**Lemma 3.2.** *We have*

*In particular,*

*Proof.* It is not hard to calculate the power series expansion

(The "!!" means double factorial.) Calculating the AM and GM means gives us

\[M(1+k, 1-k) = M(1, \sqrt{1-k^2}).\]Hence, applying proposition 2.1 for \(b/a = \sqrt{1-k^2}\), we get

\[a F_S(a, b) = \frac{\pi}{2 M(1, b/a)} \\ F_S(a, b) = \frac{\pi}{2 M(a, b)},\]with the last equality following from the identity \(M(ca, cb) = cM(a, b)\). \(\blacksquare\)