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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!

  • 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?2012-05-10
  • 0
    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.2012-05-10
  • 0
    The closed form is given by $$J(x)=\frac{2 \left(-(-1+x) \text{EllipticE}\left[-\frac{4 x}{(-1+x)^2}\right]+(1+x) \text{EllipticK}\left[-\frac{4 x}{(-1+x)^2}\right]\right)}{\sqrt{(-1+x)^2} (1+x)}.$$2012-05-10
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    In Maple (which I think uses a different convention for the elliptic integrals) the closed form is $$2\,{\it EllipticK} \left( 2\,{\frac {\sqrt {x}}{x+1}} \right) \left( x+1 \right) ^{-1}-2\,{\it EllipticE} \left( 2\,{\frac {\sqrt {x}}{x+1} } \right) \left( x-1 \right) ^{-1}$$2012-05-10
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    ... and we can also write it as a Taylor series (converging for $|x| < 1$): $$ \sum _{k=0}^{\infty }2\,{\frac {\pi \, \left( \left( 2\,k \right) ! \right) ^{2}{16}^{-k} \left( 2\,k+1 \right) {x}^{2\,k}}{ \left( k! \right) ^{4}}} $$2012-05-10
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    For $|x|<1$ this can also be written as $\dfrac{4 \it{EllipticE}(x)}{1-x^2}$.2012-05-10
  • 1
    @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](http://math.stackexchange.com/questions/143555/direct-evaluation-of-complete-elliptic-integral#comment330891_143555).2012-05-10

1 Answers 1

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Let's assume $-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} 
  • 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 $E(-m)=\sqrt{1+m}\,E\left(\frac{m}{1+m}\right)$.2012-05-10
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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.2012-05-10
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    Of course, there ought to be a direct way of getting that expression...2012-05-10