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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?

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


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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Try putting $x = 0$ into each, I think you get $0$ for one equation and something strictly negative for the other one... – Ben Millwood Sep 18 '12 at 12:19
Mathematica says $$(x^2-1)\frac{d}{dx}x\frac{dE(x)}{dx}=\frac{1}{4}(1+x)E(x)$$. – Jon Sep 18 '12 at 13:28
Do not care, now I checked Wikipedia article. – Jon Sep 18 '12 at 13:50
up vote 2 down vote accepted

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:


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


$$\frac{dK}{dx}=\left [\int_0^{\frac{\pi}{2}} \frac{dt}{\sqrt{(1-x^2 \sin^2 t)^3}}-\int_0^{\frac{\pi}{2}} \frac{dt}{\sqrt{1-x^2 \sin^2 t}}\right ]\frac{1}{x}$$ 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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How do you show the second equation in the system? – joriki Sep 19 '12 at 8:27
@Martin Gales: I could not see an easy proof of second relation either. First one is very easy. If you know easy proof of second one, Could you please edit your answer and show us . – Mathlover Sep 19 '12 at 8:46
@Mathlover An edit is made – Martin Gales Sep 21 '12 at 8:26
@joriki An edit is made – Martin Gales Sep 21 '12 at 8:27
In what sense is that last equation obvious? – joriki Sep 21 '12 at 10:07

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. – Mathlover Sep 20 '12 at 13:16

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