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Math Help - Help integrating complex function!

  1. #1
    Junior Member platinumpimp68plus1's Avatar
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    Help integrating complex function!

    I'm stuck... its been awhile since I've done integrals.

    So the integral is \int\frac{dz}{z} along the straight line segment from 1 to i. I parameterized the curve as \gamma:[0,1]\rightarrow C, defined as \gamma(t)=(-t+1)+it and then used the property that dividing 1 by z is the equivalent of z^{-1}. So I have (with the constant (-1+i) omitted):

    )\int(\frac{-t+1}{(-t+1)^2+t^2})-i\frac{t}{(-t+1)^2+t^2}dt\\=\int(\frac{-t+1}{(-t+1)^2+t^2})dt-i\int\frac{t}{(-t+1)^2+t^2}dt

    I have no clue how to integrate these... is there some trick I'm forgetting? any help?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by platinumpimp68plus1 View Post
    I'm stuck... its been awhile since I've done integrals.

    So the integral is \int\frac{dz}{z} along the straight line segment from 1 to i. I parameterized the curve as \gamma:[0,1]\rightarrow C, defined as \gamma(t)=(-t+1)+it and then used the property that dividing 1 by z is the equivalent of z^{-1}. So I have:

    \int(\frac{-t+1}{(-t+1)^2+t^2})-i\frac{t}{(-t+1)^2+t^2}dt\\=\int(\frac{-t+1}{(-t+1)^2+t^2})dt-i\int\frac{t}{(-t+1)^2+t^2}dt

    I have no clue how to integrate these... is there some trick I'm forgetting? any help?
    \int\frac{-t+1}{(-t+1)^2+t^2}dt=\int\frac{-t+1}{t^2-2t+1+t^2}dt=\int\frac{-t+1}{2t^2-2t+1}dt. Let z=2t^2-2t+1\implies dz=2t-2..
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  3. #3
    Junior Member platinumpimp68plus1's Avatar
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    Quote Originally Posted by Drexel28 View Post
    \int\frac{-t+1}{(-t+1)^2+t^2}dt=\int\frac{-t+1}{t^2-2t+1+t^2}dt=\int\frac{-t+1}{2t^2-2t+1}dt. Let z=2t^2-2t+1\implies dz=2t-2..
    I got kind of excited until I noticed its actually dz=4t-2... but that lead me on this train of thought:

    (Integrating the real term):

    z=2t^2-2t+1 \implies dz=4t-2
    -t+1= \frac{-1}{4} (4t-4)=\frac{-1}{4} (4t-2-2)=\frac{-1}{4} (4t-2)+\frac{1}{2}

    So:

    =\frac{-1}{4}\int\frac{4t-2}{2t^2-2t+1}dt+\frac{1}{2}\int<br />
\frac{1}{2t^2-2t+1}dt

    Which should be...

    =-\frac{1}{4}\ln(2t^2-2t+1)+\frac{1}{2}\tan^-1(2x-1)

    yay/nay?
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