MHB Please check Residue theorem excercise

[ognik]

Show $ \int_{-\infty}^{\infty}\frac{1}{x^4 - 2 cos 2 \theta + 1} \,dx = \frac{\pi}{2sin \theta} $

I know I want to use the residue theorem for $ \int_{0}^{\pi}\frac{1}{z^4 - 2 cos 2 \theta z^2 + 1} \,dz $, and have found the 4 poles ($ \pm e^{\pm i \theta } $).

I chose the upper semi-circle (closed path), |z|=R, which leaves only 2 poles interior, $ \pm e^{i \theta} $

I can show that the 'return path' integral around the semi-circle tends to 0 as R tends to $\infty$, so I can use the residue theorem for the path along the real axis. All good to here I think?

Using $ Res[f, {z}_{0}] = \lim_{{z}\to{{z}_{0}}} \frac{(z-{z}_{0})}{ (z+e^{i \theta})(z-e^{i \theta})(z+e^{-i \theta})(z-e^{-i \theta}) } $ I get
$ Res[f, e^{i \theta}] = \frac{1}{4i e^{i \theta}} \frac{1}{sin(2\theta)}$ and $ Res[f, -e^{i \theta}] = \frac{-1}{4i e^{i \theta}} \frac{1}{sin(2\theta)}$, but those residues cancel, so something is wrong (I was kinda hoping they would add). I've checked and re-checked, getting nowhere...

Could someone please just:
1) Confirm if my approach is OK
2) Check my 2 residues

That should be enough for me to finish this tomorrow (Sleepy) ...

 

[Opalg]

Show $ \int_{-\infty}^{\infty}\frac{1}{x^4 - 2 cos 2 \theta + 1} \,dx = \frac{\pi}{2sin \theta} $

I know I want to use the residue theorem for $ \int_{0}^{\pi}\frac{1}{z^4 - 2 cos 2 \theta z^2 + 1} \,dz $, and have found the 4 poles ($ \pm e^{\pm i \theta } $).

I chose the upper semi-circle (closed path), |z|=R, which leaves only 2 poles interior, $\pm e^{i \theta} $

No. The two poles inside the contour are $e^{i\theta}$ and $-e^{-i\theta}$ (on the assumption that $0<\theta<\pi$).

 

[ognik]

Thanks, but that tells me I have learned something wrong. Yes, $ 0 < \theta < \pi $

$ e^{i \theta} $ is in quadrant 1, but I thought $ -e^{-i \theta} $ would be in quad 3, because $ -e^{-i \theta} = -(cos(-i \theta) +i sin(-i \theta)) = -cos(i \theta) - i sin(i \theta) $, so both real and imaginary components are negative, therefore quad 3?

I know I must have this wrong, but why? How should I have 'plotted' these poles?

 

[Opalg]

Thanks, but that tells me I have learned something wrong. Yes, $ 0 < \theta < \pi $

$ e^{i \theta} $ is in quadrant 1, but I thought $ -e^{-i \theta} $ would be in quad 3, because ${\color{red} -e^{-i \theta} = -(cos(-i \theta) +i sin(-i \theta))} = -cos(i \theta) - i sin(i \theta) $, so both real and imaginary components are negative, therefore quad 3?

I know I must have this wrong, but why? How should I have 'plotted' these poles?

Too many $i$'s there, for a start. It should be $-e^{-i \theta} = -(\cos(-\theta) +i \sin(- \theta)) = -\cos(\theta) + i \sin(\theta) $, because $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$.

 

[ognik]

Oops, 'i' feel rather embarrassed ...Thanks Opalg.

Then $ Res[f,e^{i \theta }] = \frac{1}{2e^{i \theta }(e^{i \theta } +e^{-i \theta }) (e^{i \theta } - e^{-i \theta }) }
= \frac{1}{4ie^{i \theta }} \frac{1}{2sin \theta cos \theta} $

Similarly, $ Res[f,-e^{-i \theta }] = \frac{1}{4ie^{-i \theta }} \frac{1}{2sin \theta cos \theta} $

Finally $ \int_{-\infty}^{\infty} f(x)\,dx = 2 \pi i (\frac{1}{4i e^{i \theta + }}\frac{1}{4i e^{-i \theta}})\frac{1}{2sin \theta cos \theta}
=\frac{\pi}{8cos \theta}\frac{1}{2sin \theta cos \theta} $ - which is $ \frac{1}{8cos^2 \theta} $ different from what it should be? I can't find a mistake?

 

[Opalg]

Finally $ \int_{-\infty}^{\infty} f(x)\,dx = 2 \pi i \Bigl(\frac{1}{4i e^{i \theta }} + \frac{1}{4i e^{-i \theta}}\Bigr)\frac{1}{2sin \theta cos \theta}
=\frac{\pi}{8cos \theta}\frac{1}{2sin \theta cos \theta} $ - which is $ \frac{1}{8cos^2 \theta} $ different from what it should be? I can't find a mistake?

$\dfrac{1}{4i e^{i \theta }} + \dfrac{1}{4i e^{-i \theta}} = \dfrac1{4i}\cdot \dfrac{e^{-i\theta} + e^{i\theta}}{e^{i\theta} e^{-i\theta}} = \dfrac1{4i}\cdot \dfrac{2\cos\theta}1 = \dfrac{\cos\theta}{2i}$

 

[ognik]

Thanks Opalg, happy that I was on the right track - but how do you persuade $ e^{\theta} e^{i\theta} $ to = 1? I haven't come across that before, tried whatever I could think of and couldn't find anything on the web ...

 

[Fernando Revilla]

- but how do you persuade $ e^{\theta} e^{i\theta} $ to = 1? I haven't come across that before, tried whatever I could think of and couldn't find anything on the web ...

Just a typo, $ e^{-i\theta} e^{i\theta}=e^0=1. $

 

[ognik]

I should have seen that, but thanks Fernando. (times like this I get quite annoyed with myself!)