Introduction to the calculation of Feynman Integrals Peter Marquard DESY CAPP 2021 1 / 92 Outline 1 Introduction 2 Mathematical prelude & basics 3 Mellin Barnes 4 Integration by parts 5 Differential equations 6 Sector decomposition 2 / 92 Introduction Outline 1 Introduction 2 Mathematical prelude & basics 3 Mellin Barnes 4 Integration by parts 5 Differential equations 6 Sector decomposition 3 / 92 Introduction Introduction Feynman integrals come in different shapes and colors one loop $ many loops many legs $ no legs many scales $ no scales for many types of diagrams many results are known especially one-loop is solved at three loops and more, massive tadpoles and massless propagators have been studied in great detail at two loops, much progress has been made for integrals relevant for 2 ! 2 scattering processes, but every new process requires a new study of the integrals involved 2 ! 3 at two loops looks promising 4 / 92 Introduction Introduction many different methods have been invented over the years to calculate the needed integrals most methods work well for certain classes but fail for others the ultimate method/tool is still missing will present here only an overview of personal selection of methods 5 / 92 Mathematical prelude & basics Outline 1 Introduction 2 Mathematical prelude & basics 3 Mellin Barnes 4 Integration by parts 5 Differential equations 6 Sector decomposition 6 / 92 Mathematical prelude & basics Gamma function Defining property zΓ(z) = Γ(z + 1) = z! Integral representation Z 1 Γ(z) = tz−1e−t dt 0 We see immediately from the properties that Γ(−n) is singular for n = 0; 1; 2;:::. The singularities are simple poles (−1)n 1 Γ(−n + ) = + O(0) n! 7 / 92 Mathematical prelude & basics Series expansion of the Gamma function In many applications one needs more than the pole of the Γ-function. This is best done by using the derivative of log Γ(z) and defines the digamma function Ψ(z) d log(Γ(z)) Γ0(z) Ψ(z) = = dz Γ(z) Therefore, Γ(z − z0) = Γ(z0) + Γ(z0)Ψ(z0)(z − z0) 1 + Γ(z )Ψ0(z ) + Γ(z )Ψ2(z ) (z − z )2 2 0 0 0 0 0 for regular points z0. 8 / 92 Mathematical prelude & basics The Digamma-function Ψ(z) The digamma function satisfies the relation 1 Ψ(z + 1) = Ψ(z) + z For positive integer values the digamma function evaluates to Ψ(1) = −γE ; Ψ(2) = 1 − γE ··· n X 1 Ψ(n + 1) = − γ k E k=1 with the Euler-Mascheroni constant γE = 0:577216 ::: 9 / 92 Mathematical prelude & basics The Digamma-function Ψ(z) cont’d For non-positive integers the digamma functions evaluates again to simple poles 1 Ψ(−n + ) = − + O(0) Γ(z) Ψ(z) 6 10 4 5 2 -2 -1 1 2 3 4 -2 -1 1 2 3 4 -2 -5 -4 -10 -6 10 / 92 Mathematical prelude & basics Schwinger Parametrization From the definition of the Γ function Z 1 Γ(z) = tz−1e−t dt 0 follows immediately the Schwinger Parametrization 1 1 1 Z 2 2 = z−1 −(M −k )t 2 2 z dt t e (−k + M ) Γ(z) 0 by performing the substitution t ! t0 = (M2 − k 2)t 11 / 92 Mathematical prelude & basics Simple example Consider simplest diagram possible, the one-loop tadpole (vacuum diagram) Z 1 I = d4k 1 −k 2 + M2 Either introduce an explicit parametrization of the measure or use the Schwinger parametrization for α = 1 Z Z 1 4 −(M2−k 2)t I1 = d k dt e 0 12 / 92 Mathematical prelude & basics Simple example Consider simplest diagram possible, the one-loop tadpole (vacuum diagram) Z 1 I = d4k 1 −k 2 + M2 Either introduce an explicit parametrization of the measure or use the Schwinger parametrization for α = 1 Z 1 Z −M2t 4 k 2t I1 = dt e d k e 0 13 / 92 Mathematical prelude & basics Simple example perform Wick rotation k0 ! ik0 with the result that 2 2 2 2 2 2 2 2 2 k = k0 − k1 − k2 − k3 ! −k0 − k1 − k2 − k3 and we get Z 1 Z −M2t 4 −k 2t I1 = i dt e d k e 0 14 / 92 Mathematical prelude & basics Simple example Z 1 Z −M2t 4 −k 2t I1 = i dt e d k e 0 doing the Gaussian integral we get Z 1 e−M2t = π2 I1 i dt 2 0 t This integral does not converge for t ! 0 . ) first need a way to give meaning to these kind of integrals. 15 / 92 Mathematical prelude & basics Dimensional regularization Most Feynman integrals are not convergent in four space time dimensions. Common way out is the use of dimensional regularization, where the four-dimensional space time is extended to d dimensions. Divergences of the integrals then become manifest as poles in d − 4. d dimensional integrals behave identical to their four-dimensional counterparts 16 / 92 Mathematical prelude & basics d-dimensional integration d-dimensional integrals have to fulfil these axioms Linearity Z Z Z dd k (af (k) + bg(k)) = a dd k f (k) + b dd k g(k) Scaling Z Z dd k f (s k) = s−d dd k f (k) Translational invariance Z Z dd k f (k + p) = dd k f (k) 17 / 92 Con: dimensional regularization regularizes both UV and IR singularities in the same way scaleless integrals vanish Z dd k(k 2)α = 0 integration by parts Z @ dd k f (k) = 0 @k µ Interchange of integrations Z Z Z Z dd p dd k f (p; k) = dd k dd p f (p; k) Mathematical prelude & basics d-dimensional integration – properties Pro: dimensional regularization regularizes both UV and IR singularities 18 / 92 scaleless integrals vanish Z dd k(k 2)α = 0 integration by parts Z @ dd k f (k) = 0 @k µ Interchange of integrations Z Z Z Z dd p dd k f (p; k) = dd k dd p f (p; k) Mathematical prelude & basics d-dimensional integration – properties Pro: dimensional regularization regularizes both UV and IR singularities Con: dimensional regularization regularizes both UV and IR singularities in the same way 18 / 92 Mathematical prelude & basics d-dimensional integration – properties Pro: dimensional regularization regularizes both UV and IR singularities Con: dimensional regularization regularizes both UV and IR singularities in the same way scaleless integrals vanish Z dd k(k 2)α = 0 integration by parts Z @ dd k f (k) = 0 @k µ Interchange of integrations Z Z Z Z dd p dd k f (p; k) = dd k dd p f (p; k) 18 / 92 Mathematical prelude & basics Gaussian integrals in d dimensions For many purposes the problem of d-dimensional integrations can be reduced to one specific integral: the Gaussian integral in d dimensions Z d −k 2 d d k e = π 2 which is the most natural generalization of the integer dimension one. 19 / 92 Mathematical prelude & basics Gaussian integrals in d dimensions For many purposes the problem of d-dimensional integrations can be reduced to one specific integral: the Gaussian integral in d dimensions d Z 2 π dd k e−Ak = 2 A which is the most natural generalization of the integer dimension one. The dependence on A follows by rescaling k ! pk A 20 / 92 Mathematical prelude & basics Simple example – Improved Z 1 I = dd k 1 −k 2 + M2 Z 1 Z −M2t d −k 2t I1 = i dt e d ke 0 Z 1 e−M2t I = iπd=2 dt 1 d=2 0 t 21 / 92 Mathematical prelude & basics Simple example – Improved Z 1 I = dd k 1 −k 2 + M2 Z Z −M2t d −k 2t I1 = i dt e d k e Z 1 d=2 −d=2 −M2t I1 = iπ dt t e 0 22 / 92 Mathematical prelude & basics Simple example – Improved Z 1 I = dd k 1 −k 2 + M2 Z Z −M2t d −k 2t I1 = i dt e d k e Z 1 d=2 −d=2 −M2t I1 = iπ dt t e 0 Z 1 d=2 2 d=2−1 −d=2 −t I1 = iπ (M ) dt t e 0 = iπd=2(M2)d=2−1Γ(−d=2 + 1) 23 / 92 Mathematical prelude & basics Simple example – Improved Z 1 I = dd k 1 −k 2 + M2 = iπd=2(M2)d=2−1Γ(−d=2 + 1) (M2)d=2−1 overall mass dimension of the integral, could be read off from the original integral Γ(−d=2 + 1) contains the real information singular for d ! 4 1 Γ(−d=2 + 1) = − + (γ − 1) E 1 + −6γ2 + 12γ − π2 − 12 12 E E 24 / 92 Mathematical prelude & basics Simple example – Improved 1 Γ(−d=2 + 1) = − + (γ − 1) E 1 + −6γ2 + 12γ − π2 − 12 12 E E not a very compact result, better to choose a suitable normalization 1 Γ(−d=2 + 1)=Γ(1 + ) = − − 1 − or 1 π2 Γ(−d=2 + 1)= exp(−γ ) = − − 1 + −1 − E 12 25 / 92 Mathematical prelude & basics Simple example – Extended Z 1 I = dd k 1 (−k 2 + M2)α 1 Z 2 Z 2 I = i dt tα−1e−M t dd ke−k t 1 Γ(α) d=2 Z 1 π α−d=2−1 −M2t I1 = i dt t e Γ(α) 0 d=2 Z 1 iπ 2 d=2−α α−d=2−1 −t I1 = (M ) dt t e Γ(α) 0 iπd=2 = (M2)d=2−αΓ(α − d=2) Γ(α) Also the integral with α = 2 is divergent 26 / 92 Mathematical prelude & basics not that simple example: One-loop propagator Consider the one-loop propagator Z 1 P = d4k 1 −k 2(−(k + q)2) introduce Feynman parameters 1 Γ(P k ) Z 1 Z 1 δ(P x − 1) Q xki −1 = i i dx ··· dx i i i i k Q 1 n P k 1 kn Γ(k ) (P x D ) i i D1 ··· Dn i i 0 0 i i i in their simplest form 1 Z 1 1 = dx 2 D1D2 0 [xD1 + (1 − x)D2] 27 / 92 Mathematical prelude & basics not that simple example: One-loop propagator Z 1 P = d4k 1 −k 2(−(k + q)2) Z Z 1 1 = 4 d k dx 2 2 2 0 [−k − 2xk · q − xq ] complete the square Z Z 1 1 = 4 P1 d k dx 2 2 2 2 2 0 [−(k + xq) + x q − xq ] shift k = k + (1 − x)q Z Z 1 1 = 4 P1 d k dx 2 2 2 2 2 0 [−k + x q − xq ] 28 / 92 Mathematical prelude & basics not that simple example: One-loop propagator Z Z 1 1 = 4 P1 d k dx 2 2 2 0 [−k + x(x − 1)q ] doing the momentum integration gives Z 1 d=2−2 d=2 h 2i P1 = iπ Γ(2 − d=2) dx x(x − 1)q 0 let’s assume q2 < 0 Z 1 d=2 2 d=2−2 d=2−2 P1 = iπ Γ(2 − d=2)(−q ) dx [x(1 − x)] 0 29 / 92 Mathematical prelude & basics not that simple example: One-loop propagator Z 1 d=2 2 d=2−2 d=2−2 P1 = iπ Γ(2 − d=2)(−q ) dx [x(1 − x)] 0 what is left is special case of the Beta-function Z Γ(a)Γ(b) B(a; b) = dt ta−1(1 − t)b−1 = Γ(a + b) Γ2(d=2 − 1) P = iπd=2Γ(2 − d=2)(−q2)d=2−2 1 Γ(d + 2) 30 / 92 Mathematical prelude & basics Cuts branch cuts appear in an Feynman integral when the particles in the loop can be produced as real particles in the propagator example before the factor (−q2)d=2−2 could have again be predicted from mass dimension of the the analytic properties of the diagram the imaginary parts then corresponds to the total cross section belonging to the respective cut much information can be gained from studying cuts of Feynman integrals the full results can be obtained from the imaginary part by dispersion integrals 31 / 92 Mathematical prelude & basics Tensor integrals So far we have only dealt with scalar integrals, i.e.