I'm trying to show that the complete elliptic integral of the second kind can be represented as: $$E(k)=\frac{\pi}2 \left(1-\sum_{n=0}^{\infty}\left(\frac{(2n-1)!!}{(2n)!!}\right)^2\frac{k^{2n}}{2n-1}\right)$$

I first used the binomial theorem to get: $$E(k)=\int_0^\frac{\pi}2\sum_{n=0}^\infty\binom{\frac{-1}2}nk^{2n}\sin^{2n}\theta d\theta$$

I then used Wallis' identity to evaluate the integral: $$\sum_{n=0}^{\infty}\binom{\frac{-1}2}nk^{2n}\left(\frac{\pi}2\right)\frac{(2n-1)!!}{(2n)!!}$$

I'm a mechanical engineer, so I'm not well-versed in real analysis. What are some tricks I could use to combine my binomial and my double factorials? Should I express them both as infinite products and combine them? All input is appreciated. Thanks!

EDIT: Let me clarify the question I'm asking. I'm really just trying to show that the complete elliptic integral of the second kind can be shown as the series above; I'm not trying to demonstrate any relationships between it and any of the other elliptic integrals/functions.

Now, I've come pretty close to determining the derivation of the series, but I've made some mistakes along the way. I will be showing each of the steps I've taken, starting with the integral form and ending with the series.

Beginning with: $$ E(k)=\int_0^{\frac{\pi}2} \sqrt{1-k^2\sin^2\theta}d\theta $$ One can convert the integrand into a binomial series: $$ E(k)=\int_0^{\frac{\pi}2} \sum_{n=0}^{\infty} \binom{\frac12}n(-1)^nk^{2n}\sin^{2n}\theta d\theta $$ Using a form of Wallis' identity, $sin^{2n}\theta d\theta$ becomes $\frac{\pi}2 \frac{(2n-1)!!}{(2n)!!}$, and therefore: $$ E(k)=\frac{\pi}2 \sum_{n=0}^{\infty} \binom{\frac12}n(-1)^nk^{2n} \frac{(2n-1)!!}{(2n)!!} $$ To eliminate the binomial coefficient, $\binom{\frac12}n$ may be (as Lucian pointed out) represented as: $$ \binom{\frac12}n=\frac{(\frac12)(\frac12-1)(\frac12-2)...(\frac12-n+1)}{n!} $$ By multiplying the numerator and denominator by $2^n(\frac12-n)$, one gets $$ \binom{\frac12}n=\frac{(2n-1)!!}{(2n)!!}\frac{(-1)^n}{\frac12-n} $$ By multiplying the numerator and denominator by -2, one gets $$ \binom{\frac12}n=\frac{(2n-1)!!}{(2n)!!}\frac{(-1)^{n+1}2}{2n-1} $$ Replacing the binomial with the above in the series, one gets: $$ E(k)=\frac{\pi}2 \sum_{n=0}^{\infty} (-1)^{2n+1}\left(\frac{(2n-1)!!}{(2n)!!}\right)^2\frac{2k^{2n}}{2n-1} $$ Assuming $n$ is an integer, the exponent of -1 is always an odd parity number, and so the summand is consistently negative. Therefore, the series may be expressed as: $$ E(k)=-\frac{\pi}2 \sum_{n=0}^{\infty} \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{2k^{2n}}{2n-1} $$ Evaluating the summand at $n=1$ yields -2, so: $$ E(k)=\frac{\pi}2\left(2-\sum_{n=0}^{\infty} \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{2k^{2n}}{2n-1}\right) $$ Needless to say, those pesky 2s are a bit of a problem. Can you see where I went wrong?

  • $\begingroup$ $\displaystyle{1/2\choose n}=-(-1)^n\frac{(2n-3)!!}{(2n)!!}$ $\endgroup$ – Lucian Jan 26 '15 at 4:03
  • $\begingroup$ Those should be the same: $$\binom{\frac12}n=\frac{(\frac12)(\frac12-1)(\frac12-2)...(frac12-n+1)}{n!}$$ $$=\frac{2^n(\frac12)(\frac12-1)(\frac12-2)...(\frac12-n)}{2^nn!(\frac12-n)}$$ $$=\frac{(1)(-1)(-3)(-5)...(2n-1)}{(2n)!!(\frac12-n)}$$ $$=\frac{(-1)^n(2n-1)!!}{(2n)!!(\frac12-n)}$$ Try using an arbitrary value for $n$ (I used 5) and then evaluate both expressions using the value. $\endgroup$ – Jeff Paul Timm Jan 26 '15 at 4:57
  • $\begingroup$ Multiplying the numerator and denominator by 2 and -1 should respectively yield: $$=\frac{(-1)^n2(2n-1)!!}{(2n)!!(1-2n)}$$ $$=\frac{(-1)^{n+1}2(2n-1)!!}{(2n)!!(2n-1)}$$ $\endgroup$ – Jeff Paul Timm Jan 26 '15 at 5:08
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    $\begingroup$ You have an extra factor of $2$. $\endgroup$ – Lucian Jan 26 '15 at 5:19
  • $\begingroup$ On my last expression? How so? If I begin with $(\frac12-n)$ in the denominator, and then multiply $n$ by $\frac22$, I have $(\frac{1-2n}{2})$. Because this fraction is in the denominator, the $2$ can be brought up to the numerator. So, instead of having $\frac1{\frac{1-2n}{2}}$, I have $\frac2{1-2n}$. $\endgroup$ – Jeff Paul Timm Jan 26 '15 at 5:31

Hint: $\displaystyle{m\choose n}=\dfrac{m(m-1)\cdots(m-n+1)}{1\cdot2\cdot3\cdots n}$ for all $m\in\mathbb C$ and $n\in\mathbb N.~$ Now let $m=-\dfrac12$ and simplify the expression $($by amplifying both numerator and denominator with $2^n$, and then regrouping the terms$)$.

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    $\begingroup$ What I got from this was $\binom{-1/2}n=\frac{(-1)^n(2n-1)!!}{(2n)!!}$. Does this seem right? $\endgroup$ – Jeff Paul Timm Jan 25 '15 at 22:03
  • $\begingroup$ Yes, that's correct. $\endgroup$ – Lucian Jan 25 '15 at 22:10
  • $\begingroup$ Awesome! I'm having a tough time seeing how I get from $E(k)=\frac{\pi}2\sum_{n=0}^\infty (-1)^n k^{2n}\left(\frac{(2n-1)!!}{(2n)!!}\right)^2$ to the form I have above. Also, the initial $n$ in the form above should be 1, not zero. $\endgroup$ – Jeff Paul Timm Jan 25 '15 at 22:14
  • $\begingroup$ The sum in your first expression equals $-\dfrac2\pi~E(k)$, and the one from your second expression $K(-k)$ $\endgroup$ – Lucian Jan 25 '15 at 22:29
  • $\begingroup$ That means the first expression is saying that $E(k)=\pi$. I don't understand. $\endgroup$ – Jeff Paul Timm Jan 25 '15 at 23:10

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