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I ran into an integral a little while ago that looks like an elliptic integral of the first kind, however I am having trouble seeing how it can be put into the standard form. I've tried messing around with some trig identities but didn't get anywhere. Perhaps there is some other definition that I'm missing. Here it is. If it is solvable by contour integration I would be open to this as well.

$$ \int_0^{\phi_0} \frac{d\phi}{\sqrt{1-a\sin{\phi}}}$$

EDIT

If it makes it any easier, let $a=2$ and $\phi_0=\frac\pi6$. By standard form I mean: $$ \int_0^{\phi_0} \frac{d\phi}{\sqrt{1-k^2\sin^2{\phi}}}=F(\phi_o,k) $$

A Mathematica calculation reveals (using the constants I mentioned):

$$\int_0^{\pi/6} \frac{d\phi}{\sqrt{1-2\sin{\phi}}}=i\left(2F(\pi/4,4)-K(1/4)\right)$$

Where $F$ is an incomplete elliptic integral of the first kind and $K$ is a complete elliptic integral of first kind. Note that Mathematica uses a different convention where $k$ is replaced by the parameter $m=k^2$. This answer leads me to believe that the integral should be split up somehow.

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  • $\begingroup$ What do you mean by standard form? $\endgroup$ – Mhenni Benghorbal Mar 1 '16 at 6:20
  • $\begingroup$ @MhenniBenghorbal see my edit $\endgroup$ – ClassicStyle Mar 1 '16 at 6:34
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    $\begingroup$ Hint: $\sin\phi = -\cos\left(\phi + \frac{\pi}{2}\right) = 2\sin^2\left(\frac{\phi}{2}+\frac{\pi}{4}\right) - 1$. $\endgroup$ – achille hui Mar 1 '16 at 6:52
  • $\begingroup$ Are you sure you put the limits of integration from $0$ to $\pi/6$ in mathematica? $\endgroup$ – Mhenni Benghorbal Mar 1 '16 at 6:56
  • $\begingroup$ Boom got it! Thanks @achillehui if you want to type out a little answer I will accept it $\endgroup$ – ClassicStyle Mar 1 '16 at 7:06
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Let $\displaystyle\;\mathcal{I} = \int_0^{\phi_1} \frac{d\phi}{\sqrt{1-a\sin\phi}}\;$ be the integral at hand. Since $$\sin\phi = -\cos\left(\phi + \frac{\pi}{2}\right) = 2 \sin^2\left(\frac{\phi}{2} + \frac{\pi}{4}\right) - 1$$ If one change variable to $\theta = \frac{\phi}{2} + \frac{\pi}{4}$, we have $$\begin{align} \mathcal{I} &= \int_{\theta_0}^{\theta_1} \frac{2d\theta}{\sqrt{(1+a)-2a\sin^2\theta}} = \frac{2}{\sqrt{1+a}}\int_{\theta_0}^{\theta_1} \frac{d\theta}{\sqrt{1-k^2\sin^2\theta}}\\ &= \frac{2}{\sqrt{1+a}}\left[ F(\theta_1,k) - F\left(\theta_0, k\right) \right] \end{align} $$ where $\displaystyle\;\theta_0 = \frac{\pi}{4}\;$, $\displaystyle\;\theta_1 = \frac{\phi_1}{2} + \frac{\pi}{4}\;$ and $\displaystyle\;k = \sqrt{\frac{2a}{1+a}}$.

If $0 < a < 1$, this is what we need.

If $a > 1$, the modulus $k$ for the elliptic integral is bigger than $1$. This is usually undesirable. Sometimes this will lead to superficial complex numbers in the expression of a real integral.

We can resolve this by another change of variable $\sin\psi = k\sin\theta$.
Let $\psi_0 = \sin^{-1}(k\sin\theta_0)$ and $\psi_1 = \sin^{-1}(k\sin\theta_1)$, we have

$$\frac{\sqrt{1+a}}{2}\mathcal{I} = \int_{\theta_0}^{\theta_1} \frac{d\sin\theta}{\sqrt{(1-\sin^2\theta)(1-k^2\sin^2\theta)}} = \int_{\psi_0}^{\psi_1} \frac{k^{-1}d\sin\psi}{\sqrt{(1-k^{-2}\sin^2\psi)(1-\sin^2\psi)}}$$ This leads to an alternate expression for the integral at hand: $$\mathcal{I} = \sqrt{\frac{2}{a}}\int_{\psi_0}^{\psi_1} \frac{d\psi}{\sqrt{1-k^{-2}\sin^2\psi}} = \sqrt{\frac{2}{a}}\left[F(\psi_1;k^{-1}) - F(\psi_0;k^{-1})\right] $$ For the test case $\phi_1 = \frac{\pi}{6}$, this give us $$\mathcal{I} = K\left(\sqrt{\frac34}\right) - F\left(\sin^{-1}\sqrt{\frac23},\sqrt{\frac34}\right)$$ Using the command EllipticK[3/4] - EllipticF[ArcSin[Sqrt[2/3]],3/4] on WA, one find $$\mathcal{I} \approx 1.078257823749821617719337499400161014432055108246412680182...$$

This matches numerically the result i*(2*EllipticF[Pi/4,4]-EllipticK[1/4]) returned by symbolic integrating the integral on WA.

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  • $\begingroup$ Great answer. Appreciate you! $\endgroup$ – ClassicStyle Mar 1 '16 at 13:34
  • $\begingroup$ This is effectively the reciprocal modulus transformation for $F$. $\endgroup$ – J. M. is a poor mathematician Mar 7 at 0:50

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