1
$\begingroup$

How can you show that the complete elliptic integral of first kind $ \displaystyle K(m)=\int_0^\frac{\pi}{2}\frac{\mathrm du}{\sqrt{1-m^2\sin^2 u}}$ that is the same as a series $K(m)=\frac{\pi}{2} \left(1+\left(\frac{1}{2}\right)^{2}m^2 +\left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}m^4 +...+ \left(\frac{(2n-1)!!}{2n!!} \right )^2m^{2n} + ... \right)$

increases in m?

Thanks

  • 0
    yeah yeah, thx, i already correct it.2012-12-11

1 Answers 1

1

You can show that the derivative with respect to $m$ is always positive:

Note that $K'(m)=\int_0^\frac{\pi}{2}\frac{m \sin^2 u\, du}{(1-m^2\sin^2 u)^{3/2}} \geq 0$ as the integrand is positive for all $0\leq m \leq 1$.

  • 0
    The numerator $m\sin u$ should read $\frac12\sin^2u$.2012-12-08