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I have the following integration to solve.

$$f(k) = \int_0^{\pi/2} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}d\theta,\quad0<k<1$$

assuming $\sin\theta = t$ which results $d\theta = \frac{dt}{\sqrt{1-t^2}}$ and when $\theta = 0, t=0$ and $\theta=\frac{\pi}{2},t=1$ so above equation can be rewritten as,

$$f(k) = \int_0^1{t^2\frac{\sqrt{1-k^2t^2}}{\sqrt{1-t^2}}}dt$$ I'm stuck in solving this further. Can somebody help me with some clues/solution to solve this further.

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  • $\begingroup$ what is $k$? are there any conditions for $k$? $\endgroup$ – Jane Feb 7 '16 at 19:21
  • $\begingroup$ $f$ does not depend on $\theta$, so it should be $f(k)$ $\endgroup$ – Yuriy S Feb 7 '16 at 19:22
  • $\begingroup$ @Jane Thanks for your comment. condition for $k$ is $0<k<1$. $\endgroup$ – user3563929 Feb 7 '16 at 19:26
  • $\begingroup$ @YuriyS. Corrected. Thanks $\endgroup$ – user3563929 Feb 7 '16 at 19:28
  • $\begingroup$ You can see the final solution in terms of elliptic integrals here $\endgroup$ – Yuriy S Feb 7 '16 at 19:43
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I will use $I(k)$ for the integral instead of $f(k)$.

$$ I(k)=\int^1_0 t^2 \frac{\sqrt{1-k^2 t^2}}{\sqrt{1-t^2}}dt $$

First, let's find some particular value, we will need it later.

$$I(1)=\frac{1}{3} $$

Now the definition for the elliptic integral of the second kind:

$$ E(k)=\int^1_0 \frac{\sqrt{1-k^2 t^2}}{\sqrt{1-t^2}}dt $$

It's easy to show that:

$$ I(k)=E(k)-\int^1_0 \sqrt{1-t^2} \sqrt{1-k^2 t^2} dt $$

Taking $k$ derivative:

$$ \frac{dI}{dk}=\frac{dE}{dk}+k \int^1_0 t^2 \frac{\sqrt{1- t^2}}{\sqrt{1-k^2t^2}}dt $$

Now let's use integration by parts for $I(k)$:

$$ I(k)=-k^2 \int^1_0 t^2 \frac{\sqrt{1- t^2}}{\sqrt{1-k^2t^2}}dt+\int^1_0 \sqrt{1-t^2} \sqrt{1-k^2 t^2} dt $$

Finally we use all three equations to show:

$$ I(k)=-I(k)+E(k)-k\frac{dI}{dk}+k\frac{dE}{dk} $$

We get a linear ODE for $I(k)$:

$$ \frac{dI}{dk}=-\frac{2}{k} I(k)+\frac{dE}{dk}+\frac{E(k)}{k} $$

Using the usual method for such equations (reference here) we get the general solution:

$$ I(k)=\frac{C_1}{k^2}+\frac{1}{k^2} \int k^2 \left(\frac{dE}{dk}+\frac{E(k)}{k} \right) dk $$

To calculate the integral we use the known formula (can be seen here):

$$ \int k E(k) dk=\frac{1}{3} \left[(1+k^2)E(k)-(1-k^2)K(k) \right] $$

Integrating by parts:

$$ \int k^2 \left(\frac{dE}{dk}+\frac{E(k)}{k} \right) dk=k^2 E(k)-\int k E(k) dk $$

Finally, the general solution:

$$ I(k)=\frac{C_1}{k^2}+\frac{-(1-2k^2)E(k)+(1-k^2)K(k)}{3k^2} $$

Now we use the value $I(1)$ we calculated earlier and the known values $E(1)=1$ and $\lim_{k \rightarrow 1} (1-k^2)K(k)=0$ (see here) to obtain the final solution:

$$I(k)=\frac{(1-k^2)K(k)-(1-2k^2)E(k)}{3k^2}$$

We can also check the result. From the original integral we can see that:

$$I(0)=\frac{\pi}{4}$$

From the solution (and using series expansions for $E$ and $K$) we get:

$$I(0)=\frac{\pi}{6k^2}(1-k^2+k^2/4-1+2k^2+k^2/4)=\frac{\pi}{4}$$

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Possible hints to perfume some kind of calculations.

Leaving apart the extrema of the integra, for the moment.

$$\int\sin^2\theta \sqrt{1 - k^2 \sin^2\theta}\ \text{d}\theta$$

Using the substitution

$$k\sin\theta = \cos\phi ~~~~~~~ \sin\theta = \frac{\cos\phi}{k} ~~~ \to ~~~ \sin^2\theta = \frac{\cos^2\phi}{k^2}$$

$$\phi = \arccos(k\sin\theta)$$

$$\text{d}\phi = \frac{- k\cos\theta}{\sqrt{1 - k^2\sin^2\theta}}\ \text{d}\theta = -\frac{k\sqrt{1 - \sin^2\theta}}{\sqrt{1 - \cos^2\phi}}\ \text{d}\theta = -\frac{k\sqrt{1 - \frac{\cos^2\phi}{k^2}}}{\sin^2\phi}\ \ \text{d}\theta$$

Thence $$\text{d}\theta = -\frac{\sin^2\phi}{\sqrt{k^2 - \cos^2\phi}}\ \text{d}\phi$$ and the integral becomes

$$- \int \frac{\sin^2\phi}{\sqrt{k^2 - \cos^2\phi}}\left(\frac{\cos^2\phi}{k^2}\right)\sin^2\phi \text{d}\phi = -\frac{1}{k^2}\int \frac{\sin^4\phi\cos^2\phi}{\sqrt{k^2 - \sin^2\phi}}\ \text{d}\phi$$

Now we can use the trigonometric reduction formula for the numerator of the integrand:

$$\sin^4\phi\cos^2\phi = \frac{1}{32}(2 - \cos(2\phi) - 2\cos(4\phi) + \cos(6\phi))$$

to get

$$ -\frac{1}{32k^2}\int\ \frac{2 - \cos(2\phi) - 2\cos(4\phi) + \cos(6\phi)}{\sqrt{k^2 - \sin^2\phi}}\ \text{d}\phi $$

Which might be splitter into four parts, and then.. who knows!

The solution, however, lies into Jacobi Elliptic Functions of the First an Second Kind.

More on Jacobi Elliptic Integrals

https://en.wikipedia.org/wiki/Elliptic_integral

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    $\begingroup$ You made a mistake when calculating $d \phi$, there should be $-k \cos \theta$ in the numerator, which will complicate the integral $\endgroup$ – Yuriy S Feb 7 '16 at 20:25
  • $\begingroup$ @YuriyS Thant's it!!! Thank you so much, I'm going to delete/edit the answer! I don't think I will be helpful but who knows.. Upvote for you $\endgroup$ – Sigma Algebra Feb 7 '16 at 20:26
  • $\begingroup$ Note, that I made a mistake myself, I edited the comment. Check it again to be sure $\endgroup$ – Yuriy S Feb 7 '16 at 20:27
  • $\begingroup$ @YuriyS Haha funny! Don't worry, I found my error the same, thanks to your hint! $\endgroup$ – Sigma Algebra Feb 7 '16 at 20:28
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Following on from @Yuriy-s, I have noticed that the integral:

$$\int k E(k) dk=\frac{1}{3} \left[(1+k^2)E(k)-(1-k^2)K(k) \right]$$

Is incorrect according to wolfram alpha. @Yuriy-s has used this to evaluate the expression, but note that wolfram alpha read this as:

$$\int k E(k^2) dk$$

Making a correction to this gives me this:

$$\int k E(k) dk=\frac{2}{45} \left[3k^2 + k-4)K(k)+(9k^2+k+4)E(k) \right] + C_2$$

where $C_2$ is a constant.

The following @Yuriy-s and subbing in various things:

$$I(k)=\frac{C_1}{k^2}+\frac{1}{k^2} \int k^2 \left(\frac{dE}{dk}+\frac{E(k)}{k} \right) dk$$

$$I(k)=\frac{C_1}{k^2}+E(k) - \frac{2}{45k^2} [(3k^2 +k -4) K(k) +(9k^2 +k +4)E(k)] + C_2$$

But there is a problem.

putting in the boundary conditions of: $$I(0)=\frac{\pi}{4}$$

$$E(0)=\frac{\pi}{2}$$ $$K(0)=\frac{\pi}{2}$$

Gives us: $$\frac{\pi}{4} = \frac{C_1}{0}+\frac{\pi}{2} - \frac{2}{0} [-4\frac{\pi}{2} +4\frac{\pi}{2}] +C_2 $$

$$C_2=\frac{-\pi}{4}$$ $$C_1=0$$

for a final result of: $$I(k)=E(k) - \frac{2}{45k^2} [(3k^2 +k -4) K(k) +(9k^2 +k +4)E(k)] - \frac{\pi}{4}$$

Unfortunately this is not the same as wolframs' final answer though:

$$I(k)=\frac{ \sqrt{1-k^2} [K(\frac{k^2}{k^2-1}) + (2k^2-1)E(\frac{k^2}{k^2-1}) ]}{3k^2}$$

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  • $\begingroup$ There are two typical conventions for elliptic integrals: Taking them to be functions of $k$, or as functions of $m=k^2$. Mathematica, and therefore Wolfram Alpha, uses the latter; the answer by Yuriy-s, by contrast, uses the former convention. That'd be my guess for why there's a seeming discrepancy. $\endgroup$ – Semiclassical Sep 1 '16 at 16:02
  • $\begingroup$ @Semiclassical-19 That seems plausible. But then why does wolfram give different results depending on whether you input the k varaible as any other symbol? wolframalpha.com/input/?i=integrate+a+EllipticE%5Ba%5D wolframalpha.com/input/?i=integrate+k+EllipticE%5Bk%5D $\endgroup$ – user209848 Sep 2 '16 at 11:47

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