While doing some research I got stuck trying to prove that the following function is decreasing
$$f(k):= k K(k) \sinh \left(\frac{\pi}{2} \frac{K(\sqrt{1-k^2})}{K(k)}\right)$$ for $k \in (0,1)$.
Here $K$ is the Complete elliptic integral of the first kind, defined by $$K(k):= \int_{0}^{1} \frac{dt}{\sqrt{1-t^2} \sqrt{1-k^2t^2}}.$$
This seems to be true, as the graph below suggests :
I really don't know much about elliptic integrals, so perhaps someone here can give some insight. Any relevant reference on elliptic integrals of the first kind is welcome.
Thank you, Malik
EDIT (2012-07-09) :
Using J.M.'s suggestion to rewrite the function $f(k)$ as $$f(k) = kK(k) \frac{1-q(k)}{2 \sqrt{q(k)}}$$ and using the derivative formulas $$K'(k) = \frac{E(k)}{k(1-k^2)} - \frac{K(k)}{k},$$ $$q'(k)=\frac{\pi^2}{2} \frac{q(k)} { K(k)^2 (1-k^2)k}$$ where $E(k)$ is the Complete elliptic integral of the second kind, I was able to calculate $f'(k)$ and reduce the problem to showing that the following function is negative for $k \in (0,1)$ :
$$g(k):= 4(1-q(k))K(k)E(k) - \pi^2 (1+q(k)).$$
Below is the graph of $g$ obtained with Maple :
EDIT (19-07-2012)
I asked the question on MathOverflow!
A few more terms for those investigating. From Maple. These coefficients are not listed in the On-line Encyclopedia of Integer Sequences.
$$ \frac{4}{\pi} \sqrt{m} \;K(4 \sqrt{m}) \sinh \biggl(\frac{\pi\; K(\sqrt{1 - 16 m})}{2\;K(4 \sqrt{m})}\biggr) \\ = 1 - m - 6 m^{2} - 54 m^{3} - 575 m^{4} - 6715 m^{5} - 83134 m^{6} - 1071482 m^{7} - \\ \quad{}\quad{} 14221974 m^{8} - 193050435 m^{9} - 2667157340 m^{10} - 37378279402 m^{11} - \\ \quad{}\quad{} 530024062361 m^{12} - 7590192561912 m^{13} - \\ \quad{}\quad{}109610113457650 m^{14} - 1594344146568120 m^{15} - \\ \quad{}\quad{}23336667998911128 m^{16} - 343468859344118109 m^{17} - \\ \quad{}\quad{}5079858166426507168 m^{18} - 75457168334744888190 m^{19} - \\ \quad{}\quad{}1125223725054635766392 m^{20} + \operatorname{O} \bigl(m^{21}\bigr) $$
added
Who knows if this is relevant? See A002849 $$ \frac{2}{\pi}K(4\sqrt{m}) = 1+4m+36m^2+400m^3+4900m^4+\dots =\sum_{n=0}^\infty \binom{2n}{n}^2m^n $$