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Thread: [SOLVED] Line integral, Cauchy's integral formula

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Line integral, Cauchy's integral formula

    Let $\displaystyle C$ be the frontier of the square whose sides are the lines $\displaystyle x=\pm 2$ and $\displaystyle y= \pm 2$. (orientation is positive)
    Calculate the following integral using Cauchy's integral formula for a function or its derivative.

    $\displaystyle \int _C \frac{e^{-z}dz}{z-\frac{i\pi}{2}}$.

    My attempt : I find confusing the "using Cauchy's integral formula for a function or its derivative.", I don't really know what they mean by that.


    So Cauchy's formula is $\displaystyle f(z)=\frac{1}{2\pi i} \int _C \frac{f(\xi)}{\xi -z}d \xi$.

    Looking at the integrand of the integral, I see that it is not defined for $\displaystyle z=i \frac{\pi}{2}$. Does that mean that $\displaystyle f$ is not analytic in this point?
    I believe the integral should equals $\displaystyle 2 \pi i$, I don't really know why.

    I need help, I find all this new theory somewhat hard to grasp due to the amount of it.
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  2. #2
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    Cauchy's Integral Formula states that if $\displaystyle f(z)$ is analytic in $\displaystyle C$ and the point $\displaystyle z$ is not on C, then:

    $\displaystyle f(z)=\frac{1}{2\pi i} \int _C \frac{f(\xi)}{\xi -z}d \xi
    $

    Ok, $\displaystyle f(z)=e^{-z}$ and that's analytic in C. Don't confuse the ENTIRE integrand with the expression for $\displaystyle f(z)$ when using this formula.

    So plugging in $\displaystyle z=\pi i/2$, then the integral is just:

    $\displaystyle 2\pi i f(\pi i/2)$ right?

    Try and get good at calculating these in Mathematica so that you can check your results. For example (where I used an equivalent circle of radius 2 via deformation of path theorem):

    Code:
    In[111]:=
    NIntegrate[(Exp[-z]/(z - Pi*(I/2)))*2*I*
        Exp[I*t] /. z -> 2*Exp[I*t], 
      {t, 0, 2*Pi}]
    N[2*Pi*I*Exp[(-Pi)*(I/2)]]
    
    Out[111]=
    6.283185307179801 - 
      2.708944180085382*^-14*I
    
    Out[112]=
    6.283185307179586
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks for the reply, however I'm extremely confused.
    Quote Originally Posted by shawsend View Post
    Cauchy's Integral Formula states that if $\displaystyle f(z)$ is analytic in $\displaystyle C$ and the point $\displaystyle z$ is not on C, then:

    $\displaystyle f(z)=\frac{1}{2\pi i} \int _C \frac{f(\xi)}{\xi -z}d \xi
    $
    but $\displaystyle z$ has to be in the interior of $\displaystyle C$, right?
    If so, then in the exercise I can consider $\displaystyle z$ as $\displaystyle i \frac{\pi}{2}$ and I notice that indeed $\displaystyle z$ is inside the square $\displaystyle C$.
    Also z is a pole I think. I don't know if it's important.


    Ok, $\displaystyle f(z)=e^{-z}$ and that's analytic in C. Don't confuse the ENTIRE integrand with the expression for $\displaystyle f(z)$ when using this formula.

    So plugging in $\displaystyle z=\pi i/2$, then the integral is just:

    $\displaystyle 2\pi i f(\pi i/2)$ right?
    Here I don't understand why you worked with the numerator of the integrand only. I know that it is analytic in $\displaystyle C$. However I think the denominator is not analytic at $\displaystyle z$, inside $\displaystyle C$, the square.


    Try and get good at calculating these in Mathematica so that you can check your results. For example (where I used an equivalent circle of radius 2 via deformation of path theorem):

    Code:
    In[111]:=
    NIntegrate[(Exp[-z]/(z - Pi*(I/2)))*2*I*
        Exp[I*t] /. z -> 2*Exp[I*t], 
      {t, 0, 2*Pi}]
    N[2*Pi*I*Exp[(-Pi)*(I/2)]]
    
    Out[111]=
    6.283185307179801 - 
      2.708944180085382*^-14*I
    
    Out[112]=
    6.283185307179586
    I never heard of such a theorem... looks interesting though.
    Before computing I'd like to understand what I'm doing. I know much less than you think.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Oh... I think I just understood what you meant :
    Thanks to CFI we have $\displaystyle 2 \pi i f(\frac{i \pi}{2})= \int _C \frac{e^{-z}dz}{z- \frac{i \pi}{2}}$.
    But $\displaystyle f(\frac{i\pi}{2})=e^{-\frac{i \pi}{2}}\Rightarrow \int _C \frac{e^{-z}dz}{z- \frac{i \pi}{2}}=2\pi i \cdot (-i)= 2\pi $.
    Fantastic, I get your result!
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