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need help with calculating this:

$$\int_{0}^{2\pi}\frac{1-x\cos\phi}{(1+x^2-2x\cos\phi)^{\frac{3}{2}}}d\phi$$

Thanks in advance!

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3  
Hi DrVs, welcome to Math.SE. It would be helpful to know some context about how this integral arose, and what you have tried so far. Could you edit your question to include this? –  Antonio Vargas May 10 '12 at 0:20
    
As a function $J(x)$ we have $J(0) = 2\pi$, $J$ has poles at $\pm 1$, and $J(x) \to 0^-$ as $|x|\to \infty$. Putting the expression in mathematica gives a closed form in terms of known elliptic integrals, so I suspect there is not closed form in terms of elementary functions. –  nullUser May 10 '12 at 2:45
    
The closed form is given by $$J(x)=\frac{2 \left(-(-1+x) \text{EllipticE}\left[-\frac{4 x}{(-1+x)^2}\right]+(1+x) \text{EllipticK}\left[-\frac{4 x}{(-1+x)^2}\right]\right)}{\sqrt{(-1+x)^2} (1+x)}.$$ –  nullUser May 10 '12 at 2:47
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For $|x|<1$ this can also be written as $\dfrac{4 \it{EllipticE}(x)}{1-x^2}$. –  Robert Israel May 10 '12 at 3:02
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@Antonio: This is the force exerted by a charged circle of radius $x$ on a charge at distance $1$ from the centre; see this comment. –  joriki May 10 '12 at 18:41

1 Answer 1

up vote 4 down vote accepted

Let's assume $-1<x<1$, for simplicity.

Changing integration variable $\phi \to 2 \phi$, and using $\cos(2\phi) = 1- 2 \sin^2(\phi)$ we get: $$\begin{eqnarray} \int_0^{2\pi} \frac{1-x \cos(\phi)}{\left(1+x^2-2x \cos(\phi)\right)^{3/2}}\mathrm{d} \phi &=& 2 \int_0^{\pi} \frac{1-x \cos(2\phi)}{\left(1+x^2-2x \cos(2\phi)\right)^{3/2}}\mathrm{d} \phi \\ &=& 4 \int_0^{\pi/2} \frac{1-x + 2 x \sin^2(\phi)}{\left((1-x)^2 + 4 x \sin^2(\phi)\right)^{3/2}} \mathrm{d} \phi \\ &=& \frac{4}{(1-x)^2} \int_0^{\pi/2} \frac{1 + \frac{2 x}{1-x} \sin^2(\phi)}{\left(1 + \frac{4 x}{(1-x)^2} \sin^2(\phi)\right)^{3/2}} \mathrm{d} \phi \end{eqnarray} $$ Letting $m=-\frac{4 x}{(1-x)^2}$ and $a= \frac{2x}{1-x}$: $$\begin{eqnarray} \int_0^{\pi/2}\frac{1 + a \cdot \sin^2(\phi)}{\left(1-m \cdot \sin^2(\phi)\right)^{3/2}} \mathrm{d}\phi &\stackrel{t = \sin^2(\phi)}{=}& \int_0^1 \frac{1+a t}{(1-m t)^{3/2}} \frac{\mathrm{d}t}{2\sqrt{t(1-t)}}\\ \end{eqnarray} $$ Using $$ \frac{1+a t}{(1-m t)^{3/2}} \frac{1}{2\sqrt{t(1-t)}}= \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{a+m}{1-m} \frac{\sqrt{t(1-t)}}{\sqrt{1-m t}}\right) + \frac{a+m}{2m(1-m)} \frac{\sqrt{1-m t}}{\sqrt{t(1-t)}} - \frac{a}{2m} \frac{1}{\sqrt{t(1-t)}\sqrt{1-m t}} $$ we arrive at $$ \int_0^1 \frac{1+a t}{(1-m t)^{3/2}} \frac{\mathrm{d}t}{2\sqrt{t(1-t)}} = \frac{a+m}{m(1-m)} E(m) - \frac{a}{m} K(m) $$ Substituting $m = -\frac{4x}{(1-x)^2}$ and $a=\frac{2x}{1-x}$ and combining terms: $$ \int_0^{2\pi} \frac{1-x \cos(\phi)}{\left(1+x^2-2x \cos(\phi)\right)^{3/2}}\mathrm{d} \phi = \frac{2}{1+x} \operatorname{E}\left(\frac{-4x}{(1-x)^2}\right) + \frac{2}{1-x} \operatorname{K}\left(\frac{-4x}{(1-x)^2}\right) $$ Here is a numerical check with Mathematica:

In[27]:= With[{x = 0.78}, {NIntegrate[(
   1 - x Cos[a])/(1 + x^2 - 2 x Cos[a])^(3/2), {a, 0, 2 Pi}],
   2 EllipticE[-4 x/(1 - x)^2]/(1 + x) + 
   2 EllipticK[-4 x/(1 - x)^2]/(1 - x)}]

Out[27]= {13.2161, 13.2161}
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1  
As a tiny but important note: computationally, the usual algorithms for the complete elliptic integrals work best if their argument is within the interval $[0,1)$. It thus makes sense, if $0 \leq x < 1$ to arrange things so that the arguments in Sasha's answer can be made to lie in the unit interval. Luckily, there are the imaginary modulus identities $K(-m)=\frac1{\sqrt{1+m}}\,K\left(\frac{m}{1+m}\right)$ and $E(-m)=\sqrt{1+m}\,E\left(\frac{m}{1+m}\right)$. –  J. M. May 10 '12 at 15:28
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Applying those identities to the last relation Sasha derived yields the expression $$\frac2{x+1}\operatorname{K}\left(\frac{4x}{(x+1)^2}\right)-\frac2{x-1} \operatorname{E}\left(\frac{4x}{(x+1)^2}\right)$$ which is essentially Robert's answer in the comments, except that Robert (and Maple) use the modulus as an argument instead of the parameter, which is what Mathematica, Sasha, and yours truly are using. –  J. M. May 10 '12 at 15:31
    
Of course, there ought to be a direct way of getting that expression... –  J. M. May 10 '12 at 15:34

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