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Evaluate integral

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Poirot

Banned
Feb 15, 2012
250
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?
 
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dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
Jan 26, 2012
890
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

Well-known member
Feb 13, 2012
1,704
... 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

Well-known member
Jan 26, 2012
890
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

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 
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Poirot

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Feb 15, 2012
250
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

Well-known member
Jan 26, 2012
890
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