MHB Solving Integral using Complex Analysis: 0 to 2pi of $\frac{1}{13+12cost}$

[Poirot1]

I wish to evalute integral from 0 to 2pi of $\frac{1}{13+12cost}$ using compex analysis. So I say $cosz=0.5(z+z^{-1})$ and change variables thus integrand becomes:

$-i\frac{1}{13z+6z^2+6}$. What do I do from here?

 

[Dustinsfl]

I wish to evalute integral from 0 to 2pi of $\frac{1}{13+12cost}$ using compex analysis. So I say $cosz=0.5(z+z{-1})$ and change variables thus integrand becomes:

$-i\frac{1}{13z+6z^2+6}$. What do I do from here?

You need to determine where the poles occur at.

Then check to see which pole(s) are in the unit circle.

Then use Residue Theory.

 

[CaptainBlack]

I wish to evalute integral from 0 to 2pi of $\frac{1}{13+12cost}$ using compex analysis. So I say $cosz=0.5(z+z^{-1})$ and change variables thus integrand becomes:

$-i\frac{1}{13z+6z^2+6}$. What do I do from here?

There is no change of variable involved, you are using an identity.

Where did the -i come from?

*The poles occur at the zeros of \(6x^2+13x+6\).
CB

 

[chisigma]

... there is no change of variable involved, you are using an identity... where did the -i come from?...

Is...

$\displaystyle z=e^{i\ t} \implies dz=i\ e^{i t}\ dt= i\ z\ dt \implies dt=\frac{d z}{i\ z}=-i\ \frac{d z}{z}$

Kind regards

$\chi$ $\sigma$

 

[CaptainBlack]

Is...

$\displaystyle z=e^{i\ t} \implies dz=i\ e^{i t}\ dt= i\ z\ dt \implies dt=\frac{d z}{i\ z}=-i\ \frac{d z}{z}$

Kind regards

$\chi$ $\sigma$

What is being used is:

\(\cos(z)=\frac{e^{iz}+e^{-iz}}{2}\)

so:

\(\cos(t)=\frac{e^{it}+e^{-it}}{2}\)

just relabelling of the dummy variable of integration.

CB

 

[chisigma]

What is being used is:

\(\cos(z)=\frac{e^{iz}+e^{-iz}}{2}\)

so:

\(\cos(t)=\frac{e^{it}+e^{-it}}{2}\)

just relabelling of the dummy variable of integration.

CB

Of course the correct solving procedure is to set $\displaystyle \cos t = \frac{z+z^{-1}}{2},\ z=e^{i\ t}$ , that conducts to the identity...

$\displaystyle \int_{0}^{2 \pi} \frac{dt}{13+12\ \cos t}= -i\ \int_{C} \frac {d z}{6 z^{2}+13 z+6}$ (1)

... where C is the unit circle. The purpose of my post was to justify the 'funny term' -i in the formula...

Kind regards

$\chi$ $\sigma$

 

[Poirot1]

Thanks , chisgma. That is the method I was trying in vain to remember. I get simple poles at z=-2/3 and -3/2 but -3/2 is outside of unit circle. I will call integrand f(z). Res(f,-2/3) =1/5 so using cauchy's formula I obtain integral is 2/5(pi).

 

[CaptainBlack]

Of course the correct solving procedure is to set $\displaystyle \cos t = \frac{z+z^{-1}}{2},\ z=e^{i\ t}$ , that conducts to the identity...

$\displaystyle \int_{0}^{2 \pi} \frac{dt}{13+12\ \cos t}= -i\ \int_{C} \frac {d z}{6 z^{2}+13 z+6}$ (1)

... where C is the unit circle. The purpose of my post was to justify the 'funny term' -i in the formula...

Kind regards

$\chi$ $\sigma$

Yes, I should have notice4d where it had come from straight off.

Cb