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My question is about the following integration using the residue theorem. $$f(x)=\int_0^{2\pi} \frac{\cos(x)}{5-4\cos(x)}dx $$

Using $$z=e^{ix},\; 2\;cos(x)= z+z^{-1},\; dz=\frac{1}{zi}$$ I am able to get the equation $$f(z)= i/2 \int_C \frac{(z^2+1)}{z(2z-1)(z-2)} $$ and the corresponding residues: $$\operatorname{Res}(z=0)=\lim_{z\rightarrow0} zf(z) = -\frac{i}{4}$$ $$\operatorname{Res}(z=0.5)=\lim_{z\rightarrow0.5} (z-0.5)f(z) = \frac{5i}{12}$$ $$\operatorname{Res}(z=2)=\lim_{z\rightarrow2} (z-2)f(z) = -\frac{5i}{12}$$ From the residue theorem, for closed curve $f(z)$ $$\int_C f(z)\,dz= 2\pi i \cdot \sum \operatorname{Res}(f(z))=\pi/2.$$ From WolframAlpha's computation magic, the actual value is $\pi/3$. I've checked my arithmetic half a dozen times and I am at my wits end. I suspect my mistake is somewhere is variable transformation of $dx$ to $dz$ but I could not figure out what.

Could anyone kindly help me with this.

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The path $C$ has $|z|=1$. So we shouldn't consider $z=2$ which is outside $C$.

Also check the signs of the residues.

$$R(0) = \frac{i}{2}\lim_{z\to0}\frac{z(z^2+1)}{z(2z-1)(z-2)} = \frac{i}{2}\frac{1}{(-1)(-2)} = \frac{i}{4}$$

and $$R(0.5) = \frac{i}{2}\lim_{z\to{0.5}}\frac{(z-0.5)(z^2+1)}{2z(z-0.5)(z-2)} = \frac{i}{4}\frac{5/4}{(1/2)(-3/2)} = -\frac{5i}{12}$$ So, $$I = 2\pi i[R(0)+R(0.5)] = 2\pi i\bigg[\frac{i}{4}-\frac{5i}{12}\bigg] = 2\pi i \bigg[\frac{-2i}{12}\bigg]= \frac{\pi}{3}$$

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  • $\begingroup$ It has worked out; never considered the path size. Thank you kindly! Was repeating the same calculations too many times over I messed the signs. $\endgroup$ – Thomas Jun 30 '19 at 13:57
  • $\begingroup$ You're welcome. $\endgroup$ – Ak. Jun 30 '19 at 13:59

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