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

    \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 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 z=i \frac{\pi}{2}. Does that mean that f is not analytic in this point?
    I believe the integral should equals 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 f(z) is analytic in C and the point z is not on C, then:

    f(z)=\frac{1}{2\pi i} \int _C \frac{f(\xi)}{\xi -z}d \xi<br />

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

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

    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 f(z) is analytic in C and the point z is not on C, then:

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


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

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

    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 C. However I think the denominator is not analytic at z, inside 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 2 \pi i f(\frac{i \pi}{2})= \int _C \frac{e^{-z}dz}{z- \frac{i \pi}{2}}.
    But 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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