one-loop

and using induction we obtain a general formula
		1		= (n) Z01 dx1	Z01 x1		dx2
	a1a2		an
				1 x1 xn 1			dxn 1
				Z0				(A.13)
						[a1x1 + a2x2 + + an(1 x1 xn 1)]n

Before closing the section let us give an example that will be useful in the self-energy case. Consider the situation with the kinematics described in Fig. (16).

k

p p

p+k

Figure 16:

We get

1

d

k

1

I =

Z0

Z

d

(2 )d [(k + p)2

m12 + i ] [k2 m22 + i ]

1

dx

d

k

1

Z0

Z

d

(2 )d [k2 + 2p k x + p2 x m12 x m22 (1 x) + i ]2

=

1

dx

ddk

1

dx Z

=

Z0

(2 )d [k2 + 2P

k M 2 + i ]2

d

k

1

1

=

Z0

dx Z

d

(A.14)

(2 )d [(k + P )2 P 2 M 2 + i ]2

where in the last line we have completed the square in the term with the loop momenta k. The quantities P and M 2 are, in this case, de ned by

P = xp	(A.15)
and
M 2 = x p2 + m12 x + m22 (1 x)	(A.16)

They depend on the masses, external momenta and Feynman parameters, but not in the loop momenta. Now changing variables k ! k P we get rid of the linear terms in k andnally obtain

I = Z0	1	Z	ddx	1
	dx					(A.17)
			(2 )d		[k2 C + i ]2

60

Im k0

x

x

Re k0

Figure 17:

where C is independent of the loop momenta k and it is given by

C = P 2 + M 2

(A.18)

Notice that the i factors will add correctly and can all be put as in Eq. (A.17).

A.3 Wick Rotation

From the example of the last section we can conclude that all the scalar integrals can be

reduced to the form	ddx		k2r
Ir;m = Z				(A.19)
	(2 )d		[k2 C + i ]m

It is also easy to realize that also all the tensor integrals can be obtained from the scalar integrals. For instance

Z	ddx			k
	(2 )d		[k2 C + i ]m
	ddx
Z				k k
	(2 )d [k2			C + i ]m

= 0
=	1	g	Z	ddx	k2	(A.20)
	d				[k2 C + i ]m

and so on. Therefore the integrals Ir;m are the important quantities to evaluate. We will consider that C > 0. The case C < 0 can be done by analytical continuation of the nal formula for C > 0.

To evaluate the integral Ir;m we will use integration in the complex plane of the variable k0 as described in Fig. 17. We can then write

Ir;m = Z	dd 1k					k2r
		Z	dk0				(A.21)
	(2 )d				hk02 j~kj2 C + i im
The function under the integral has poles for
	k0 = q					i
				j~kj2 + C			(A.22)

61

has shown in Fig. 17. Using the properties of functions of complex variables (Cauchy theorem) we can deform the contour, changing the integration from the real to the imaginary axis plus the two arcs at in nity. This can be done because in deforming the contour we do not cross any pole. Notice the importance of the i prescription to be able to do this. The contribution from the arcs at in nity vanishes in dimension su ciently low for the integral to converge, as we assume in dimensional regularization. We have then changed the integration along the real axis into an integration along the imaginary axis in the plane of the complex variable k0. If we write

+1

+1

k0 = ikE0

com

Z1

dk0

! i Z1 dkE0

(A.23)

and k

2

0

)

2

~

2

0

2

~

2

2

0 ~

= (k

jkj

= (kE )

jkj

kE

, where kE = (kE

; k) is an euclidean

vector. By this we mean that is we calculate the scalar product using the euclidean metric diag(+; +; +; +),

2	0		2	~		2		(A.24)
kE	= (kE )			+ jkj
We can them write		ddkE					kE2r
	r m
Ir;m = i( 1)	Z							(A.25)
		(2 )d [kE2					+ C]m

where we do not need the i because the denominator is strictly positive (C > 0). This procedure is known as Wick Rotation. We note that the Feynman prescription for the propagators that originated the i rule for the denominators is crucial for the Wick rotation to be possible.

A.4 Scalar integrals in dimensional regularization

We have seen in the last section that the scalar integrals to be calculated with dimensional regularization had the general form of Eq. (A.25). We are now going to nd a general formula for Ir;m. We begin by writing

Z	Z
ddkE =	dk			k	d 1 d d 1	(A.26)

q

0 2 j~ j2

where k = (kE ) + k is the length of the vector kE in the euclidean space in d dimensions and d d 1 is the solid angle that generalizes spherical coordinates in that euclidean space. The angles are de ned by

kE = k(cos 1; sin 1 cos 2; sin 1 sin 2; sin 1 sin 2 cos 3; : : : ; sin 1 sin d 1 )

We can then write

Z		sin 1d 2 d 1 Z0					2
	d d 1 = Z0								d d 1
Using now						m+1
								)
	sin m d = p ( 2
	Z0
					(	m+2		)
						2

(A.27)

(A.28)

(A.29)

62

where (z) is the gamma function (see section A.6) we get

d

Z

d d 1 = 2

2

( d2 )

The integration in k is done using the result

1

p

a

p+1

Z0

dx (xn + an)q = ( 1)

n sin( p+1n ) ( p+12 q + 1)

x

q 1

p+1 nq

(

n )

and we nally get

r m

d

(m r

d

Ir;m = iCr m+ d2

( 1)

(r + 2 )

2 )

d

(4 ) 2

( d2 )

(m)

(A.30)

(A.31)

(A.32)

Before ending the section we note that the integral representation for Ir;m, Eq. (A.19), is

valid only for d < 2(m r) to ensure convergence when k ! 1. However the nal form in Eq. (A.32) can be analytically continued for all values of d except for those where the function (m r d=2) has poles, that is for (see section A.6),

m r	d	6= 0; 1; 2; : : :	(A.33)
	2

For the application in dimensional regularization it is convenient to rewrite Eq. (A.32) using the relation d = 4 . we get

I	r;m	= i	( 1)r m		4
			(4 )2		C

	C2+r m (2 + r						(A.34)
			2 ) (m r 2 +			2 )
	2
		(2 2 )			(m)

A.5 Tensor integrals in dimensional regularization

We are frequently faced with the task of evaluating the tensor integrals of the form of Eq. (A.7),

T^n 1 p Z	ddk		k 1		k p	(A.35)
	(2 )d		D0D1		Dn 1

The rst step is to reduce to one common denominator using the Feynman parameterization technique. The result is,

1

1 x1 xn 1

ddk

k 1

k p

T^n 1 p = (n) Z0

dx1 Z0

dxn 1

Z

[k2 + 2k

M 2

+ i ]n

(2 )d

P

1

1 x1 xn 1

= (n) Z0

dx1 Z0

dxn 1 In 1 p

(A.36)

where we have de ned

ddk

k 1

k p

In 1 p Z

+ i ]n

(A.37)

(2 )d [k2 + 2k

P M 2

63

that we call, from now on, the tensor integral. In principle all these integrals can be written in terms of scalar integrals. It is however convenient to have a general formula for them. This formula can be obtained by noticing that

@

1

= n

2k

(A.38)

@P

[k2 + 2k P M 2 + i ]n

[k2 + 2k P M 2 + i ]n+1

Using the last equation one can write the nal result

i (4 ) =2

1 dt

@

@

In 1 p =

Z0

tn 3+ =2

e t C

(A.39)

16 2 (n)

(2t)p

@P 1

@P p

where C = P 2 + M 2.

After doing the derivatives the remaining integrals can be done

using the properties of the function (see section A.6). Notice that P , M 2 and therefore also C depend not only in the Feynman parameters but also in the exterior momenta. The advantage of having a general formula is that it can be programmed [10] and all the integrals can then be obtained automatically.

A.6 function and useful relations

The function is de ned by the integral

(z) = Z01 tz 1e tdt	(A.40)
or equivalently
Z01 tz 1e tdt = z (z)	(A.41)
The function (z) has the following important properties

(z + 1) = z (z)

(n + 1) = n!

(A.42)

Another related function is the logarithmic derivative of the function, with the properties,

	(z) =	d		ln (z)			(A.43)
		dz
	(1) =				1		(A.44)
	(z + 1) = (z) +						(A.45)
					z

where is the Euler constant. The function (z) has poles for z = 0; 1; 2; . Near the pole z = m we have

64

(z) =

( 1)m

1

+

( 1)m

(m + 1) + O(z + m)

(A.46)

m! m + z

m!

From this we conclude that when ! 0

2

1)n

2

=

+ (1) + O( )

( n + ) =

(

+

(n + 1) + 1

(A.47)

2

n!

and

2

2

(1 + ) = 1 +

2 +

!

+

;

! 0

(A.48)

6

2!

Using these results we can expand our integrals in powers of and separate the divergent and nite parts. For instance for the one of the integrals of the self-energy,

i

4

2 (1 +

)

2

I0;2 =

2

(4 )2

C

i

2

=

+ ln 4 ln C + O( )

16 2

=

i

[ ln C + O( )]

(A.49)

16 2

where we have introduced the notation

=

2

+ ln 4

(A.50)

for a combination that will appear in all expressions.

A.7 Explicit formulas for the 1{loop integrals

Although we have presented in the previous sections the general formulas for all the integrals that appear in 1{loop, Eqs. (A.34) and (A.39), in practice it is convenient to have expressions for the most important cases with the expansion on the already done. The results presented below were generated with the Mathematica package OneLoop [10] from the general expressions. In these results the integration on the Feynman parameters has still to be done. This is in general a di cult problem and we have presented in section 3 an alternative way of expressing these integrals, more convenient for a numerical evaluation.

A.7.1 Tadpole integrals

With the de nitions of Eqs. (A.34) and (A.39) we get

I0;1	=		i	C(1 + ln C)			(A.51)
			16 2
I1	=	0					(A.52)
			i	1			(A.53)
I1	=					C2 g (3 + 2 2 ln C)
			16 2		8

65

where for the tadpole integrals

P = 0 ; C = m2

(A.54)

because there are no Feynman parameters and there is only one mass. In this case the above results are nal.

A.7.2 Self{Energy integrals

For the integrals with two denominators we get,

I0;2

I2

I2

I2

=	i		( ln C)			(A.55)
	16 2
=	i	( + ln C)P				(A.56)
	16 2
=	i	1			Cg (1 + ln C) + 2( ln C)P P	(A.57)
	16 2	2
=	i	1			Cg (1 + ln C)P Cg (1 + ln C)P	(A.58)
	16 2	2

Cg (1 + ln C)P (2 P P 2 ln CP P )P

where, with the notation and conventions of Fig. (15), we have

P = x r1 ; C = x2 r12 + (1 x) m20 + x m21 x r12

A.7.3 Triangle integrals

For the integrals with three denominators we get,

I0;3	=	i		1
		16 2 2C
I3	=	i	1		P
		16 2		2C
I3	=	i	1		Cg ( ln C) 2P P
		16 2 4C
I3	=	i	1		Cg ( + ln C)P + Cg ( + ln C)P
		16 2		4C
					+ Cg ( + ln C)P + 2P P P
I3	=	i	1		C2 (1 + ln C) g g + g g + g g
		16 2		8C

+2C ( ln C) g P P + g P P + g P P + g P P

(A.59)

(A.60)

(A.61)

(A.62)

(A.63)

(A.64)

(A.65)

(A.66)

(A.67)

66

+g P P + g P P 4P P P P (A.68)

where

P	=	x1 r1 + x2 r2	(A.69)
C	=	x12 r12 + x22 r22 + 2x1 x2 r1 r2 + x1 m12 + x2 m22	(A.70)
		+(1 x1 x2) m02 x1 r12 x2 r22	(A.71)

A.7.4 Box integrals

I0;4	=	i	1								(A.72)
		16 2		6C2
I4	=	i		1	P						(A.73)
		16 2 6C2
4		16 2		12C	2
I	=	i		1		Cg		2P P			(A.74)
		i
I4	=		1			C (g P + g P + g P ) 2P P P					(A.75)
		16 2		12C	2
I4	=	i	1		2	C2 ( ln C)			g g + g g	+ g g	(A.76)
		16 2		12C
		2C g P P + g P P + g P P + g P P									(A.77)
		+ g P P + g P P + 4P P P P									(A.78)

where

P	=	x1 r1 + x2 r2 + x3 r3	(A.79)
C	=	x12 r12 + x22 r22 + x32 r32 + 2x1 x2 r1 r2 + 2x1 x3 r1 r3 + 2x2 x3 r2 r3	(A.80)
		+x1 m12 + x2 m22 + x3 m32 + (1 x1 x2 x3) m02	(A.81)
		x1 r12 x2 r22 x3 r32	(A.82)

A.8 Divergent part of 1{loop integrals

When we want to study the renormalization of a given theory it is often convenient to have expressions for the divergent part of the one-loop integrals, with the integration on the Feynman parameters already done. We present here the results for the most important cases. These divergent parts were calculated with the help of the package OneLoop [10].

67

A.8.1 Tadpole integrals

Div hI0;1i	=		i		m2			(A.83)
			16 2
Div hI1 i	=	0		i	1			(A.84)
Div hI1 i	=						m4 g	(A.85)
			16 2			4

A.8.2 Self{Energy integrals

h i

Div I0;2

h i

Div I2

h i

Div I2

h i

Div I2

=	i
	16 2
	i				1
=						r1
	16 2				2
=	i	1		(3m12 + 3m02 r12)g + 4r1 r1
	16 2		12
=	i				1		(4m12 + 2m22 r12) (g r1
	16 2				24

(A.86)

(A.87)

(A.88)

+ g r1 + g r1 ) (A.89)

+ 6 r1 r1 r1

(A.90)

A.8.3 Triangle integrals

h i

Div I0;3

h i

Div I3

h i

Div I3

h i

Div I3

h i

Div I3

= 0

(A.91)

= 0

(A.92)

=

i

1

g

(A.93)

16 2

4

i

1

=

g (r1 + r2 ) + g (r1 + r2 ) + g (r1 + r2 )

(A.94)

16 2

12

=

i

1

(2m12 + 2m22 + 2m32) g g + g g + g g

(A.95)

16 2

48

+g h2r1 r1 + r1 r2 + (r1

$ r2)i + g h2r1 r1 + r1 r2 + (r1

$ r2)i

(A.96)

+g h2r1 r1 + r1 r2 + (r1

$ r2)i + g h2r1 r1 + r1 r2 + (r1

$ r2)i

(A.97)

+g h2r1 r1 + r1 r2 + (r1

$ r2)i + g h2r1 r1 + r1 r2 + (r1

$ r2)i

(A.98)

+ r12 + r1 r2

r22

g g + g g + g g

(A.99)

68

A.8.4 Box integrals
Div hI0;4i	=	Div hI4 i = Div hI4 i = Div hI4 i = 0						(A.100)
Div hI4 i	=		i		1		g g + g g + g g	(A.101)
			16 2	24

A.9 Useful results for PV integrals

Although the PV approach is intended primarily to be used numerically there are situations where one wants to have explicit results. These can be useful to check cancellation of divergences or because in some simple cases the integrals can be done analytically. We note that as our conventions for the momenta are the same in sections 3 and A.7 one can read immediately the integral representation of the PV in terms of the Feynman parameters just by comparing both expressions, not forgetting to take out the i=(16 2) factor. For instance, from Eq. (3.21) for C and Eq. (A.64) for I3 we get

2		1			1 x1		x1x2
C12(r12; r122 ; r22; m02; m12; m22) = (3)		Z0	dx1	Z0		dx2		(A.102)
	4						C
with
C = x12 r12 + x22 r22 + x1 x2 (r12 + r22 r122 ) + x1 m12 + x2 m22
+(1 x1 x2) m02 x1 r12 x2 r22								(A.103)

A.9.1 Divergent part of the PV integrals

The package LoopTools provides ways to numerically check for the cancellation of divergences. However it is useful to know the divergent part of the Passarino-Veltman integrals. Only a small number of these integrals are divergent. They are

h

i

Div A0(m02)

=

m02

(A.104)

Div hB0

(r102 ; m02; m12)i

=

(A.105)

Div hB1

(r102 ; m02; m12)i

=

1

(A.106)

2

Div hB00

(r102 ; m02; m12)i

=

1

3m02 + 3m12 r102

(A.107)

12

Div hB11

(r102 ; m02

; m12)i

=

1

(A.108)

3

Div hC00(r102 ; r122 ; r202 ; m02; m12

; m22)i

=

1

(A.109)

4

Div hC001 (r102 ; r122 ; r202 ; m02; m12

; m22)i

=

1

(A.110)

12

69