Differential Equations and Feynman Integrals

Abstract

The role of differential equations in the process of calculating Feynman integrals is reviewed. An example of a diagram is given for which the method of differential equations was introduced, the properties of the inverse-mass-expansion coefficients are shown, and modern methods based on differential equations are discussed.

Appendix: Massive Part of J 1(q 2, m 2) in Eq. (5)

In this appendix we consider the following diagrams

(43)

The IBP relations for the internal loop of the diagram produce two equations:

$$\displaystyle \begin{aligned} \begin{array}{rcl} & &\displaystyle (d-1-2\alpha) \, I_2^{(\alpha)}(q^2,m^2) = \alpha \, J_2^{(\alpha+1)}(q^2,m^2) - m^2 \, \alpha \, I_2^{(\alpha+1)}(q^2,m^2) \, ,\quad {} \end{array} \end{aligned} $$

(44)

(45)

where

$$\displaystyle \begin{aligned} J_2^{(\alpha)}(q^2,m^2) = T_{0,\alpha}(m^2) \, L_{1,1}(q^2) \, - S^{(1,2)}(q^2,m^2) \, . {} \end{aligned} $$

(46)

We note that T _0,2(m ² = 0) = 0 in dimensional regularization and

$$\displaystyle \begin{aligned} T_{0,2}(m^2) L_{1,1}(q^2) = \frac{1}{(4\pi)^{d}} \, \frac{R(0,2)A(1,1)}{m^{2(2-d/2)}q^{2(2-d/2)}} \, , {} \end{aligned} $$

(47)

where R(α ₁, α ₂) and A(α ₁, α ₂) are given in Eqs. (13) and (12), respectively.

The IBP relations for internal triangles of the diagram $I_2^{(1)}(q^2,m^2)$ produce two additional equations:

(48)

(49)

Using Eqs. (45) and (49) in the combination: 2 ×(45) + (49) , we have

(50)

So, we have for the mass-dependent part of J ₁(q ², m ²), see Eq. (5),

(51)

i.e. the mass-dependent combination is expressed through the diagram $I_2^{(1)}(q^2,m^2)$ and its derivative.

Using Eq. (44) one obtains

$$\displaystyle \begin{aligned} \left[d-2-\alpha - m^2 \frac{d}{dm^2}\right] \, I_2^{(\alpha)}(q^2,m^2) = \alpha \, J_2^{(\alpha+1)}(q^2,m^2) \, , {} \end{aligned} $$

(52)

i.e. diagram $I_2^{(\alpha )}(q^2,m^2)$ obeys the differential equation with the inhomogeneous term $J_2^{(\alpha +1)}(q^2,m^2)$ having a very simple form: it contains only one-loop diagrams. We see that the last term in $J_2^{(\alpha )}(q^2,m^2)$, see Eq. (46), is expressed through the massive one loop integral $M_{\alpha _1,\alpha _2}(q^2,m^2)$:

(53)

Indeed,

$$\displaystyle \begin{aligned} S^{(1,\alpha)}(q^2,m^2) = \, A(1,1) \, M_{2-d/2,\alpha}(q^2,m^2) \, . {} \end{aligned} $$

(54)

The one-loop diagram M _2−d∕2,α(q ², m ²) can be evaluated by one of some effective methods, for example, by Feynman parameters.

We would like to note that $I_2^{(1)}(q^2,m^2)$ satisfies Eq. (52) with α = 1 that is not of the type of (35). But the integral $I_2^{(2)}(q^2,m^2)$ satisfies Eq. (52) with α = 2 and is of the type of (35). So, it is convenient to rewrite (51) with $I_2^{(2)}(q^2,m^2)$ in its r.h.s.:

(55)

Now we should compare the IBP-based equations for $J_2^{(2)}(q^2,m^2)$ and $J_2^{(3)}(q^2,m^2)$ obtained in the right-hand sides of (55) and (52), respectively, with Eq. (35). Since $J_2^{(3)}(q^2,m^2)=-(d/dm^2) \, J_2^{(2)}(q^2,m^2)$, consider only $J_2^{(2)}(q^2,m^2)$.

So, we should prepare the IBP-based equations for the massive one-loop diagrams M _ε,2(q ², m ²) and M _ε,3(q ², m ²). Applying the IBP procedure with massive distinguished line to M _ε,2(q ², m ²), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} (-3\varepsilon) M_{\varepsilon,2}(q^2,m^2) & =&\displaystyle \varepsilon \bigl[M_{1+\varepsilon,1}(q^2,m^2) - (q^2+m^2) \, M_{1+\varepsilon,2}(q^2,m^2)\bigr]\\ & &\displaystyle -4m^2 \, M_{\varepsilon,3}(q^2,m^2) \, . {} \end{array} \end{aligned} $$

(56)

The corresponding applications of the IBP procedure with massless distinguished line to M _1+ε,1(q ², m ²) and M _1+ε,2(q ², m ²) leads to the following results:

$$\displaystyle \begin{aligned} \begin{array}{rcl} & &\displaystyle (1-4\varepsilon) M_{1+\varepsilon,1}(q^2,m^2) = M_{\varepsilon,2}(q^2,m^2) - (q^2+m^2) \, M_{1+\varepsilon,2}(q^2,m^2) \, ,\quad {} \end{array} \end{aligned} $$

(57)

$$\displaystyle \begin{aligned} \begin{array}{rcl} & &\displaystyle -4\varepsilon M_{1+\varepsilon,2}(q^2,m^2) = 2 \, M_{\varepsilon,3}(q^2,m^2) - 2 \, (q^2+m^2) \, M_{1+\varepsilon,3}(q^2,m^2). {} \end{array} \end{aligned} $$

(58)

The last equation has the following form

$$\displaystyle \begin{aligned} \left[-4\varepsilon - (q^2+m^2) \,\frac{d}{dm^2} \, \right] \, M_{1+\varepsilon,2}(q^2,m^2) = - \,\frac{d}{dm^2} \, M_{\varepsilon,2}(q^2,m^2). {} \end{aligned} $$

(59)

Putting (57) to (56), we have after little algebra

$$\displaystyle \begin{aligned} \begin{array}{rcl} -4\varepsilon(1-3\varepsilon) M_{\varepsilon,2}(q^2,m^2) & =&\displaystyle -2\varepsilon(1-4\varepsilon)\, (q^2+m^2) \, M_{1+\varepsilon,2}(q^2,m^2)\big] \\ & &\displaystyle -4(1-4\varepsilon)\,m^2 \, M_{\varepsilon,3}(q^2,m^2) \, , {} \end{array} \end{aligned} $$

(60)

which transforms to

$$\displaystyle \begin{aligned} \begin{array}{rcl} -\left[4\varepsilon(1-3\varepsilon) + 2(1-4\varepsilon) \frac{d}{dm^2} \right] M_{\varepsilon,2}(q^2,m^2) = -2\varepsilon(1-4\varepsilon)(q^2+m^2) M_{1+\varepsilon,2}(q^2,m^2)\\ {} \end{array} \end{aligned} $$

(61)

So, Eqs. (59) and (61) can be considered as a system of equations having a form similar to Eq. (35).