I'm trying to evaluate this integral using the Residue theorem (It doesn't look to hard) $$\int_{0}^{2\pi}\frac{ dx}{(a+b\cos(x))^2}$$ (where $a>b>0$ ) I do the standard substitution of $z=e^x$ and get $dx=-idz/z$ and $\cos(x)=\frac{z+z^{-1}}{2} $, put it in my integral and, with a little algebra, get

$$ -4i \oint_{|z|=1} \frac{zdz}{(bz^2+2az+b)^2} \,$$

But when I try find the poles I end up getting:

$$ z_{1,2}=\frac{-2a\pm \sqrt{4a^2-4b^2}}{2b} \overset{\mathrm{c=a/b}}{=}-c\pm \sqrt{c^2-1}$$

but now I can't find which pole is inside the contour and which isn't becouse:

for $z_1$

$ -c+\sqrt{c^2-1}<1$


$c^2-1<1+2c+c^2 $


and since our $c>1$ this pole is inside the contour. On the other hand for $z_2$ we have


$-\sqrt{c^2-1}<c+1\implies \sqrt{c^2-1}>-c-1 $

since $c>1>0$ this is equal to saying $-5<6$ or $-6<5$ it will always be true! Is that a problem (is it a sign of a mistake) or is the pole at $z_2$ also inside the contour?

  • $\begingroup$ $-c-\sqrt{c^2-1}<-1$, so it's outside $\endgroup$ – user8268 Jan 31 '18 at 19:07
  • $\begingroup$ so my condition should actually be $|-c\pm \sqrt{c^2-1}|<1$ ? $\endgroup$ – Alexandar Solženjicin Jan 31 '18 at 19:29
  • $\begingroup$ Have a look at page 46 of my notes for alternative approaches. I like to regard such integral as a multiple of the area enclosed by some ellipse. $\endgroup$ – Jack D'Aurizio Jan 31 '18 at 19:39
  • $\begingroup$ Wow, thank you very much :) $\endgroup$ – Alexandar Solženjicin Jan 31 '18 at 19:47

$z_2\not\in B(0,1)$. Indeed, if $z_2\in B(0,1)$ then $$|z_2|<1$$ $$c+\sqrt{c^2-1}<1$$ $$1+\sqrt{c^2-1}<c+\sqrt{c^2-1}<1$$ $$\sqrt{c^2-1}<0$$ which is a contradiction.

  • $\begingroup$ what did you do in step 3? $\endgroup$ – Alexandar Solženjicin Feb 1 '18 at 15:37
  • $\begingroup$ as $c=a/b$ and $a>b>0$ entonces $c>1$ $\endgroup$ – Mauricio Ruiz Feb 1 '18 at 15:43
  • $\begingroup$ I agree, but how does $1+\sqrt{c^2-1}<c+\sqrt{c^2-1}<1$ imply that if $1+\sqrt{c^2-1}<1$ then $c+\sqrt{c^2-1}<1$ ? $\endgroup$ – Alexandar Solženjicin Feb 1 '18 at 15:46
  • $\begingroup$ we have $$c+\sqrt{c^2-1}<1$$. As $c=a/b$ then $c>1$, then $$1+\sqrt{c^2-1}<c+\sqrt{c^2-1}$$. By transitivity $$1+\sqrt{c^2-1}<1$$. So $$\sqrt{c^2-1}<0$$ $\endgroup$ – Mauricio Ruiz Feb 1 '18 at 15:53
  • $\begingroup$ aaa, ok, I missed the point of what you were trying to show, thankyou. $\endgroup$ – Alexandar Solženjicin Feb 1 '18 at 16:11

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