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Homework Help: Complex analysis

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data

    I=∫_2^∞ (1/(x(x-2)^.5)) dx

    using the calculus of residues. be sure to choose an appropriate contour and to explain what happens on each part of that contour.

    2. Relevant equations

    transform to a complex integral

    I= ∮ (1/(z(z-2)^.5)) dz

    3. The attempt at a solution

    i don't even know where to really begin. i keep ending up dividing by zero
  2. jcsd
  3. Dec 8, 2011 #2
    First make sure you have it clearly expressed as in:


    So how about use a key-hole contour that begins above the real-axis at x=2, travels down it to infinity, loops all the way round to right below the real axis at infinity, travles down the axis to 2 again, then loops around 2. That's a closed contour and I'll write the contour integral around that as:


    Now, can you analyze the integral over each leg of that contour (I count 4 distinct legs, two horizontal ones, that big circular one, and that real small one around 2). Keep in mind that the horizontal legs are over a branch-cut of the square root function so on top, it's one value of the branch say [itex]\sqrt{z-2}[/itex] and on bottom, it's the other value, [itex]-\sqrt{z-2}[/itex].
    Last edited: Dec 8, 2011
  4. Dec 9, 2011 #3
    ok i did so, so it the contour breaks up into for parts, Jo + JI + Jtop+ Jbottom where Jo means the outer part of the contour and JI is the inner part to the contour. Jo=JI=0. now i just do the calculus of residues on the Jtop and Jbottom, whic are

    [itex]\int\frac{1}{x\sqrt{x-2}} dx[/itex]


    [itex]\int\frac{-1}{x\sqrt{x-2}} dx[/itex]

    this since the bounds are flipped, the negative sign can be cancelled and the bounds are the same allowing me to add them,

    2[itex]\int\frac{1}{x\sqrt{x-2}} dx[/itex]

    i know that there is only one sing. in my contour which is z= 0. so

    2[itex]\int\frac{1}{x\sqrt{x-2}} dx[/itex]=2*pi*i Ʃ residues

    residue= lim[itex]z\rightarrow 0[/itex] [itex]\frac{2}{\sqrt{z^{2}-2}}[/itex]
  5. Dec 9, 2011 #4
    thanks btw
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