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Calculating integral using residue

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the integral
    ∫[itex]\frac{1}{(a+bcos^2(ϕ))^2}[/itex]dϕ from 0 to 2π a,b >0

    2. Relevant equations

    Residue theorem
    Cauchy's integral formula

    3. The attempt at a solution

    The first thing I did was attempt to find the poles of the integral and use residue theorem to solve the integral. I've considered transforming the cosine part into it's exponential form. The problem is dealing with unknown variables a,b as I'm not sure how to deal with them. Does anyone have any ideas?
     
    Last edited: Oct 16, 2013
  2. jcsd
  3. Oct 15, 2013 #2
    There is no problem with having a and b in the integral; your final answer will simply be expressed in terms of a and b. For example, the value of the integral might be ##2\pi ab## or ##\sqrt(ab) \pi/2## but such answers are perfectly valid. (Note: it's unlikely that either of these is the answer - I just chose some random expressions to show that the final result will [probably] depend on a and b).


    EDIT: Another "hint" I can give you is to just be very careful when doing the simplifying algebra. These problems can get pretty nasty in terms of algebra (not hard, but messy) and if you make a mistake it can be a pain to find it, so take your time and go through each step slowly.
     
  4. Oct 15, 2013 #3
    I do understand that there's no problem with it. The thing I'm wondering is whether I'm using the correct methods. Right now I'm simply trying to identify the simple poles by trying to solve the denominator, then trying to use residues to compute the integral.
     
  5. Oct 16, 2013 #4
    My apologies - I must have misinterpreted your original post; I thought you were concerned with that fact that you had unknown variables a and b.

    I would proceed with this problem in the manner you have described (that is, find the poles and apply the residue theorem or Cauchy's integral formula). I recommend the substitution ##z = e^{i\theta}## from which you should be able to write ##d\theta## and ##cos\theta## and solve for the simple poles and proceed from there.
     
  6. Oct 16, 2013 #5
    My apologizes, it's is actually cos^2(ϕ). Edited

    So i would consider using the power reduction to get (1+cos(2ϕ))/2. I can use exponential formula for cos, so now my terms [itex]\frac{b}{4}[/itex]e-2iϕ+ [itex]\frac{b}{4}[/itex]z+[itex]\frac{b}{2}[/itex]+a=0. Not sure where I can go from here to find the poles, I can rearrange to get a and b to the other side, and have the exponential terms to the left.

    [itex]\frac{a}{b}[/itex]=[itex]\frac{1}{2}[/itex]([itex]\frac{1}{2}[/itex]e-2iϕ+[itex]\frac{1}{2}[/itex]e2iϕ+1)

    Actually simplyfing as much as possible on the denomintor:
    (a+[itex]\frac{1}{4}[/itex]b(e-iϕ+e)2)2 which looks better!

    Using the suggested substituion then (a+[itex]\frac{1}{4}[/itex]b(z+[itex]\frac{1}{z}[/itex])2)2 and dϕ=[itex]\frac{dz}{iz}[/itex] From here im not sure on how to find the poles
     
    Last edited: Oct 16, 2013
  7. Oct 16, 2013 #6
    You're getting there! From here all you have to do is solve it - just keep going! You can simplify that expression by expanding the inner bracket and writing everything with respect to a common denominator so that you would have something like ##\frac{some stuff}{polynomial in z}##.

    $$\int_{0}^{2\pi} \frac{1}{(a +bcos^{2}\phi)^{2}} d\phi$$

    You can make my suggested substitution immediately. ##z = e^{i\phi}## so ##dz = ie^{i\phi} d\phi = iz d\phi \rightarrow d\phi = \frac{1}{iz} dz##. From Euler's formula we know ##cos\phi = \frac{e^{i\phi} + e^{-i\phi}}{2}## which we can write as ##cos\phi = \frac{z +\frac{1}{z}}{2} = \frac{z^{2} + 1}{2z}## so ##cos^{2}\phi = \left( \frac{e^{i\phi} + e^{-i\phi}}{2}\right)^{2} = \left( \frac{z^{2} + 1}{2z} \right )^{2}##

    $$\int_{0}^{2\pi} \frac{1}{(a +bcos^{2}\phi)^{2}} d\phi = \int_{0}^{2\pi} \left( \frac{1}{a + b( \frac{z^{2} + 1}{2z})^{2}} \right)^{2} \frac {1}{iz} dz$$

    What I have here should be equivalent to what you've done already expressed in a slightly different manner. From here you should be able to simplify things even further so that you end up with a polynomial in z in the denominator (looks like it'll be of order 4) which you can then factor to find the poles and then apply the Residue theorem. Note: I'm pretty sure what I've done is correct, but I may have made an algebra error somewhere - so if you're getting something different at the end don't worry about it too much. I'll check my work and get back to you later with corrections if Ive made any algebra errors.

    Good luck with the rest of the problem!

    EDIT: I fixed my expression; forgot to square it at some point. Hopefully that's correct now (please be more careful than me!)
     
    Last edited: Oct 16, 2013
  8. Oct 16, 2013 #7
    Often when you have a messy integral, you can check your work at each step numerically before going to the next step. For example, the two integrals above, I use Mathematica:

    Code (Text):

    In[12]:= a = 2;
    b = 3;
    NIntegrate[1/(a + b Cos[t]^2)^2, {t, 0, 2 \[Pi]}]
    NIntegrate[(1/(a + b ((z^2 + 1)/(2 z))))^2 1/(I z) I Exp[I t] /.
      z -> Exp[I t], {t, 0, 2 \[Pi]}]

    Out[14]= 0.695421

    During evaluation of In[12]:= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in t near {t} = {2.30701}. NIntegrate obtained 2181.77 +7.73309*10^-10 I
     and 1522.4410650043299` for the integral and error estimates. >>

    Out[15]= 2181.77 + 7.73309*10^-10 I
     
    That means likely there is an error.
     
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