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I would like to evaluate the following: $$\frac{\partial }{\partial \beta }\int _0^{\cos ^{-1}(\beta )}\text{dx} \sqrt{\beta +\cos (x)}$$

given that $0\leq\beta\leq1$

basically I'd like to find the area under this curve:

curve

and then see the rate of change with respect to the parameter $\beta$

I understand this will probably be some kind of elliptic integral, but it must have a simple analytic form i would imagine.

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  • $\begingroup$ Have you tried the Leibniz Integral Rule? mathworld.wolfram.com/LeibnizIntegralRule.html EDIT: unfortunately, the first integral when you apply the rule still gives you an elliptic integral. I don't think it's possible to express this in terms of elementary functions. $\endgroup$ – Deepak Jun 5 '14 at 1:32
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You're definitely correct about this being an elliptic integral. However, while the results given by user152166 are certainly valid, Mathematica doesn't do a great job of simplifying such elliptic integrals in general---in fact, the derivative turns out to just be the complete elliptic integral of the first kind!

To begin, I think there are are two misstatements in the question as stated. For one, the problem as stated is in fact sensible for $\beta\in[-1,1]$ and I will assume that range. Secondly, based upon your plot the upper limit of integration should be $\cos^{-1}(-\beta)$ so that the integrand vanishes at this endpoint. With this in mind, the boundary terms in the Leibniz Integral Rule cited in the comments by Deepok both vanish and so

$$\frac{\partial}{\partial \beta}\int_0^{\cos^{-1}{\beta}}dx\sqrt{\beta+\cos x} = \int_0^{\cos^{-1}{\beta}}dx\frac{\partial}{\partial \beta}\!\left(\sqrt{\beta+\cos x}\right) =\frac{1}{2} \int_0^{\cos^{-1}{\beta}}\!\frac{dx}{\sqrt{\beta+\cos x}}.$$

To simplify this integral, note that the double-angle identity allows the integrand to be written as $$\beta+\cos x = \beta +1 -2\sin^2{\frac{x}{2}} = 2u-2\sin^2{\frac{x}{2}}$$ where I have introduced $u=(1+\beta)/2\in[0,1]$. The substitution $\sin{(x/2)} = u^{1/2} \sin\theta$ then produces the result $$\frac{1}{\sqrt{2}}\int_0^{\pi/2}\frac{d\phi}{\sqrt{1-u \sin^2\phi}}.$$

But this integral is the definition of the complete elliptic integral of the first kind $K(u)$, and so the derivative simplifies to $\dfrac{1}{\sqrt{2}}K\left(\dfrac{1+\beta}{2}\right)$. Finally, this same substitution (along with minor algebra) gives the original integral as $$\sqrt{8}\left[E\left(\dfrac{1+\beta}{2}\right)-\frac{1-\beta}{2}K\left(\dfrac{1+\beta}{2}\right)\right]$$ where $E(u) = \int_{0}^{\pi/2}d\phi\sqrt{1-u \sin^2\phi}$ is the complete elliptic integral of the second kind.

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  • $\begingroup$ Great!! But when you move the d/db past the integral sign into the integrand, should you take some derivative on the integral bounds which also depend on b ie arccos(b)? Maybe this is a silly question $\endgroup$ – AimForClarity Jul 12 '14 at 15:31
  • $\begingroup$ @AimForClarity: Those are the boundary terms to which I was referring. The lower one vanishes because it doesn't depend on $\beta$. The upper one vanishes because the integrand vanishes there (hence the importance of the correct upper limit). I can edit the answer a bit if needed. $\endgroup$ – Semiclassical Jul 12 '14 at 17:33
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Mathematica 5.0 gives

$$I(\beta)=\int _0^{\cos ^{-1}(\beta )}\text{dx} \sqrt{\beta +\cos (x)}=2(1+\beta)^{1/2}E\left(\frac{\cos^{-1}\beta}{2},\frac{2}{1+\beta}\right)$$

where $E(x,k)$ is an elliptical integral of second kind.

The derivative you are seeking for is given by:

$$\frac{\partial I(\beta)}{\partial \beta }=-\frac{(2\beta(1-\beta^2))^{1/2}}{1-\beta^2}+(1+\beta)^{1/2}F\left(\frac{\cos^{-1}\beta}{2},\frac{2}{1+\beta}\right)$$

Where $F(x,k)$ is an elliptical integral of first kind.

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