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Thread: Complex integral, 2 problems

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

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

    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

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

    so f does not have a zero at zero.

    So to calculate the residue at 0 we have

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

    So the residue is

    $\displaystyle \displaystyle \lim_{z \to 0}\frac{d}{dz}z^2g(z)=\lim_{z\to 0}\frac{d}{dz}\cos(e^{z})=-\sin(1)$
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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 \displaystyle \int_{-\infty}^\infty \frac{\sin x}{x}dx $ by integrating the complex function $\displaystyle \displaystyle f(z)=\frac{e^{iz}}{z} $.
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    I have noticed that you have been calculating a lot of residues this help.

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

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

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

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

    This gives that

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

    This gives that

    $\displaystyle \text{Res}(f,z=a)=\lim_{z\to a}\frac{1}{(k-1)!}\frac{d^{k-1}}{dz^{k-1}}(z-a)^{k}f(z)$
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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 $\displaystyle \cos (e^{-z})$ but it doens't change the fact that $\displaystyle f(0) \neq 0$. 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 $\displaystyle z_0$ and if $\displaystyle \lim _{z \to z_0} (z-z_0) f(z) =\pm \infty$ then $\displaystyle z_0$ is either a pole of order greater than 1 or an essential singularity. On the other hand, if $\displaystyle \lim _{z \to z_0} (z-z_0) f(z) =0$ then f has a removable singularity at $\displaystyle z_0$. While if $\displaystyle \lim _{z \to z_0} (z-z_0) f(z) =k$ with $\displaystyle k \in \mathbb{C}$ and $\displaystyle k \neq 0$, then f has a pole of order 1 at $\displaystyle z_0$.
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  6. #6
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    For this problem we had a function of the form

    $\displaystyle \displaystyle \frac{\cos(e^{-z})}{z^2}$ 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 $\displaystyle \frac{a_0}{z^2}$So it has a pole of order 2.
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