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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

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  • $\begingroup$ The expansion of the integral you wrote has terms for every odd power of $m$ as well. $\endgroup$ – Did Dec 8 '12 at 20:56
  • $\begingroup$ Im sorry, can you explain again? something wrong on the expansion? $\endgroup$ – JHughes Dec 8 '12 at 21:58
  • $\begingroup$ Yes, something was definitely wrong with the expansion... But it seems you saw the problem since you made the necessary correction. Note that this makes the accepted answer, which addresses (incorrectly) the original version of your question, a little odd. $\endgroup$ – Did Dec 8 '12 at 23:42
  • $\begingroup$ yeah yeah, thx, i already correct it. $\endgroup$ – JHughes Dec 11 '12 at 1:24
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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 $0\leq m \leq 1$.

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  • $\begingroup$ The numerator $m\sin u$ should read $\frac12\sin^2u$. $\endgroup$ – Did Dec 8 '12 at 23:23

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