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$E(x)=\int_0^{\frac{\pi}{2}} \sqrt{1-x^2 \sin^2 t}\, dt$ Where $E(x)$ is complete elliptic integral of the second kind.

$u=\sin t$

$E(x)=\int_0^{1} \frac{\sqrt{1-x^2 u^2}}{\sqrt{1-u^2}}\, du$

$\frac{dE(x)}{dx}=-x\int_0^{1} \frac{u^2}{\sqrt{1-u^2}\sqrt{1-x^2 u^2}} \, du$

$\frac{d}{dx}(x\frac{dE(x)}{dx})=-2x\int_0^{1} \frac{u^2}{\sqrt{1-u^2}\sqrt{1-x^2 u^2}} \, du-x^2\int_0^{1} \frac{xu^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du$

$\frac{d}{dx}(x\frac{dE(x)}{dx})=\int_0^{1} \frac{-2xu^2(1-x^2 u^2)-x^3u^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du$

$(x^2-1)\frac{d}{dx}(x\frac{dE(x)}{dx})=\int_0^{1} \frac{(-2xu^2+x^3u^4)(x^2-1)}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du \tag1$

$xE(x)=\int_0^{1} \frac{x\sqrt{1-x^2 u^2}}{\sqrt{1-u^2}}\, du \tag 2$

According to Wikipedia, Equation 1 and 2 are equal but I could not prove it. Could you please help me to prove that?

$(x^2-1) \frac{d}{dx}(x \frac{dE(x)}{dx})=xE(x)$ http://en.wikipedia.org/wiki/Elliptic_integral

EDIT:

If $(x^2-1) \frac{d}{dx}(x \frac{dE(x)}{dx})=xE(x)$ is true, then

$(x^2-1) \frac{d}{dx}(x \frac{dE(x)}{dx})-xE(x)=0$

$\int_0^{1} \frac{(-2xu^2+x^3u^4)(x^2-1)}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du-\int_0^{1} \frac{x\sqrt{1-x^2 u^2}}{\sqrt{1-u^2}}\, du=0$ must be. And then

$\int_0^{1} \frac{(-2xu^2+x^3u^4)(x^2-1)-x (1-x^2 u^2)^2}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$

$-x\int_0^{1} \frac{1-2u^2+x^2u^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$

If Wikipedia differential equation is true ,

$\int_0^{1} \frac{1-2u^2+x^2u^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$

$\int_0^{1} \frac{1-u^2}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du-\int_0^{1} \frac{u^2(1-x^2u^2)}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$

$\int_0^{1} \frac{\sqrt{1-u^2}}{(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=\int_0^{1} \frac{u^2}{\sqrt{1-u^2}\sqrt{1-x^2 u^2}} \, du$ must be true too. Now I need to prove that last equation. Any idea how to proceed? Thanks a lot for advice.

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    Do not care, now I checked Wikipedia article.2012-09-18

2 Answers 2

2

Let's call on the help of E's older brother, $K(x)$. That is the complete elliptic integral of the first kind:

$K(x)=\int_0^{\frac{\pi}{2}} \frac{dt}{\sqrt{1-x^2 \sin^2 t}}$ Together, they form the following system:

$x\frac{dE}{dx}=E-K$

$x\frac{dK}{dx}=\frac{E}{1-x^2}-K$ which is easy to prove.

Placing them into the diff-equation we get

$(x^2-1)\frac{d}{dx}(x\frac{dE(x)}{dx})=(x^2-1)\left(\frac{dE}{dx}-\frac{dK}{dx}\right )$ $=\frac{x^2-1}{x}\left(E-K-\frac{E}{1-x^2}+K\right )$

$=\frac{x^2-1}{x}\frac{1-x^2-1}{1-x^2}E=xE $
Edit: Here is the proof of

$x\frac{dK}{dx}=\frac{E}{1-x^2}-K$

$\frac{dK}{dx}=\left [\int_0^{\frac{\pi}{2}} \frac{dt}{\sqrt{(1-x^2 \sin^2 t)^3}}-\int_0^{\frac{\pi}{2}} \frac{dt}{\sqrt{1-x^2 \sin^2 t}}\right ]\frac{1}{x}$ But

$\int_0^{\frac{\pi}{2}} \frac{dt}{\sqrt{(1-x^2 \sin^2 t)^3}}=\frac{1}{1-x^2}\int_0^{\frac{\pi}{2}} \sqrt{1-x^2 \sin^2 t}\, dt$ The last result follows from the obvious equality:

$\frac{1-x^2}{\sqrt{(1-x^2 \sin^2 t)^3}}=$

$=\sqrt{1-x^2 \sin^2 t}-x^2\frac{d}{dt}\left (\frac{\sin t \cos t}{\sqrt{1-x^2 \sin^2 t}}\right )$

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    @MartinGales : Really the last equality is great and beautiful. It is certainly the key for elliptic integrals. I wonder how you got that result. I wish to know which sense and methods were used to get such eqaution. Thanks a lot for your answer.2012-09-22
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The derivative of the complete elliptic integral of the second kind$E(x)$ is given by,

$ E'(x) = {\frac {{\it E} \left( x \right) }{x}}-{\frac {{\it K} \left( x \right) }{x}} \,,$ where $K(x)$ is the complete elliptic integral of the first kind. Multiplying the above equation by $x$ gives $ x E'(x) = {\it E}(x) - {\it K(x) } $

$\Rightarrow (x E'(x))' = {\frac {{\it E} \left( x \right) }{x}}-{\frac {{\it E} \left( x \right) }{ \left( 1-{x}^{2} \right) x}} =\frac{x E(x)}{x^2-1} \,.$

Multiplying both sides of the last equation by $ (x^2-1) $ yields the desired result.

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    :Thank you for answer. $x\frac{dE}{dx}=E-K$ which is easy to prove. But As you see in previous answer, I wondered how we can get $x\frac{dK}{dx}=\frac{E}{1-x^2}-K$. I tried to proof it but I could not see a way how . Could you please edit how to proof $x\frac{dK}{dx}=\frac{E}{1-x^2}-K$. If we have $x\frac{dK}{dx}=\frac{E}{1-x^2}-K$, I can proof the diff equation easily.2012-09-20