TOPICS
Search

Elliptic Integral Singular Value--k_1

The first singular value k_1 of the elliptic integral of the first kind K(k), corresponding to

K^'(k_1)=K(k_1),
(1)

is given by

k_1=1/(sqrt(2))
(2)
k_1^'=1/(sqrt(2)).
(3)

The value K(k_1) is given by

K(1/(sqrt(2)))=int_0^1(dt)/(sqrt((1-t^2)(1-1/2t^2))),
(4)

which can be transformed to

K(1/(sqrt(2)))=sqrt(2)int_0^1(dt)/(sqrt(1-t^4)).
(5)

Let

u=t^4
(6)
du=4t^3dt
(7)
=4u^(3/4)dt
(8)
dt=1/4u^(-3/4)du,
(9)

then

K(1/(sqrt(2)))=(sqrt(2))/4int_0^1u^(-3/4)(1-u)^(-1/2)du
(10)
=(sqrt(2))/4B(1/4,1/2)
(11)
=(Gamma(1/4)Gamma(1/2))/(Gamma(3/4))(sqrt(2))/4,
(12)

where B(a,b) is the beta function and Gamma(z) is the gamma function. Now use

Gamma(1/2)=sqrt(pi)
(13)

and

1/(Gamma(1-x))=(sin(pix))/piGamma(x),
(14)

so

1/(Gamma(3/4))=1/(Gamma(1-1/4))=(sin(pi/4))/piGamma(1/4)=1/(pisqrt(2))Gamma(1/4).
(15)

Therefore,

K(1/(sqrt(2)))=(Gamma^2(1/4)sqrt(pi)sqrt(2))/(4pisqrt(2))=(Gamma^2(1/4))/(4sqrt(pi)).
(16)

Now consider

E(1/(sqrt(2)))=int_0^1sqrt((1-1/2t^2)/(1-t^2))dt.
(17)

Let

t^2=1-u^2
(18)
2tdt=-2udu
(19)
dt=-1/tudu
(20)
=u(1-u^2)^(-1/2)du,
(21)

so

E(1/(sqrt(2)))=int_0^1sqrt((1-1/2(1-u^2))/(1-(1-u^2)))u(1-u^2)^(-1/2)du
(22)
=int_0^1(sqrt(1/2(1+u^2)))/uu(1-u^2)^(-1/2)du
(23)
=1/(sqrt(2))int_0^1sqrt((1+u^2)/(1-u^2))du.
(24)

Now note that

(1/(sqrt(1-u^4))+(u^2)/(sqrt(1-u^4)))^2=(1+u^2)/(1-u^2),
(25)

so

E(1/(sqrt(2)))=1/(sqrt(2))int_0^1sqrt((1+u^2)/(1-u^2))du
(26)
=1/(sqrt(2))int_0^1(1/(sqrt(1-u^4))+(u^2)/(sqrt(1-u^4)))du
(27)
=1/2K(1/(sqrt(2)))+1/(sqrt(2))int_0^1(u^2du)/(sqrt(1-u^4)).
(28)

Now let

t=u^4
(29)
dt=4u^3du,
(30)

so

int_0^1(u^2du)/(sqrt(1-u^4))=1/4int_0^1t^(1/2)t^(-3/4)(1-t)^(-1/2)dt
(31)
=1/4int_0^1t^(-1/4)(1-t)^(-1/2)dt
(32)
=1/4B(3/4,1/2)=(Gamma(3/4)Gamma(1/2))/(4Gamma(5/4)).
(33)

But

[Gamma(5/4)]^(-1)=[1/4Gamma(1/4)]^(-1)
(34)
Gamma(3/4)=pisqrt(2)[Gamma(1/4)]^(-1)
(35)
Gamma(1/2)=sqrt(pi),
(36)

so

int_0^1(u^2du)/(sqrt(1-u^4))=1/4(pisqrt(2)·4sqrt(pi))/(Gamma^2(1/4))=(sqrt(2)pi^(3/2))/(Gamma^2(1/4))
(37)
E(1/(sqrt(2)))=1/2K+(pi^(3/2))/(Gamma^2(1/4))
(38)
=(Gamma^2(1/4))/(8sqrt(pi))+(pi^(3/2))/(Gamma^2(1/4))
(39)
=1/4sqrt(pi/2)[(Gamma(1/4))/(Gamma(3/4))+(Gamma(3/4))/(Gamma(5/4))].
(40)
(41)

Summarizing (◇) and (41) gives

K(1/(sqrt(2)))=(Gamma^2(1/4))/(4sqrt(pi))
(42)
K^'(1/(sqrt(2)))=(Gamma^2(1/4))/(4sqrt(pi))
(43)
E(1/(sqrt(2)))=(Gamma^2(1/4))/(8sqrt(pi))+(pi^(3/2))/(Gamma^2(1/4))
(44)
E^'(1/(sqrt(2)))=(Gamma^2(1/4))/(8sqrt(pi))+(pi^(3/2))/(Gamma^2(1/4)).
(45)

Explore with Wolfram|Alpha

Cite this as:

Weisstein, Eric W. "Elliptic Integral Singular Value--k_1." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/EllipticIntegralSingularValuek1.html

Subject classifications