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Example Integral From Particle Physics

Jim FitzSimons

December 24, 1998






An integral that occurs in particle physics might be interesting to reduce. It is
ó
õ
c


b
((d-t)(c-t)(t-b)(t-a))1/2 dt
t
,     (1)
a special complete case of I(1,1,1,1,-1). For the integral to be well defined, one assumes a<b<c<d. It is symmetric in some peculiar combinations of a,b,c,d, and I once tried unsuccessfully to understand why.

1  Numerical Example

For a numerical example I will pick random rational values of a,b,c,d.
a=65/168,b=29/47,c=116/129,d=147/82     (2)
Use the values 2 in integral 1 and numerically integrate.
ó
õ
c


b
((d-t)(c-t)(t-b)(t-a))1/2 dt
t
= 0.0254192442518     (3)
This numeric result will be used to check the symbolic results.

2  Symbolic Expansion

To simplify the equations we will define integral 1
I(m)= ó
õ
c


b
((d-t)(c-t)(t-b)(t-a))1/2 dt
t
.     (4)
Using a similar notation we will define:
I(e0)= ó
õ
c


b
dt
((d-t)(c-t)(t-b)(t-a))1/2
,     (5)
I(3 e1)= ó
õ
c


b
(t-a)5/2 dt
((d-t)(c-t)(t-b)1/2
,     (6)
I(2 e1)= ó
õ
c


b
(t-a)3/2 dt
((d-t)(c-t)(t-b))1/2
,     (7)
I(e1)= ó
õ
c


b
(t-a)1/2  dt
((d-t)(c-t)(t-b))1/2
,     (8)
I(-e1)= ó
õ
c


b
dt
((d-t)(c-t)(t-b))1/2 (t-a)3/2
,     (9)
I(-e5)= ó
õ
c


b
dt
((d-t)(c-t)(t-b)(t-a))1/2  t
.     (10)
A partial fraction expansion of the integral 4 gives this result:
I(m) = I(3 e1)-(d+c+b-2 a) I(2 e1)+(d (c+b-a)
    +(c-a) (b-a)) I(e1)+d c b a I(-e5)-d c b I(e0)     (11)
The integral I(3 e1) and I(2 e1) are not basic integrals listed in my table of integrals. They need to be expanded in terms of basic integrals. First expand I(2 e1).
I(2 e1)=
(d+c+b-3 a) I(e1)-(d-a) (c-a) (b-a) I(-e1)
2
.     (12)
Expand I(3 e1).
I(3 e1) = 3/4 (d+c+b-3 a) I(2 e1) -1/2 (d (c+b-2 a)+c (b-2 a)
    -1/4 a (2 b-3 a)) I(e1) +1/4 (d-a) (c-a) (b-a) I(e0)     (13)
Substitute for I(2 e1) in equation 13 using equation 12.
I(3 e1) = -3/8 (d+c+b-3 a) (d-a) (c-a) (b-a) I(-e1) +1/8 (3 d2
    +2 d (c+b-5 a)+3 c2+2 c (b-5 a)+3 b2-10 b a +15 a2) I(e1)
    +1/4 (d-a) (c-a) (b-a) I(e0)     (14)
Substitute for I(2 e1) and I(3 e1) in equation 11 using equation 12 and equation 14.
I(m) =
(d+c+b+a) (d-a) (c-a) (b-a) I(-e1) +(4 (d a+c b)-(b-a-d+c)2) I(e1)
8
   
+d c b a I(-e5)-
(d (c (3 b+a)+a (b-a))+a (c-a) (b-a)) I(e0)
4
.
    (15)

3  Test Symbolic Expansion

I will use a numeric example to test the symbolic expansion. Evaluate then basic integrals I(-e1), I(e1), I(-e5), and I(e0).
I(-e1) = 15.5551588727087865579456027522749969092112909472,     (16)
I(e1) = 1.88326583323058247475968437222597866805394080504,     (17)
I(-e5) = 7.07042777739491562524941501012375160488888670702,     (18)
I(e0) = 5.20183713283531436045024182994033810742118035674     (19)
Substitute the values of I(-e1), I(e1), I(-e5), and I(e0) into equation 15.
I(m)=0.02541924425179893774029322781478333663469179658     (20)
The test of the symbolic value in equation 20 matches the numerical integration value in equation 3.

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