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So I've had a crack at this contour integration question and have somehow managed to get a complex solution for a real integral... I've gone through my working a number of times but can't seem to find the mistake, so was hoping someone here could help.

Evaluate the integral

$$ I=\int^\pi_{-\pi} \frac{\,d\theta}{a+b\cos\theta+c\sin\theta} $$

where $a$, $b$, $c$ are real positive constants such that $a^2>b^2+c^2>0$

My Attempt:

Consider the complex function


This has simple poles when denominator equals zero, i.e. at $z=z_1$ and $z=z_2$, where

$$z_1=\frac{-a+\sqrt{a^2-(b^2+c^2)}}{b-ic} \ \text{and}\ \ z_2=\frac{-a-\sqrt{a^2-(b^2+c^2)}}{b-ic}$$

by the quadratic formula. See that $|z_2|>|z_1|\,\,\,\,\,\,(*)$.

Note that $|z_1||z_2|=|z_1 z_2|=\left|\frac{b^2+c^2}{(b-ic)^2}\right|=1$

Referring back to $(*)$ we see that $|z_1|<1$ and $|z_2|>1$.

$\text{Res}[f(z), z_1]=lim_{z\to z_1}[(z-z_1)f(z)]=\frac{1}{z_1-z_2}=\frac{b-ic}{2 \sqrt{a^2-b^2-c^2}}$

Since $z_1$ is the only pole enclosed by the contour $|z|=1$, by the residue theorem:

$$\oint_{|z|=1}f(z)dz=2\pi i \text{Res}[f(z), z_1]\,\,\,\,\,\,(**)$$

Along the contour $|z|=1$, we can write $z=e^{i\theta}$ $dz=ie^{i\theta}d\theta$ with $\theta$ in $[-\pi,\pi]$


$$\oint_{|z|=1}f(z)dz=\int^\pi_{-\pi}\frac{ie^{i\theta}}{ae^{i\theta}+\frac{1}{2}b(e^{2i\theta}+1)+\frac{1}{2i}c(e^{2i\theta}-1)}d\theta$$ $$=i\int^\pi_{-\pi}\frac{1}{a+\frac{1}{2}b(e^{i\theta}+e^{-i\theta})+\frac{1}{2i}c(e^{i\theta}-e^{-i\theta})}d\theta=iI$$

Returning to $(**)$, we see that

$$I=2\pi\text{Res}[f(z), z_1]=\frac{b-ic}{\sqrt{a^2-b^2-c^2}}\pi$$

Why is my result complex? I'd appreciate it if someone could point out where I went wrong.

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up vote 5 down vote accepted

I can't tell what you're doing in your residue calculation, but I get the following for the residue of the pole inside the unit circle:

$$\frac{2}{2 \sqrt{a^2-b^2-c^2}}$$

I get this using the formula for the residue at a simple pole $z_0$ of a function $f(z)/g(z)$:

$$\text{Res}_{z=z_0} \frac{f(z)}{g(z)} = \frac{f(x_0)}{g'(z_0)}$$

This means that the integral is $$\frac{2 \pi}{\sqrt{a^2-b^2-c^2}}$$

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I was using this result for the residue at a simple pole:… and writing $f(z)$ as $\frac{1}{(z-z_1)(z-z_2)}$ – IchiCC Apr 30 '13 at 1:28
@IchiCC: That is correct of course but far more error-prone. – Ron Gordon Apr 30 '13 at 6:21
Hmm I see... I've not actually seen a residue being calculated using the method you posted - up until now I've always either taken the limit, or expressed the function as its Laurent series. Can you see where I've gone wrong in my calculation? – IchiCC Apr 30 '13 at 9:23
@IchiCC: it is actually quite straightforward to prove for a simple pole using the definition of a derivative. See this for example. – Ron Gordon Apr 30 '13 at 9:28
You also left off the constant factors in front of the original polynomial in your residue calculation - that explains why you still have the complex factor and are missing a factor of $2$. – Ron Gordon Apr 30 '13 at 9:30

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