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.
To simplify the equations we will define integral 1
I(m)=
⌠ ⌡
c
b
________________ √(d−t)(c−t)(t−b)(t−a)
dt
t
.
(4)
Using a similar notation we will define:
I(e0)=
⌠ ⌡
c
b
dt
________________ √(d−t)(c−t)(t−b)(t−a)
,
(5)
I(3 e1)=
⌠ ⌡
c
b
(t−a)5/2dt
____________ √(d−t)(c−t)(t−b)
,
(6)
I(2 e1)=
⌠ ⌡
c
b
(t−a)3/2dt
____________ √(d−t)(c−t)(t−b)
,
(7)
I(e1)=
⌠ ⌡
c
b
___ √t−a
dt
____________ √(d−t)(c−t)(t−b)
,
(8)
I(−e1)=
⌠ ⌡
c
b
dt
____________ √(d−t)(c−t)(t−b)
(t−a)3/2
,
(9)
I(−e5)=
⌠ ⌡
c
b
dt
________________ √(d−t)(c−t)(t−b)(t−a)
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)+dcbaI(−e5)−dcbI(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)=
34
(d+c+b−3 a) I(2 e1)−
12
(d (c+b−2 a)+c (b−2 a)
−
14
a (2 b−3 a)) I(e1)+
14
(d−a) (c−a) (b−a) I(e0)
(13)
Substitute for I(2 e1) in equation 13 using equation 12.
I(3 e1)=
−
38
(d+c+b−3 a) (d−a) (c−a) (b−a) I(−e1)+
18
(3 d2
+2 d (c+b−5 a)+3 c2+2 c (b−5 a)+3 b2−10 ba+15 a2) I(e1)
+
14
(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.
