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Homework Help: Checking an Integral

  1. Mar 5, 2008 #1
    [SOLVED] Checking an Integral

    I evaluated the integral below using Cauchy's Formula but I get a different answer than what the book has. The integral is

    [tex]\int_0^{2\pi} \frac{d\theta}{a + b\cos \theta}[/tex]

    where a > b > 0. Here's what I did: I make the subsitution [itex]z = e^{i\theta}[/itex] so that [itex]a + b\cos\theta = a +b/2(z + z^{-1})[/itex] and [itex]d\theta = dz/(iz)[/itex]. Thus the integral becomes

    [tex]\int_{|z|=1} \frac{2 dz}{ibz^2 + i2az +ib}[/tex]

    which I can rewrite as

    [tex]\int_{|z|=1} \frac{2 dz}{(z - p)(z - q)}[/tex]

    where [itex]p = -a/b + \sqrt{b^2 - a^2}/(ib)[/itex] and [itex]q = -a/b - \sqrt{b^2 - a^2}/(ib)[/itex]. Now I can apply Cauchy's Formula by letting f(z) = 2/(z - p) so that the above integral equals [itex]2\pi i f(q) = 2\pi b / \sqrt{b^2 - a^2}[/itex]. Now according to the book, the answer [itex]\pi/\sqrt{a^2 - b^2}[/itex]. I rechecked my algebra and I don't see anything wrong. Is the book wrong?
     
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  3. Mar 5, 2008 #2

    HallsofIvy

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    An obvious point: you say that a> b but you are working with [itex]\sqrt{b^2- a^2}[/itex]. Are you taking into account the fact that that is imaginary?
     
  4. Mar 5, 2008 #3
    Does it matter? I think that a > b > 0 was given so the square root is guaranteed not be 0.
     
  5. Mar 6, 2008 #4

    Gib Z

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    Let [tex]\theta = 2 \arctan x[/tex] and check whose answer it matches up with.
     
  6. Mar 6, 2008 #5
    If [itex]\theta = 2 \arctan x[/itex], what is [itex]\cos \theta[/itex]?
     
  7. Mar 7, 2008 #6

    Gib Z

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    Well, [tex]\tan (2\arctan x) = \frac{2x}{1-x^2}[/tex] by using a double angle identity and inverse function relation. So then draw a right angled triangle, with 2arctan x as another angle. Then label the opposite side 2x and the adjacent side 1-x^2, and use the Pythagorean theorem for the hypotenuse. Then, read off the cosine of the angle.
     
  8. Mar 7, 2008 #7
    OK. With that substitution, the integral becomes

    [tex]\int_0^\infty \frac{1 + x^2}{(a-b)x^2 + (a+b)x} \, dx[/tex]

    Now how do I solve this integral?
     
  9. Mar 8, 2008 #8

    Gib Z

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    To keep it tidy, let a-b = u, a+b = v. Then mess around abit with polynomial division, partial fractions and the like.
     
  10. Mar 8, 2008 #9

    Gib Z

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    To keep it tidy, let a-b = u, a+b = v. Then mess around abit with polynomial division, partial fractions and the like.
     
  11. Mar 8, 2008 #10

    Gib Z

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    To keep it tidy, let a-b = u, a+b = v. Then mess around abit with polynomial division, partial fractions and the like.
     
  12. Mar 8, 2008 #11
    Using your u and v, I get that the integrand is

    [tex]\frac{u-vx}{(ux)^2 + uvx} + \frac{1}{u}[/tex]

    I've been trying to integrate the first term to no avail.
     
  13. Mar 8, 2008 #12
    e(ho0n3, if you look at an older post of me:

    https://www.physicsforums.com/showthread.php?t=204639

    you will find a very powerful substitution for this kind of problems. The answer can be found very fast once you understand it. This method also works for sin functions, study it, you will certainly appreciate the genius Sommerfeld was. Now the answer you gave is wrong. It should be:

    [tex]I=\frac{2\cdot \pi}{\sqrt{a^2-b^2}}[/tex]

    Or you made a typo error or the book is wrong.
     
  14. Mar 8, 2008 #13
    That is an interesting method. Now I have three separate answers: my answer, the books and yours. What I would like to know is: what is wrong with my method?
     
  15. Mar 8, 2008 #14

    Gib Z

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    [tex]\frac{1+x^2}{ux^2+ vx} = \frac{1}{ux^2+vx} + \frac{x}{ux + v} = \frac{1}{u (x^2 + \frac{v}{u} x)} + \left(\frac{1}{u}\right) \left(\frac{ux + (v - v)}{ux+v} \right)= \left(\frac{1}{u}\right)\left(\frac{1}{x^2+\frac{v}{u}x}\right) + \frac{1}{u} \left( 1 - \frac{v}{ux+v}\right)[/tex] would have seemed the easiest method.
     
    Last edited: Mar 8, 2008
  16. Mar 9, 2008 #15
    It seems I am having database errors when I try to submit a post. I need to split it up in two parts. Here's part 1

    Indeed, three different methods which all give the same result I posted earlier.
    [tex]I=\frac{2\cdot \pi}{\sqrt{a^2-b^2}}[/tex]
    So, let's look at the problems.

    The substitution in itself is correct, however if you actually apply it, you get:
    [tex]\frac{2}{i}\int_{C}\frac{dz}{bz^2+2az+b}[/tex]
    Now applying the residue theorem on the one singularity which is inside the unit circle, you arrive at the correct result.

    After applying this substitution I got the following integral:
    [tex]4\int_{0}^{\pi}\frac{dx}{(a+b)+(a-b)x^2}[/tex]
    This can be integrated to become an atan function and applying the limits of integration again gives you the correct result.
     
  17. Mar 9, 2008 #16
    This is part 2

    I don't know how you got here.

    Rewriting the integral as:
    [tex]\frac{1}{a}\int_{0}^{2\pi}\frac{d\theta}{1+\frac{b}{a}cos(\theta)}[/tex]
    And applying the substitution:
    [tex]\frac{b}{a}=\epsilon[/tex]
    [tex]1+\epsilon cos(\theta)=\frac{1- \epsilon^2}{1-\epsilon cos(\gamma)}[/tex]
    You get the following integral:
    [tex]\frac{1}{a\sqrt{1-\epsilon^2}}\int_{0}^{2\pi}d\gamma[/tex]
    Which again gives the same result.

    Hope this clears it up.
     
  18. Mar 9, 2008 #17

    Gib Z

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    Omg he got the substitution wrong >.<" Sigh that would have been much easier.
     
  19. Mar 9, 2008 #18
    Would you believe that I know nothing of the residue theorem yet?

    Going back to my first post, the method I'm using is Cauchy's formula, which doesn't seem to be working. If I let f(z) = 2/(z - q) I get the negative of what I got in the first post. I'm guessing that it is because both 2/(z - p) and 2/(z - q) are not analytic inside the circle |z| = 1 which means I'm not allowed to use Cauchy's formula.

    Both p and q are in |z| <= 1 so regardless of the f I choose, it will not be analytic in |z| <= 1. No wonder my method isn't working.

    Now why would the author state a problem that can't be worked out using Cauchy's formula?
     
  20. Mar 9, 2008 #19
    I gave the following explanation for solving your original post.

    Now you state that you don't know the residue theorem? Strange, then I would say that this exercise is a little bit too soon for your knowledge. Don't be scared that you will never learn how to solve it, residues will be taught shortly after Cauchy. Anyway, one of the zero's is inside the unit circle. That's something you should be able to see.

    There were also two other ways for solving this integral. They are equally valid. Especially the one of Sommerfeld. Have you looked at it yet to see how powerful it is? Have you found the error in the third way for solving the integral? So, you should be able to solve it without the residue theorem.
     
  21. Mar 9, 2008 #20
    If one of the roots is inside the unit circle, then so is the other one. I understand this now.

    The Sommerfeld one is pretty cool. However, I don't understand why you can write [itex]1 - \epsilon \cos \theta[/itex] some other fraction involving [itex]\cos \gamma[/itex]. I messed up in the arctan substitution because I suck at algebra but I'm sure it works.

    Thank for your help coomast.
     
  22. Mar 10, 2008 #21
    No, one is inside, the other is not. Try p.e. a=10 and b=1, you get:

    [tex]z_1\approx -0.05[/tex]
    [tex]z_2\approx -19.95[/tex]

    Check this out.

    That is exactly why people are today still talking about how Sommerfeld got it. It is nothing more then a presentation of his brilliant mind.

    Try it, again, this is not so difficult. Post your problems if you are in need of help.
     
  23. Mar 10, 2008 #22
    Starting over:

    By letting [itex]z = e^{i\theta}[/itex],

    [tex]\int_0^{2\pi} \frac{d\theta}{a + b\cos\theta} = \int_{|z|=1} \frac{2dz}{(z-p)(z-q)}[/tex]

    where

    [tex]p = -\frac{a - \sqrt{a^2 - b^2}}{b} \text{ and } q = -\frac{a + \sqrt{a^2 - b^2}}{b}[/tex]

    To find if p and q are inside |z| = 1, I must determine how |p| and |q| compare to 1.

    [tex]|p| = \frac{a - \sqrt{a^2 - b^2}}{b} \mbox{ and } |q| = \frac{a - \sqrt{a^2 - b^2}}{b} [/tex]

    Let D = [itex]\sqrt{a^2 - b^2}[/itex].

    |p| ? 1
    (a - D)/b ? 1
    a - D ? b
    a - b ? D
    (a - b)2 ? D2
    a2 - 2ab + b2 ? a2 - b2
    2b2 - 2ab ? 0
    2b(b - a) ? 0

    Since b - a is negative because a > b > 0, then 2b(b - a) is negative so ? is actually < and so |p| < 1. p is inside the unit circle.

    |q| ? 1
    (a + D)/b ? 1
    a + D ? b
    D ? b - a
    D2 ? (b - a)2
    a2 - b2 ? b2 - 2ab + a2
    0 ? 2b(b - a)

    In this case ? is > so |q| > 1 so q is outside the unit circle. Thus, I can apply Cauchy's formula by letting f(z) = 2/(z - q) so

    [tex]\int_{|z|=1} \frac{f(z) \, dz}{z - p}[/tex]

    This is equal to [itex]2i\pi f(p) = 2i \pi b/D[/tex]. Hmm...I messed up again. Argh.
     
    Last edited: Mar 10, 2008
  24. Mar 10, 2008 #23
    Hello e(ho0n3, I'll come back to it tomorrow. It's late and I need to get up early for work.
     
  25. Mar 11, 2008 #24
    Hello e(ho0n3,

    One of the zero's is inside the unit circle. However the integral you rewrote, is not correct, a factor b has been forgotten. So, let's start from the beginning. Because you haven't studied residues yet I will give the complete solution using complex analysis, however it would be nice if you studied the other methods as well. Using the substitution:

    [tex]z=e^{i\theta}[/tex]

    the integral becomes:

    [tex]\int_{0}^{2\pi}\frac{d\theta}{a+b\cdot cos(\theta)}= \int_{C}\frac{dz}{iz[a+\frac{b}{2}(z+\frac{1}{z})]} = \frac{2}{ib}\int_{C}\frac{dz}{z^2+2\frac{a}{b}z+1}[/tex]

    The zero's of the denominator:

    [tex]z^2+2\frac{a}{b}z+1=(z-z_1)(z-z_2)[/tex]

    With:

    [tex]z_{1,2}=-\frac{a}{b}\left[1 \pm \sqrt{1-\left(\frac{b}{a}\right)^2}\right][/tex]

    Now, you have the following because it was stated in the original post that [tex]a>b>0[/tex], so [tex]\frac{b}{a}<1[/tex] thus [tex]\left(\frac{b}{a}\right)^2<1[/tex] thus [tex]1-\left(\frac{b}{a}\right)^2<1[/tex] thus [tex]\sqrt{1-\left(\frac{b}{a}\right)^2}<1[/tex] thus:

    [tex]1+\sqrt{1-\left(\frac{b}{a}\right)^2}>1[/tex]
    [tex]1-\sqrt{1-\left(\frac{b}{a}\right)^2}<1[/tex]

    and:

    [tex]-\frac{a}{b}\left[1+\sqrt{\left(\frac{b}{a}\right)^2}\right]<-1[/tex]
    [tex]-\frac{a}{b}\left[1-\sqrt{\left(\frac{b}{a}\right)^2}\right]>-1[/tex]

    From which you can see that only the second one is inside the unit circle. Now the residue theorem states that the integral around a closed loop is the sum of the residues of the poles inside the contour times [tex]2\pi i[/tex]. The residue of the pole inside the unit circle is calculated as (you wil see this after you studied Cauchy in this regard):

    [tex]\lim_{z\rightarrow z_2}\frac{z-z_2}{(z-z_1)(z-z_2)}=\frac{1}{z_2-z_1}[/tex]

    Therefore the integral is:

    [tex]I=2\pi i \frac{2}{ib} \frac{1}{z_2-z_1}=\frac{2\pi}{\sqrt{a^2-b^2}}[/tex]

    So, it is normally not the intention to give results like this, however if one reads through the complete post from the beginning it is understandable what I do here. Please come back to it and look at it again after you have studied residues in class. And moreover it is solvable in the reals for which two methods were given, look at these before going to an exam :-)
     
  26. Mar 11, 2008 #25
    Sweet. Thanks a lot for working it out in detail. I will definitely be reviewing this problem when I learn the residue theorem.
     
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