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Math Help - Residue Theorem

  1. #1
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    Residue Theorem

    Calculate the following integral using Residue Theorem

    \int \frac{dz}{(1+z^2)^2}

    from \infty to -\infty
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  2. #2
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    Hello,

    Let f(z)=\frac{1}{(1+z^2)^2}

    Obviously, i and -i are double poles, because 1+z^2=(z+i)(z-i)

    So, using the general formula for a pole of order n : \text{Res}_a f(z)=\frac{1}{(n-1)!} \lim_{z\to a}\frac{\partial^{n-1}}{\partial z^{n-1}} (z-a)^n f(z)
    we easily get the residue of f at i (the one that will be useful for the computation) :
    \text{Res}_{z=i} f(z)=\lim_{z\to i} \frac{d}{dz} (z-i)^2 f(z)=\lim_{z\to i} \frac{d}{dz} \frac{1}{(z+i)^2}
    which is easy to compute.

    Spoiler:
    the residue is -\frac i4




    Now, use the semicircle contour :


    And keep the poles with positive imaginary part. So only i.

    And then \int_{-\infty}^\infty f(z) ~dz=2i\pi \text{Res}_{z=i} f(z)
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  3. #3
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    ok. so for

    Res(f,i) = \lim_{z\to i}  \frac{d}{dz}(z-i)^2f(z)
    = \lim_{z\to i}  \frac{d}{dz} \frac {1}{z+i}^2
    =\lim_{z\to i}  \frac{-(2z+2i)}{(z+i)^4}
    = \frac {-4i}{16i^4}<br />
= \frac{i}{4}

    and

    Res(f,-i) = \lim_{z\to -i}  \frac{d}{dz}(z+i)^2f(z)
    = \lim_{z\to -i}  \frac{d}{dz} \frac {1}{z-i}^2
    =\lim_{z\to -i}  \frac{-(2z-2i)}{(z-i)^4}
    = \frac {4i}{16i^4}<br />
= \frac {-i}{4}

    Am I right to say that you took semi contour and the positive side of the imaginary because of the -infinity to infinity?

    The last part would be

    \int_{-\infty}^\infty f(z) ~dz=2\pi i Res(f,i)
    = 2 \pi i * \frac {i}{4}  = \frac {-\pi}{2}

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  4. #4
    Moo
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    Hey...

    i^4=(i^2)^2=(-1)^2=1

    So = \frac {-4i}{16i^4}<br />
= {\color{red}-}\frac{i}{4}<br />

    You got wrong for the second one too.


    Isn't it surprising to get a negative answer for the integral of a positive function ?

    Am I right to say that you took semi contour and the positive side of the imaginary because of the -infinity to infinity?
    Yes. If we had the integral from infinity to -infinity, it would have been more logic to take the semi circle and the negative side of the imaginary.
    But the result is actually the same since \int_{-\infty}^\infty =-\int_\infty^{-\infty}

    But more commonly, we take the upper semi circle


    I hope I'm clear :s
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  5. #5
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    Ok. thanks for the clearance..

    Yeap. I got a careless calculation error there, thanks for pointing that out.
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