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I want to show that $\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt {1 - k^2\sin^2{x}}}\;{dx} = \frac{\pi}{2}\sum_{n \ge 0}k^{2n}\left({\frac{{1 \cdot 3 \cdots \left( {2n - 1} \right)}} {{2 \cdot 4 \cdots \cdot 2n}}} \right)$, where $ -1 < k < 1$.

Here is what I did:

$$\displaystyle \begin{aligned}\int_0^{\frac{\pi }{2}} \frac{1}{{\sqrt {1 - k^2\sin^2{x}}}}\;{dx} & = \int_{0}^{\pi/2}\sum_{n \ge 0} \frac{k^{2n}}{2^{2n}}\binom{2n}{n}\sin^{2n}{x}\;{dx} \\& = \sum_{n \ge 0}\int_{0}^{\pi/2} \frac{k^{2n}}{2^{2n}}\binom{2n}{n}\sin^{2n}{x}\;{dx} \\& = \frac{\pi}{2} \sum_{n \ge 0} ~ k^{2n} \bigg(\frac{1}{2^{2n}}\binom{2n}{n}\prod_{1 \le r \le n}\frac{2r-1}{2r} \bigg) \\& = \frac{\pi}{2} \sum_{n \ge 0} ~ k^{2n} \bigg(\prod_{1 \le r \le n}\frac{2r-1}{2r} \cdot \prod_{1 \le r \le n}\frac{2r-1}{2r} \bigg) \\& = \frac{\pi}{2}\sum_{n \ge 0}k^{2n}\bigg(\prod_{1 \le r \le n}\frac{2r-1}{2r}\bigg)^2.\end{aligned}$$

However, there no power on the coefficients in the given series, so they obviously don't match, and I couldn't whatsoever discern a mistake in my calculations. Thanks in advance.

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Are you sure that the result you're trying to get is correct? The is an elliptic integral of the 1st kind, but the sum just yields a simple square root. (Type the following into Wolfram|Alpha: Integrate[1/Sqrt[1 - k^2*Sin[x]^2], {x, 0, Pi/2}] and Sum[k^(2 n) (2 n - 1)!!/ (2 n)!!, {n, 0, Infinity}]) – Simon Jun 10 '11 at 4:08
Oh... you're missing a squared in your first sum – Simon Jun 10 '11 at 4:14
@Simon, yes, you're right of course. It turns out it's the given sum that's missing the square. Thanks for the wolfram link and codes. – Lyrebird Jun 10 '11 at 4:25
Sorry, @Asaf, where did you get the factor of $1/3$? What does it mean? – Luboš Motl Jun 10 '11 at 5:27
up vote 2 down vote accepted

In the very first step, $(2n$ choose $n)$ should be $(-1/2$ choose $n)$ because the exponent is $1/2$ and you're apparently using a standard Taylor expansion for $(1+X)^M$ for $M=-1/2$ and $X=-k^2\sin^2 x$. You should also add $(-1)^n$.

The combinatorial factor $(-1/2$ choose $n)$ instantly produces the product of odd numbers over the product of even numbers. It is equal to $(-1/2)(-3/2)\dots (-n+1/2) / n!$. Absorbing $(-1)^n$, you get $(1/2)(3/2)\dots (n-1/2)/n!$.

Oh, your description is actually equivalent.

Indeed, you will get another copy of $(1/2)(3/2)\dots (n-1/2)/n!$, together with the $\pi/2$ factor, from the integral $\int_0^{\pi/2} dx\,\sin^2 x$, so your result is correct, and the original formula is only missing the second power of the big ratio and needs to be corrected.

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@Luboš Motl I used the transformation: $$ \binom{-\frac{1}{2}}{n}= \frac{(-1)^{n}}{2^{2n}}\binom{2n}{n} $$. – Lyrebird Jun 10 '11 at 3:47
Oh, I eventually noticed that it's actually the same thing. In that case, I am sure that your result is correct. In the original formula, the whole big ratio in the parentheses should be squared, otherwise it's correct. – Luboš Motl Jun 10 '11 at 3:53
Oh, thank you! That's a relief! – Lyrebird Jun 10 '11 at 3:55
It's just a small digit "2" that someone missed, but you didn't. Maybe they thought it was a mark for a footnote. ;-) – Luboš Motl Jun 10 '11 at 3:56
Be careful, if it is an exam problem and they will mechanically compare the result with a wrong official template, they may declare your correct answer incorrect and you will have to defend yourself - for which, I believe, you have all the weapons. – Luboš Motl Jun 10 '11 at 5:26

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