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    Hello I can change the integral into a circle integral etc

    but don't know how to deal with the P?

    Any help thanks.



    Integral d(theta) / ( 1- 2pcos(theta) + p^2 ) with limits 2pi,0
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    No replies... damn what happened to TSR... I remember when I did A-Level maths people used to reply very very quickly...

    Guess all the maths geniuses have moved on from TSR
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    (Original post by CHEN20041)
    No replies... damn what happened to TSR... I remember when I did A-Level maths people used to reply very very quickly...

    Guess all the maths geniuses have moved on from TSR
    Probably would have helped if your original post wasn't missing the actual integral.

    Attitude now isn't helping either...
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    (Original post by DFranklin)
    Probably would have helped if your original post wasn't missing the actual integral.

    Attitude now isn't helping either...
    I attached an image and also entered a tinypic link, it doesn't seem to have gone through though. The integral was written but it wasn't written very clearly. I edited it now hopefully it is more clear.

    Sorry if it came across that way but that was not my intention, it was just an observation.

    Obviously I'm on a site where people give help to people like myself for free, I didn't mean to come of as self entitled, sorry about that.
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    (Original post by CHEN20041)
    Obviously I'm on a site where people give help to people like myself for free, I didn't mean to come of as self entitled, sorry about that.
    Fair enough - in a bit of a grump mood myself at the minute.
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    (Original post by CHEN20041)
    Hello I can change the integral into a circle integral etc

    but don't know how to deal with the P?

    Any help thanks.



    Integral d(theta) / ( 1- 2pcos(theta) + p^2 ) with limits 2pi,0
    Complete the square in the denominator to get (p-\cos \theta)^2 + \sin^2 \theta in the denominator
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    (Original post by Zacken)
    Complete the square in the denominator to get (p-\cos \theta)^2 + \sin^2 \theta in the denominator
    I assume he wants to do it by contour though.

    (Original post by CHEN20041)
    ..
    If z = e^{i\theta} then z + 1/z = 2 cos theta.
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    (Original post by CHEN20041)
    I attached an image and also entered a tinypic link, it doesn't seem to have gone through though. The integral was written but it wasn't written very clearly. I edited it now hopefully it is more clear.

    Sorry if it came across that way but that was not my intention, it was just an observation.

    Obviously I'm on a site where people give help to people like myself for free, I didn't mean to come of as self entitled, sorry about that.
    Because you are a new user, if you post links they will need to be approved first (to prevent spam etc.). I approved your post but I think you may have deleted the image.

    As you post more your posts won't need to approved. Best to use an original account instead of starting a new one...
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    (Original post by Notnek)
    Because you are a new user, if you post links they will need to be approved first (to prevent spam etc.). I approved your post but I think you may have deleted the image.

    As you post more your posts won't need to approved. Best to use an original account instead of starting a new one...
    Oh okay thanks.

    (Original post by Zacken)
    Complete the square in the denominator to get (p-\cos \theta)^2 + \sin^2 \theta in the denominator
    (Original post by DFranklin)
    I assume he wants to do it by contour though.

    If z = e^{i\theta} then z + 1/z = 2 cos theta.

    DM I get it now, thanks.
 
 
 
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