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The goal is to evaluate $\displaystyle\int^{\pi}_{-\pi}\frac{1}{1+\sin^2\!\theta}\,\mathrm d\theta\;$ using the Cauchy residue theorem.

Using the substitution $z=e^{i\theta}$ and $\;\sin\theta =\dfrac{z-1/z}{2i}\;$ with $\mathrm d\theta=\dfrac{\mathrm dz}{iz}\;,\;$ I rewrite the integral as

$\displaystyle\int_{|z|=1}\frac{\mathrm dz}{\left[1+\left(\frac{z-1/z}{2i}\right)^2\right]iz}=\int_{|z|=1}\frac{\mathrm dz}{\left(1+\frac{z^2-2+\frac{1}{z^2}}{-4}\right)iz}=$ $\displaystyle\int_{|z|=1}\frac{\mathrm dz}{\left(-4z+z^3-2z+\frac{1}{z}\right)\frac{-i}{4}}=\int_{|z|=1}\frac{\mathrm dz}{\left(z^4-6z^2+1\right)\frac{-i}{4z}}=\int_{|z|=1}\frac{(-4z/i)\mathrm dz}{z^4-6z^2+1}$

I can't spot a mistake in my algebra, though my brain often gets stuck and I can't find the simplest errors. Do you see any error?

My problem is that I can't factor the denominator $z^4-6z^2+1$. I want to do this so that I can find out what the residues are and apply the Cauchy residue theorem.

Please show me the algebraic steps to get this integral in my desired form.(where residue theorem is easily applied). Please also point out any mistakes I have made.

Angelo
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violet
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    Note that $z^4 - 6z^2 + 1$ is a quadratic in $w = z^2$. This should let you find the four roots and then you can use the residue theorem. –  Dec 01 '22 at 22:30
  • Note that in an in-between step you have denominator $$z\left(1+\left(\frac{z-1/z}{2i}\right)^2 \right)$$. Putting this to be equal to $0$ you get $5$ roots, which are solution of the equation $z(z^2 \pm 2z -1) =0$. From these $5$ simple poles only $3$ lie inside the contour, Finding residue at these give the correct answer.

    $z=0, \sqrt{2}-1, -\sqrt{2}+1$

     – Lavneet Dec 02 '22 at 02:23

1 Answers1

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Alternatively,$I=\int_{-\pi}^\pi \frac{d\theta}{1+\sin^2\theta} =2\int_0^\pi \frac{d\theta}{1+\sin^2\theta}\ =2\int_0^\pi \frac{d\theta}{1+\frac{(1-\cos 2\theta)}{2}}$
$=4\int_0^\pi \frac{d\theta}{3-\cos 2\theta}$
$=2\int_0^{2\pi}\frac{dt}{3-\cos\ t}$
( where $2\theta=t\implies d\theta=dt/2)$ $=4\oint_{|z|=1}\frac{dz}{i(6z-z^2-1)}$
(where $e^{i\theta}=z\implies d\theta=\frac{dz}{iz})$

Can you proceed from here?

Although, your last integral can also be simplified for a quadratic polynomial in the denominator of the integrand as follows:
Let $z^2=w\implies 2zdz=dw$
so that the modified contour is |w|=1 traced counterclockwise with winding number $p=2$ which forces you to multiply your final answer by factor $p=2$.

Nitin Uniyal
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