- The elliptic integrals K and \(K ^ { \prime }\) defined for \(0 < k < 1\) by
\[ K (k) = \int_ {0} ^ {1} \frac {d x}{((1 - x ^ {2}) (1 - k ^ {2} x ^ {2})) ^ {1 / 2}} \quad \text {and} \quad K ^ {\prime} (k) = \int_ {1} ^ {1 / k} \frac {d x}{((x ^ {2} - 1) (1 - k ^ {2} x ^ {2})) ^ {1 / 2}} \]
satisfy various interesting identities. For instance:
- Show that if \(\tilde { k } ^ { 2 } = 1 - k ^ { 2 }\) and \(0 < \tilde { k } < 1\) , then
\[ K ^ {\prime} (k) = K (\tilde {k}). \]
[Hint: Change variables \(x = ( 1 - \tilde { k } ^ { 2 } y ^ { 2 } ) ^ { - 1 / 2 }\) in the integral defining \(K ^ { \prime } ( k ) . ]\)
- Prove that if \(\tilde { k } ^ { 2 } = 1 - k ^ { 2 }\) , and \(0 < \tilde { k } < 1\) , then
\[ K (k) = \frac {2}{1 + \tilde {k}} K \left(\frac {1 - \tilde {k}}{1 + \tilde {k}}\right). \]
[Hint: Change variables \(x = 2 t / ( 1 + \tilde { k } + ( 1 - \tilde { k } ) t ^ { 2 } ) . ]\)
- Show that for \(0 < k < 1\) one has
\[ K (k) = \frac {\pi}{2} F (1 / 2, 1 / 2, 1; k ^ {2}), \]
where \(F\) the hypergeometric series. [Hint: This follows from the integral representation for \(F\) given in Exercise 9, Chapter 6.]
Solution
Write \(k'^2=1-k^2\), with \(0<k'<1\).
For (a), in \(K'(k)\) make the substitution \[ x=(1-k'^2y^2)^{-1/2}. \] As \(x\) runs from \(1\) to \(1/k\), \(y\) runs from \(0\) to \(1\). A direct calculation gives \[ dx=\frac{k'^2y}{(1-k'^2y^2)^{3/2}}\,dy, \] \[ x^2-1=\frac{k'^2y^2}{1-k'^2y^2}, \qquad 1-k^2x^2=\frac{k'^2(1-y^2)}{1-k'^2y^2}. \] Hence \[ K'(k)=\int_0^1\frac{dy}{\sqrt{(1-y^2)(1-k'^2y^2)}}=K(k'). \]
For (b), set \[ x=\frac{2t}{1+k'+(1-k')t^2}. \] This maps \([0,1]\) onto \([0,1]\). Put \[ \kappa=\frac{1-k'}{1+k'}. \] Straight substitution and simplification yield \[ \frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} =\frac{2}{1+k'}\, \frac{dt}{\sqrt{(1-t^2)(1-\kappa^2t^2)}}. \] Therefore \[ K(k)=\frac2{1+k'}K\left(\frac{1-k'}{1+k'}\right). \]
For (c), Euler’s beta-integral representation of the hypergeometric function gives, for \(|u|<1\), \[ {}_2F_1\left(\frac12,\frac12;1;u\right) =\frac1\pi\int_0^1 t^{-1/2}(1-t)^{-1/2}(1-ut)^{-1/2}\,dt. \] Set \(u=k^2\) and then \(t=x^2\). Since \(dt=2x\,dx\), \[ {}_2F_1\left(\frac12,\frac12;1;k^2\right) =\frac2\pi\int_0^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} =\frac2\pi K(k). \] Thus \[ K(k)=\frac\pi2\,{}_2F_1\left(\frac12,\frac12;1;k^2\right). \]