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Math Help - Integral

  1. #1
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    Integral

    Hey guys.
    Any idea about how to solve this integral?

    10x.
    Attached Thumbnails Attached Thumbnails Integral-1.jpg  
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  2. #2
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    Quote Originally Posted by asi123 View Post
    Hey guys.
    Any idea about how to solve this integral?

    10x.
    Hello, this can be solved analytically letting \theta=2\arctan(\phi)


    NOTE Take what I am about to say with extreme caution...I am a novice at this and just learning so wait for confirmation!

    I think this integral can also be solved using Residues by considering the function

    f(z)=\frac{i}{(z-a)(az-1)} and taking

    \int_{|z|=1}f(z)~dz
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  3. #3
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    Well, I think you are right with the second idea because we exactly learning how to work with Cauchy's integral formula so it's got to be it
    My question due is how can you break it into f(z)=\frac{i}{(z-a)(az-1)}?

    Thanks a lot.
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    Quote Originally Posted by Mathstud28 View Post
    [snip]
    NOTE Take what I am about to say with extreme caution...I am a novice at this and just learning so wait for confirmation!

    I think this integral can also be solved using Residues by considering the function

    f(z)=\frac{i}{(z-a)(az-1)} and taking

    \int_{|z|=1}f(z)~dz
    The integral can be solved this way. Details available upon request.
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  5. #5
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    Quote Originally Posted by asi123 View Post
    Well, I think you are right with the second idea because we exactly learning how to work with Cauchy's integral formula so it's got to be it
    My question due is how can you break it into f(z)=\frac{i}{(z-a)(az-1)}?

    Thanks a lot.
    Substitute z = e^{it}:

    1. dt = \frac{dz}{iz} = \frac{-i \, dz}{z}.

    2. \cos t = \frac{e^{it} + e^{-it}}{2} = \frac{z + \frac{1}{z}}{2} = \frac{z^2 + 1}{2z}.

    Simplify. The integrand becomes \frac{i}{az^2 - z - a^2 z + a} which factorises in the way Mathstud gave.
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    Thanks a lot guys.
    I have a question due, is this substitute z = e^{it} usually works?
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    Quote Originally Posted by asi123 View Post
    Thanks a lot guys.
    I have a question due, is this substitute z = e^{it} usually works?
    Integration around the contour |z| = 1, that is, making the substitution z = e^{it}, is typically used to evaluate a real integral of the form \int_0^{2 \pi} F(\cos t, \sin t) \, dt.
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  8. #8
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    I'm having a problem with the boundaries due.
    I'm getting an integral from 1 to 1, is this normal?
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  9. #9
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    Quote Originally Posted by asi123 View Post
    I'm having a problem with the boundaries due.
    I'm getting an integral from 1 to 1, is this normal?
    You're integrating around the unit circle. You need to consider singularites of f(z) that lie inside this circle and apply Cauchy's Integral Formula.
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    As I was saying, I'm completely new with this Cauchy's integral formula so just to make sure I uploaded my solution to this problem and marked the place where I got to yesterday thanks to you.
    Can someone please give me the OK?

    Again, thanks a lot.
    Attached Thumbnails Attached Thumbnails Integral-scan0003.jpg  
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  11. #11
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    Quote Originally Posted by asi123 View Post
    As I was saying, I'm completely new with this Cauchy's integral formula so just to make sure I uploaded my solution to this problem and marked the place where I got to yesterday thanks to you.
    Can someone please give me the OK?

    Again, thanks a lot.


    And you can check it by giving a an appropriate value and then comparing your answer with what a CAS gives you.

    You should now consider the case |a| > 1 (check your answer using the above mentioned way of checking).
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