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How Can I evaluate this integral ? $$\int_{0}^{2 \pi} \frac{\sin\phi}{(\sin\theta-\sin\phi)^2+2a(1-\cos(\theta -\phi))} d\phi$$

I have tried to write the denominator as $$\sin( (\theta-\phi)/2 )[2\cos((\theta+\phi)/2)+4a\sin((\theta-\phi)/2)]$$

And writing the numerator as $$\sin(\phi)=2\sin((\phi-\theta)/2)\cos((\phi - \theta)/2)\cos(\theta)+\sin(\theta)(1-2\sin^2((\phi-\theta)/2)$$ Where $$\theta , \phi \in[0,2 \pi]$$

Quanto
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Mahmoud Hassan
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  • Did you try to work it out using a complex curve integral? I ask, since you use those tags. If so, where did you get stuck in that attempt? – mickep May 06 '19 at 19:45
  • Yes I did try using $z=e^{i \phi}$ but I didn't get the result . – Mahmoud Hassan May 11 '19 at 16:04
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    I know there is an accepted answer to this question, but doesn't this integral diverge at $\phi=\theta$? The denominator goes as $C/(\theta-\phi)^2$ when expanded around $\phi$ which is not integrable or regularizable using principal value type prescriptions. – DinosaurEgg Jun 05 '20 at 07:35
  • For $\theta=0$, $\int_{0}^{2 \pi} \frac{\sin \phi}{\sin^2\phi+4a\sin^2\frac\phi2 } d\phi$ is undefined. – Quanto Jul 04 '22 at 18:30

1 Answers1

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Here $\theta=t$, $\phi=x$. Your integral is $$q(a,t)=\int_0^{2\pi}\frac{\sin x}{(\sin t-\sin x)^2+2a(1-\cos(t-x))}dx$$ Note that $$\sin\alpha-\sin\beta=2\sin\left(\frac{\alpha-\beta}2\right)\cos\left(\frac{\alpha+\beta}2\right)$$ and $$1-\cos(\alpha-\beta)=2\sin^2\left(\frac{\alpha-\beta}2\right)$$ So we have $$\begin{align} q(a,t)=&\frac14\int_0^{2\pi}\frac{\sin x}{\sin^2(\frac{t-x}2)(2a+\cos^2(\frac{t-x}2))}dx\\ =&\frac14\int_0^{2\pi}\frac{\sin x}{\sin^2(\frac{x}2-\frac{t}2)\cos^2(\frac{x}2-\frac{t}2)+2a\sin^2(\frac{x}2-\frac{t}2)}dx\\ \overset{2u=x-t}=&\frac12\int_{-t/2}^{\pi-t/2}\frac{\sin(2u+t)}{\sin^2(u)\cos^2(u)+2a\sin^2(u)}du\\ =&\frac12\int_{-t/2}^{\pi-t/2}\frac{c\sin(2u)+s\cos(2u)}{\cos^2(u)\sin^2(u)+2a\sin^2(u)}du\\ =&-\frac12\int_{-t/2}^{\pi-t/2}\sec^2(u)\frac{s\tan^2(u)-2c\tan(u)-s}{\tan^2(u)(2a\tan^2(u)+2a+1)}du\\ \overset{x=\tan(u)}=&-\frac12\int_{x_1}^{x_2}\frac{sx^2-2cx-s}{x^2(2ax^2+2a+1)}dx \end{align}$$ Where $s=\sin(t)$, $c=\cos(t)$, $x_1=\arctan(-t/2)$, and $x_2=\arctan(\pi-t/2)$. At this point we preform partial fractions to get $$q(a,t)=-\frac1{2(2a+1)}\left[\int_{x_1}^{x_2}\frac{4acx+(4a+1)s}{2ax^2+2a+1}dx-2c\int_{x_1}^{x_2}\frac{dx}{x}-s\int_{x_1}^{x_2}\frac{dx}{x^2}\right]$$ Can you take it from here?

(12/9/19) Addendum:

I finally got around to the full simplification. It is horrible.

We have $$\begin{align} -2(2a+1)q(a,t)&=2\cos(t)\left[\ln\left|\frac{\arctan(t/2)}{\arctan(\pi-t/2)}\right|+2\ln\left|\frac{\arctan^2(\pi-t/2)+1+\frac1{2a}}{\arctan^2(t/2)+1+\frac1{2a}}\right|\right]\\ &+\frac{\sin(t)}{\arctan(\pi-t/2)}+\frac{\sin(t)}{\arctan(t/2)}\\ &+\frac{2\sqrt{2}}{\sqrt{a(2a+1)}}\arctan\left(\arctan(\pi-t/2)\sqrt{\frac{2a}{2a+1}}\right)\\ &+\frac{2\sqrt{2}}{\sqrt{a(2a+1)}}\arctan\left(\arctan(t/2)\sqrt{\frac{2a}{2a+1}}\right). \end{align}$$ Which begs the question: How on Earth did you come up with this integral??

clathratus
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