Cryptography Reference
In-Depth Information
Consider now the integral
∞
dx
e
3
)
.
Substitute
x
=
℘
(
z
). The denominator becomes
℘
(
z
)
2
=
−℘
(
z
) (recall
that
℘
(
z
)
<
0) and the limits of integration are from
z
=
ω
2
/
2 to 0. Adjusting
the direction of integration and the sign yields
ω
2
/
2
4(
x
−
e
1
)(
x
−
e
2
)(
x
−
e
3
dz
=
ω
2
2
.
0
Therefore,
ω
2
=
∞
e
3
dx
(
x
e
3
)
.
−
e
1
)(
x
−
e
2
)(
x
−
The change of variables
e
3
−
(
e
3
−
e
2
)
t
+
e
3
+
(
e
3
−
e
2
)
e
1
)(
e
3
−
e
1
)(
e
3
−
x
=
t
+1
(plus a lot of algebraic manipulation) changes the integral to
1
2
dt
√
e
3
−
e
1
+
√
e
3
−
(1
ω
2
=
k
2
t
2
)
,
e
2
−
t
2
)(1
−
−
1
where
k
=
√
e
3
−
e
1
−
√
e
3
−
e
2
√
e
3
−
e
1
+
√
e
3
−
e
2
.
(9.19)
Since the integrand is an even function, we can take twice the integral over
the interval from 0 to 1 and obtain
1
4
√
e
3
− e
1
+
√
e
3
− e
2
dt
(1
− t
2
)(1
− k
2
t
2
)
.
ω
2
=
0
This integral is called an
elliptic integral
(more precisely, an elliptic integral
of the first kind). It is usually denoted by
K
(
k
)=
1
0
dt
(1
− t
2
)(1
− k
2
t
2
)
.
In the following, we'll see how to compute
K
(
k
) numerically very accurately
and quickly, but first let's find an expression for
ω
1
.
When
z
runs along the vertical line from
ω
2
/
2to
ω
2
/
2+
ω
1
/
2, the function
℘
(
z
) takes on real values (see Exercise 9.6) from
e
3
to
e
2
, and its derivative
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