Results 1 to 8 of 8

Math Help - [SOLVED] Line integral, Cauchy's integral formula

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    [SOLVED] Line integral, Cauchy's integral formula

    The problem is the same as in this thread : http://www.mathhelpforum.com/math-he...l-formula.html, with the exception of the integrand : \frac{\cosh z}{z^4}.
    My attempt : I see the singularity at z=0.
    So \int _C \frac{\cosh z}{z^4}dz=2\pi i f(0) with f(z)=\cosh z.
    As \cosh 0=1, the integral is worth 2 \pi i.


    Edit : If the integrand is \frac{z}{2z+1}, I get the result -\pi i. Am I right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Quote Originally Posted by arbolis View Post
    T
    So \int _C \frac{\cosh z}{z^4}dz=2\pi i f(0) with f(z)=\cosh z.
    As \cosh 0=1, the integral is worth 2 \pi i.
    No because that's not in the form \oint \frac{f(z)}{z-a}. However it is in the form of \oint \frac{f(z)}{(z-a)^k} and there is an extention of Cauchy's Integral Formula which covers this case.


    If the integrand is \frac{z}{2z+1}, I get the result -\pi i. Am I right?
    That's not in the form \oint \frac{f(z)}{z-a}dz but you can write it as:

    \oint \frac{\frac{z}{2}}{z+1/2}dz
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by shawsend View Post
    No because that's not in the form \oint \frac{f(z)}{z-a}. However it is in the form of \oint \frac{f(z)}{(z-a)^k} and there is an extention of Cauchy's Integral Formula which covers this case.


    That's not in the form \oint \frac{f(z)}{z-a}dz but you can write it as:

    \oint \frac{\frac{z}{2}}{z+1/2}dz
    Ah thanks a lot. It was very useful information.

    EDIT: I'm searching for the extension of the CIF but without success. If you have a link, feel free to share.
    Thanks again.
    Last edited by arbolis; September 15th 2009 at 05:01 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Quote Originally Posted by arbolis View Post
    Ah thanks a lot. It was very useful information.

    EDIT: I'm searching for the extension of the CIF but without success. If you have a link, feel free to share.
    Thanks again.
    Arbolis, that's the next section after the introduction of Cauchy's Integral Formula. It's called "Differentiability of Cauchy-Type Integrals". It's in any text on Complex Analysis. My favorite is "Basic Complex Analysis" by Marsden and Hoffman.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Another attempt!

    Quote Originally Posted by shawsend View Post
    Arbolis, that's the next section after the introduction of Cauchy's Integral Formula. It's called "Differentiability of Cauchy-Type Integrals". It's in any text on Complex Analysis. My favorite is "Basic Complex Analysis" by Marsden and Hoffman.
    Ok thanks for sharing, I think I now know the formula.

    For \oint \frac{\frac{z}{2}}{z+1/2}dz, I get that it equals -\frac{ i \pi}{2}.


    Now comes \int _C \frac{\cosh z}{z^4}dz.
    The formula states that f^{n}(z)=\frac{n!}{2\pi i} \int _C \frac{f(\xi)}{(\xi -z)^{n+1}} d \xi. I notice that in the exercise z=0 and n=3.
    Thus I have f^{(3)}(0)=\frac{3!}{2 \pi i} \int _C \frac{f(\xi)}{\xi ^4} d\xi = \frac{3}{\pi i} \int _C \frac {\cosh z}{z^4} dz = \cosh (0) ^{(3)}= \sinh 0 =0... I don't think it's possible.
    What did I do wrong?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Quote Originally Posted by arbolis View Post
    Thus I have f^{(3)}(0)=\frac{3!}{2 \pi i} \int _C \frac{f(\xi)}{\xi ^4} d\xi = \frac{3}{\pi i} \int _C \frac {\cosh z}{z^4} dz = \cosh (0) ^{(3)}= \sinh 0 =0... I don't think it's possible.
    What did I do wrong?
    That's ill-formed: \cosh(0)^{(3)} is poor syntax. Rather, if f(z)=\cosh(z), then f^{(3)}(z)=\sinh(z) and therefore:

    \sinh(0)=\frac{3!}{2 \pi i} \int _C \frac{f(\xi)}{\xi ^4} d\xi and that implies the integral is zero.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by shawsend View Post
    That's ill-formed: \cosh(0)^{(3)} is poor syntax. Rather, if f(z)=\cosh(z), then f^{(3)}(z)=\sinh(z) and therefore:

    \sinh(0)=\frac{3!}{2 \pi i} \int _C \frac{f(\xi)}{\xi ^4} d\xi and that implies the integral is zero.
    Thanks for the reply and sorry for the syntax. So my result was right?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Quote Originally Posted by arbolis View Post
    Thanks for the reply and sorry for the syntax. So my result was right?
    Yes. It's very easy to integrate it numerically in Mathematica:

    Code:
    NIntegrate[(Cosh[z]/z^4)*I*Exp[I*t] /. 
       z -> Exp[I*t], {t, 0, 2*Pi}]
    And this gives zero.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cauchy's Integral Formula
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: May 11th 2011, 11:54 AM
  2. cauchy integral formula
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: June 2nd 2010, 11:00 PM
  3. [SOLVED] Application of Cauchy's Integral Formula
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: February 16th 2010, 06:33 PM
  4. [SOLVED] Special Case of Cauchy's Integral Formula
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: February 16th 2010, 06:01 PM
  5. [SOLVED] Line integral, Cauchy's integral formula
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 15th 2009, 02:28 PM

Search Tags


/mathhelpforum @mathhelpforum