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Math Help - Complex integral, 2 problems

  1. #1
    MHF Contributor arbolis's Avatar
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    Complex integral, 2 problems

    1)I must calculate \int _{|z|=1} \frac{\cos (e^{-z})}{z^2} dz. I'm not sure if I should see it as the real part of \int _{|z|=1} \frac{e^ {i (e^{-z})}}{z^2} dz.
    Anyway the problem is obviously when z=0.
    I get an infinite residue: Res(f,z=0)=\lim _{z \to 0} \frac{\cos (e^{-z})}{z}=+\infty.

    2)I'd like to get an example of a real integral where when I use complex analysis methods and I choose a contour of integration, I fall over a singularity on the real axis.

    Thanks in advance.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Notice that

    f(z)=\cos(e^z)  \implies f(0)=\cos(0)

    so f does not have a zero at zero.

    So to calculate the residue at 0 we have

    g(z)=\frac{\cos(e^z) }{z^2}

    So the residue is

    \displaystyle \lim_{z \to 0}\frac{d}{dz}z^2g(z)=\lim_{z\to 0}\frac{d}{dz}\cos(e^{z})=-\sin(1)
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by arbolis View Post
    2)I'd like to get an example of a real integral where when I use complex analysis methods and I choose a contour of integration, I fall over a singularity on the real axis.

    Thanks in advance.
    Try  \displaystyle \int_{-\infty}^\infty \frac{\sin x}{x}dx by integrating the complex function  \displaystyle f(z)=\frac{e^{iz}}{z} .
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    I have noticed that you have been calculating a lot of residues this help.

    Suppose you have a complex function f(z) that is holomorphic except for a finite number of poles. Now if a is a pole of order k then f(z) has a Laurent expansion at a that looks like

    \displaystyle f(z)=\sum_{n=-k}^{\infty}a_{n}(z-a)^{n}
    Then

    g(z)=(z-a)^{k}f(z)=\sum_{n=-k}^{\infty}a_{n}(z-a)^{n+k}
    The residue is the coefficient at a_{-1} to recover this from g(z) we take take k-1 derivatives to get

    \displaystyle  \frac{d^{k-1}}{dz^{k-1}}g(z)=(k-1)!\sum_{n=-1}^{\infty}a_{n}(z-a)^{n+1}

    This gives that

    \displaystyle  \frac{d^{k-1}}{dz^{k-1}}g(a)=\lim_{z \to a}(k-1)!\sum_{n=-1}^{\infty}a_{n}(a-a)^{n+1}=(k-1)!a_{-1}

    This gives that

    \text{Res}(f,z=a)=\lim_{z\to a}\frac{1}{(k-1)!}\frac{d^{k-1}}{dz^{k-1}}(z-a)^{k}f(z)
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  5. #5
    MHF Contributor arbolis's Avatar
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    Thanks chip, I'll try it.
    Thanks TheEmptySet, that was very useful. However the function f(z) should be \cos (e^{-z}) but it doens't change the fact that f(0) \neq 0. Also, how did you know that g(z) had a pole of order 2? Is it because it gave me infinity when I thought it was of order 1?
    Correct me if I'm wrong in the following: If a function f has a singularity at z_0 and if \lim _{z \to z_0} (z-z_0) f(z) =\pm \infty then z_0 is either a pole of order greater than 1 or an essential singularity. On the other hand, if \lim _{z \to z_0} (z-z_0) f(z) =0 then f has a removable singularity at z_0. While if \lim _{z \to z_0} (z-z_0) f(z) =k with k \in \mathbb{C} and k \neq 0, then f has a pole of order 1 at z_0.
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  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    For this problem we had a function of the form

     \displaystyle \frac{\cos(e^{-z})}{z^2} the numerator does not have a zero at z=0 and is analytic so the constant term of its power series must be non zero and the denominator has a zero or order 2. The first term in its Laurent series must look like  \frac{a_0}{z^2}So it has a pole of order 2.
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