of $(\Delta)^{2-d/2}$ (see equation $(3)$)? – Hunter Feb 12 '14 at 13:20
• We cannot obtain an analytical expression of the integral (3). For $d=4$, (1) is log divergent. – soliton Feb 12 '14 at 13:29
• The $\int_0^1d x$ is copyed from (2). The two $I$s are not the same in your comments. – soliton Feb 12 '14 at 13:43
Although the answer given by soliton is sufficient, I've found a way to explicitly evaluate this integral (in case anybody might be interested). Let us start from equation $(2)$ in the original message: $$I = \int\limits^1_0 \mathrm{d} x \int \frac{d^d p'_\text{E}}{(2 \pi)^d} \frac{1}{\left[p_\text{E}'^2 + q_\text{E}^2 x(1-x) + m^2 \right]^2} \tag{2}$$ This can be written as: $$I = \int\limits_0^1 \mathrm{d}x \int \frac{\mathrm{d}^d p_{\text{E}}'}{(2\pi)^d} \int\limits^\infty_0 \mathrm{d} u_1 \; e^{- u_1 \left( p_{\text{E}}'^2 + q_\text{E}^2 x(1-x) + m^2 \right)} \int\limits^\infty_0 \mathrm{d} u_2 \; e^{- u_2 \left( p_{\text{E}}'^2 +q_\text{E}^2 x(1-x) + m^2 \right)}$$ The integration over $p_{\text{E}}'$ becomes an ordinary Gaussian integral: $$\int \mathrm{d}^d p_{\text{E}}' \; e^{-p_{\text{E}}'^2(u_1+u_2)} = \left( \frac{\pi}{u_1+u_2} \right)^{d/2}$$ and so: $$I = \frac{1}{(4 \pi)^{d/2}} \int\limits_0^1 \mathrm{d}x \int\limits^\infty_0 \mathrm{d} u_1 \int\limits^\infty_0 \mathrm{d} u_2 \; \left(\frac{1}{u_1+u_2}\right)^{d/2} e^{- \left( q_\text{E}^2 x(1-x) + m^2 \right)\left(u_1+u_2 \right)} \tag{3}$$ We can make the following substitution: $$\begin{array}{cc} u_1 = sx \; ,& u_2 = s(1-x) \end{array}$$ where $0 < s < \infty$ (and of course $0 < x < 1$), such that: $$u_1+u_2=s \; \; \; \text{and} \; \; \; J = \left|\frac{\partial(u_1,u_2)}{\partial(x,s)} \right| = s$$ Thus, equation $(3)$ becomes: $$I = \frac{1}{(4 \pi)^{d/2}} \int\limits_0^1 \mathrm{d}x \int\limits^\infty_0 \mathrm{d} s \; s^{1-d/2} e^{- \left( q_\text{E}^2 x(1-x) + m^2 \right)s}$$ Now, we make the substitution: $$w = \left( q_\text{E}^2 x(1-x) + m^2 \right)s \Rightarrow \mathrm{d} s = \left( q_\text{E}^2 x(1-x) + m^2 \right)^{-1} \mathrm{d} w$$ such that: \begin{aligned} I & = \frac{1}{(4 \pi)^{d/2}} \int\limits_0^1 \mathrm{d}x \int\limits^\infty_0 \mathrm{d} w \; w^{1-d/2} e^{- w} \left( q_\text{E}^2 x(1-x) + m^2 \right)^{d/2-2} \\& = \frac{1}{(4 \pi)^{d/2}} \Gamma(2-d/2) \int\limits_0^1 \mathrm{d}x \; \left( q_\text{E}^2 x(1-x) + m^2 \right)^{d/2-2} \end{aligned}