A fully symbolic procedure reduces a large class of elliptic
integrals to symmetric standard integrals, which can be computed
numerically by an efficient method based on the duplication theorem
and a polynomial in elementary symmetric functions.
The reduction is performed in two stages. In the first stage a
general integral is reduced to a set of basic integrals, and an
integral table containing some results of the first stage is presented.
In the second stage the basic integrals are expressed in terms of
symmetric standard integrals.
The integrals to be reduced have the form
|
I(m)= |
ó õ
|
x
y
|
|
h Õ
i=1
|
(ai + bi t)-1/2 |
n Õ
j=1
|
(aj + bj t)mj dt = |
ó õ
|
x
y
|
|
r(t)
s(t)
|
dt , x > y, |
| (1) |
where
|
s(t)= |
h Õ
i=1
|
(ai + bi t)1/2, r(t)= |
n Õ
j=1
|
(aj + bj t)mj, n ³ h, |
| (2) |
and mj is an integer. The a's and b's may be complex, but the
integral is assumed to be well defined, possibly as a Cauchy principal
value. The line segment with endpoints ai +bi x and ai +bi y,
1 £ i £ h is assumed to lie in the complex plane cut along the
negative real axis, and square roots have their principal value.
The integral is elliptic if h=3 (the cubic case) or h=4 (the
quartic case). The reduction may involve also algebraic functions
of the form
|
A(m) = |
r(x)
s(x)
|
- |
r(y)
s(y)
|
. |
| (3) |
We write
|
m=(m1,¼,mn)= |
n å
i=1
|
mi ei , |
| (4) |
where ei is an n-tuple with 1 in the ith position and 0's
elsewhere. We define also 0
= (0,¼,0). In the cubic case
the basic integrals are
|
I(0); I(-ej), 1 £ j £ n . |
| (5) |
In the quartic case they are
|
I(0); I(-ej), 1 £ j £ n ; I(ei), 1 £ i £ 4 . |
| (6) |
Basic integrals of type I(-ei), 1 £ i £ h , are not
linearly independent, nor are those of type I(ei), 1 £ i £ 4 .
The first stage of reduction and the resulting table of integrals are
simplified by defining
|
|
^
I
|
(m) = I(m)/B, |
^
A
|
(m)=A(m)/B, B= |
n Õ
j=1
|
bjmj, rij = |
ai
bi
|
- |
aj
bj
|
, |
| (7) |
where rij is assumed to be finite and nonzero so that different
linear factors in the integrand of I(m) are not proportional.
^
I(m) is first reduced to integrals in which m has at most
one nonzero component, and the reduction is then completed by using two
recurrence relations. See J. Symbolic Comp. 28(1999)739-753 for the
corresponding relations in unsimplified notation and for the expression
of basic integrals in terms of symmetric elliptic integrals.
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