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Improper integrals: singularity on REAL axis (complex variab

  1. Feb 6, 2015 #1
    Hello everyone! I'm having some troubles when I try to solve improper integrals exercises that have singularities on the real axis. I have made a lot of exercises where singularities are inside a semicircle in the upper half side, but I don't know how to solve them when the singularities are on the real axis.
    I read some books but I think they are not very good explained (at least, I can't understood them).

    This is the exercise:
    [tex]\int_{-\infty}^{\infty} \frac{cos(2\pi x)}{x^2-1} dx[/tex]

    Using complex variable, I have:
    [tex]f(z) =\frac{exp(i2\pi z)}{z^2-1}[/tex]
    so there are 2 singularities:
    [tex]z_1 = -1[/tex] and [tex]z_2 = 1[/tex]
    I use a curve [tex]C[/tex] that is holomorphic inside it, because both singularities are out of it. Of course, I can divide [tex]C[/tex] in 6 curves: [tex]C_R[/tex] that is the "roof" of the curve and, using Jordan's Lemma, I can prove that
    [tex]\int_{C_R}^{ } \frac{exp(i2\pi z)}{z^2-1} dx = 0[/tex]
    but I don't know what do I have to do now. I saw in some places they said that the Residue Theorem over the semicircle around the singularities was something like [tex]-i\pi\sum{}{}Res[f(z), z_k][/tex] but I didn't understand why.

    I hope you can help me, because I don't know what can I do.
    Thanks!!
     
  2. jcsd
  3. Feb 7, 2015 #2

    Svein

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    Science Advisor

    This situation is covered in detail in Ahlfors' excellent book "Complex Analysis". It goes somewhat like this:

    Let [itex]C[/itex] be the curve consisting of a half circle in the upper half plane with center in 0 (radius R) and the real axis - except for two half circles in the lower half plane around -1 and +1 with radius ε. Now let us consider what happens around z = 1 (the case around z = -1 is handled the same way).

    For ε<0.1 we can write [itex]f(z)=\frac{e^{2\pi i z}}{z^{2}-1}=\frac{0.5}{z-1}+A(z)[/itex], where [itex]A(z)[/itex] is analytic at z = 1. Now we have [itex]\int_{0}^{R}f(z)dz =\int_{0}^{1-\epsilon}f(z)dz + \int_{-\pi}^{0}0.5\cdot id\phi + \int_{1+\epsilon}^{R}f(z)dz[/itex]. Let ε→0, and we end up with [itex]\int_{0}^{R}f(z)dz +i \frac{\pi}{2}[/itex]. Do the same with z = -1, and you end up with the formula you "didn't understand".
     
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