Integrals of trigonometric functions over [o,2pi]

[jwhite2531]

Homework Statement

∫dθ/(1+βcosθ)^2 ; -1<β<1
θ=0 to 2pi

Homework Equations

The Attempt at a Solution

attempt solution:

1) make substitution:
dθ=dz/iz
Z=e^iθ
cosθ=1/2(Z+1/z)

2) substitute:

1/i*dz/(β+Z(1+(β^2)/2)+((3βZ^2)/2)+((β^2)Z^3)/4)+((β^2))/4Z)

3) Next ?

3a)Find the poles ?
We don't know how...

3b)Compute the residues

3c)Calculate integral
:)

 

[jackmell]

That's a tough one I think jwhite. Need to know how to find the poles to calculate the residue. First write it clearly:

[tex]\int_0^{2\pi}\frac{dt}{(1+b\cos(t))^2}[/tex]

and doing the [itex]z=e^{it}[/itex] substitution, I get:

[tex]\int_0^{2\pi}\frac{dt}{(1+b\cos(t))^2}=-i\mathop\oint\limits_{|z|=1}\frac{dz}{z}\frac{4z^2}{\big(2z+bz^2+b\big)^2}=-i\mathop\oint\limits_{|z|=1}\frac{4z}{\big(2z+bz^2+b\big)^2}[/tex]

Now, you can figure when that denominator is zero to find the poles and then figure which ones are in the unit circle when [itex]-1<b<1[/itex]. Note when you factor it (don't forget to factor out the b first), and the factors are squared, that means the poles are second order. You'll need to know how to compute the residue of a second-order pole. For example, if it were:

[tex]\frac{4z}{(z-z1)^2(z-z2)^2}[/tex]

then the residue at for example z2 would be:

[tex]\mathop\text{Res}\limits_{z=z2}\left\{\frac{4z}{(z-z1)^2(z-z2)^2}\right\}=\frac{d}{dz}\left(\frac{4z}{(z-z1)^2}\right)\biggr|_{z=z2}[/tex]

 

[jwhite2531]

Thank you very much Jackmell, really appreciate it.