Example Integral From Particle Physics

L^{A} T_{E} X

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

\int_{b}^{c} \frac{\sqrt{(d - t) (c - t) (t - b) (t - a)} 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.

\int_{b}^{c} \frac{\sqrt{(d - t) (c - t) (t - b) (t - a)} dt}{t} = 0.0254192442518

(3)

This numeric result will be used to check the symbolic results.

To simplify the equations we will define integral 1

I (m) = \int_{b}^{c} \frac{\sqrt{(d - t) (c - t) (t - b) (t - a)} dt}{t} .

(4)

Using a similar notation we will define:

I (e_{0}) = \int_{b}^{c} \frac{dt}{\sqrt{(d - t) (c - t) (t - b) (t - a)}},

(5)

I (3 e_{1}) = \int_{b}^{c} \frac{(t - a)^{\frac{5}{2}} dt}{\sqrt{(d - t) (c - t) (t - b)}},

(6)

I (2 e_{1}) = \int_{b}^{c} \frac{(t - a)^{\frac{3}{2}} dt}{\sqrt{(d - t) (c - t) (t - b)}},

(7)

I (e_{1}) = \int_{b}^{c} \frac{\sqrt{t - a} dt}{\sqrt{(d - t) (c - t) (t - b)}},

(8)

I (- e_{1}) = \int_{b}^{c} \frac{dt}{\sqrt{(d - t) (c - t) (t - b)} (t - a)^{\frac{3}{2}}},

(9)

I (- e_{5}) = \int_{b}^{c} \frac{dt}{\sqrt{(d - t) (c - t) (t - b) (t - a)} t} .

(10)

A partial fraction expansion of the integral 4 gives this result:

\begin{matrix} I (m) = & I (3 e_{1}) - (d + c + b - 2 a) I (2 e_{1}) + (d (c + b - a) \\ + (c - a) (b - a)) I (e_{1}) + d c b a I (- e_{5}) - d c b I (e_{0}) & (11) \end{matrix}

The integral

I (3 e_{1})

and

I (2 e_{1})

are not basic integrals listed in my table of integrals. They need to be expanded in terms of basic integrals. First expand

I (2 e_{1})

I (2 e_{1}) = \frac{(d + c + b - 3 a) I (e_{1}) - (d - a) (c - a) (b - a) I (- e_{1})}{2} .

(12)

Expand

I (3 e_{1})

\begin{matrix} I (3 e_{1}) = & \frac{3}{4} (d + c + b - 3 a) I (2 e_{1}) - \frac{1}{2} (d (c + b - 2 a) + c (b - 2 a) \\ - \frac{1}{4} a (2 b - 3 a)) I (e_{1}) + \frac{1}{4} (d - a) (c - a) (b - a) I (e_{0}) & (13) \end{matrix}

Substitute for

I (2 e_{1})

in equation 13 using equation 12.

\begin{matrix} I (3 e_{1}) = & - \frac{3}{8} (d + c + b - 3 a) (d - a) (c - a) (b - a) I (- e_{1}) + \frac{1}{8} (3 d^{2} \\ + 2 d (c + b - 5 a) + 3 c^{2} + 2 c (b - 5 a) + 3 b^{2} - 10 b a + 15 a^{2}) I (e_{1}) \\ + \frac{1}{4} (d - a) (c - a) (b - a) I (e_{0}) & (14) \end{matrix}

Substitute for

I (2 e_{1})

and

I (3 e_{1})

in equation 11 using equation 12 and equation 14.

\begin{matrix} I (m) = & \frac{(d + c + b + a) (d - a) (c - a) (b - a) I (- e_{1}) + (4 (d a + c b) - (b - a - d + c)^{2}) I (e_{1})}{8} \\ + d c b a I (- e_{5}) - \frac{(d (c (3 b + a) + a (b - a)) + a (c - a) (b - a)) I (e_{0})}{4} . & (15) \end{matrix}

I will use a numeric example to test the symbolic expansion. Evaluate then basic integrals

I (- e_{1})

I (e_{1})

I (- e_{5})

, and

I (e_{0})

\begin{matrix} I (- e_{1}) & = & 15.5551588727087865579456027522749969092112909472, & (16) \\ I (e_{1}) & = & 1.88326583323058247475968437222597866805394080504, & (17) \\ I (- e_{5}) & = & 7.07042777739491562524941501012375160488888670702, & (18) \\ I (e_{0}) & = & 5.20183713283531436045024182994033810742118035674 & (19) \end{matrix}

Substitute the values of

I (- e_{1})

I (e_{1})

I (- e_{5})

, and

I (e_{0})

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.