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Math Help - Help with integration (complex analysis)

  1. #1
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    Help with integration (complex analysis)

    I'd like to know what is wrong in the following steps:


    This will be 0, and I know for sure that this should be 2 PI i. What am I doing wrong?

    PS: I know I can use other techniques such as residues and such, but I want to understand what I'm doing wrong when trying to integrate by hand.
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  2. #2
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    Quote Originally Posted by devouredelysium View Post
    I'd like to know what is wrong in the following steps:


    This will be 0, and I know for sure that this should be 2 PI i. What am I doing wrong?

    PS: I know I can use other techniques such as residues and such, but I want to understand what I'm doing wrong when trying to integrate by hand.
    It would only be 0 if the limits were equal, and that's not very likely, is it?
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  3. #3
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    Quote Originally Posted by Mush View Post
    It would only be 0 if the limits were equal, and that's not very likely, is it?
    I mean, from 0 to 2PI.
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    Quote Originally Posted by devouredelysium View Post
    I mean, from 0 to 2PI.

    What is the COMMON value of e^{2\pi i}=e^0?

    Tonio


    \textit{Addition to the above}: Ok, so by the above you get 0, but in may all depend on what value of Log(z) you're choosing: remember that Log(z) is multivalued and the argument of its value is determined only up to a multiple of 2\pi i, so it may be this the reason why you think (or you were given or you read in the solutions sheet) that the value is 2\pi i. It can as well be any multiple of this, too.

    Tonio
    Last edited by tonio; October 28th 2009 at 07:28 AM.
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  5. #5
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    Quote Originally Posted by tonio View Post
    What is the COMMON value of e^{2\pi i}=e^0?

    Tonio
    From my understanding, e^i 2PI = e^0, or not?
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    Quote Originally Posted by devouredelysium View Post
    From my understanding, e^i 2PI = e^0, or not?
    Yes, and they're both equal to 1, and hence you're right that the integral is equal to 0. What makes you 'know for sure' that it's supposed to be 2 i pi?
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  7. #7
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    Quote Originally Posted by Mush View Post
    Yes, and they're both equal to 1, and hence you're right that the integral is equal to 0. What makes you 'know for sure' that it's supposed to be 2 i pi?
    Well, it is known in complex analysis that functions of that kind have 2 PI i as integral. If I try to directly calculate it in mathematica, it will give me 2 PI i, too.
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    Quote Originally Posted by devouredelysium View Post
    Well, it is known in complex analysis that functions of that kind have 2 PI i as integral. If I try to directly calculate it in mathematica, it will give me 2 PI i, too.

    It is "known"? What are "those" functions that have that value "as integral"? And what "integral", exactly?

    Can you be more specific and, perhaps, give some references to these claims?

    Tonio
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  9. #9
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    Quote Originally Posted by tonio View Post
    It is "known"? What are "those" functions that have that value "as integral"? And what "integral", exactly?

    Can you be more specific and, perhaps, give some references to these claims?

    Tonio
    Ok, that was it. Thanks!
    Last edited by devouredelysium; October 28th 2009 at 10:07 AM.
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  10. #10
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    Quote Originally Posted by devouredelysium View Post
    What I am trying to do is calculate the integral over a closed curve with radius 2, centered at the origin. The function which I am trying to integrate over there is 1/(z+1). So I defined g(t) = 2exp(it) from 0 to 2PI as the integration path and that gave the integrals I have shown before.
    Oh, I think I see now...sometimes posters change a question they're given and they don't add information relevant: of course, the complex integral you want indeed equals 2\pi i and this is due to the fact I added as \textit{addition} to my second post: after revolving around a circle around a zero of z+1, indeed the integral \oint\limits_{|z|=2}\frac{dz}{z+1}=2\pi i since the value of the primitive function Log(z+1) (complex logarithm, of course) adds 2\pi i to its argument!

    Tonio
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  11. #11
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    Quote Originally Posted by tonio View Post
    Oh, I think I see now...sometimes posters change a question they're given and they don't add information relevant: of course, the complex integral you want indeed equals 2\pi i and this is due to the fact I added as \textit{addition} to my second post: after revolving around a circle around a zero of z+1, indeed the integral \oint\limits_{|z|=2}\frac{dz}{z+1}=2\pi i since the value of the primitive function Log(z+1) (complex logarithm, of course) adds 2\pi i to its argument!

    Tonio
    That log argument thing was it. Thanks!
    Last edited by devouredelysium; October 28th 2009 at 10:08 AM.
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