Example Integral From Particle Physics

Jim FitzSimons

December 24, 1998

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

$\begin{equation} \int_b^c{{\sqrt{(d-t)(c-t)(t-b)(t-a)}\,dt}\over t} \end{equation}$

(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.

$\begin{equation} \int_b^c{{\sqrt{(d-t)(c-t)(t-b)(t-a)}\,dt}\over t} \end{equation}$

= 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) =

$\begin{equation} \int_b^c{{\sqrt{(d-t)(c-t)(t-b)(t-a)}\,dt}\over t} \end{equation}$

(4)

Using a similar notation we will define:

I(e₀) =

$\begin{equation} \int_b^c{dt\over \sqrt{(d-t)(c-t)(t-b)(t-a)}} \end{equation}$

(5)

I(3 e₁) =

$\begin{equation} \int_b^c{{(t-a)^{5\over 2}\,dt}\over \sqrt{(d-t)(c-t)(t-b)}} \end{equation}$

(6)

I(2 e₁) =

$\begin{equation} \int_b^c{{(t-a)^{3\over 2}\,dt}\over \sqrt{(d-t)(c-t)(t-b)}} \end{equation}$

(7)

I(e₁) =

$\begin{equation} \int_b^c{{\sqrt{t-a}\,dt}\over \sqrt{(d-t)(c-t)(t-b)}} \end{equation}$

(8)

I(-e₁) =

$\begin{equation} \int_b^c{dt\over {\sqrt{(d-t)(c-t)(t-b)}\,(t-a)^{3\over 2}}} \end{equation}$

(9)

I(-e₅) =

$\begin{equation} \int_b^c{dt\over {\sqrt{(d-t)(c-t)(t-b)(t-a)}\,t}} \end{equation}$

(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)

(12)

Expand I(3 e₁).

I(3 e₁) =

(d+c+b-3 a) I(2 e₁)-

(d (c+b-2 a)+c (b-2 a)

a (2 b-3 a)) I(e₁)+

(d-a) (c-a) (b-a) I(e₀)

(13)

Substitute for I(2 e₁) in equation 13 using equation 12.

I(3 e₁) =

(d+c+b-3 a) (d-a) (c-a) (b-a) I(-e₁)+

(3 d²

+2 d (c+b-5 a)+3 c²+2 c (b-5 a)+3 b²-10 b a+15 a²) I(e₁)

(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)

+d c b a I(-e₅)-

(d (c (3 b+a)+a (b-a))+a (c-a) (b-a)) I(e₀)

(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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On 31 Dec 1998, 12:00.