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Homework Help: Calculating an integral threw residium question

  1. Jan 14, 2010 #1
    i need to calculate this integral [tex]f(z)=\frac{z}{e^{2\pi iz^2}-1}\\[/tex]

    in this area
    [tex]\gamma _r=\left \{ |z|=r \right \},r>2[/tex]

    i need to find the points which turn to zero in the denominator
    and non zero in the numerator.
    i got two such points
    [tex]z=\pm \sqrt{n}[/tex]
    by using this formula
    [tex]res(\sqrt{a})=\frac{p(a)}{q(a)'}[/tex]
    [tex]res(\sqrt{n})=\frac{1}{4\pi i}[/tex]
    [tex]res(-\sqrt{n})=\frac{1}{4\pi i}[/tex]

    the third point is z=0 but for it we have both numerator and denominator 0
    i calculated the residium for it by [tex]res(f(x),a)=\lim_{x->a}(f(x)(x-a))[/tex] formula
    but then
    my prof says some stuff that involves the area
    he says that my points are 0 +1 -1 +2^(0.5) -2^(0.5) etc.. because the denominator goes to zero
    for each point have a residiu and i need to sum the residiums inside.
    but here the area is not defined
    its not like (by radius 3)

    i dont know what point are inside the area

    ??
     
  2. jcsd
  3. Jan 16, 2010 #2
    anyone?>?>
     
  4. Jan 16, 2010 #3
    sorry there is a mistake
    the area is
    [tex]
    \gamma _r=\left \{ |z|=r \right \},n<r^2<n+1
    [/tex]

    and the integral is from plus to minus infinity
     
    Last edited: Jan 16, 2010
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