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Math Help - Though contour integral

  1. #1
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    Though contour integral

    I'm trying to crack this problem but I can't seem to get it right.


    It's the third exercise from from chapter 2 of "Complex Analysis" by Stein & Shakarchi.

    I've drawn the sector (contour C) here (sorry for the crappy MS-paint picture):



    As there are no poles within the sector I set \oint_{C}f(z)dz = 0, so that leaves
    0 = \int_{0}^{A}e^{-Ax}dx + \int_{0}^{w}iAe^{-A^2e^{i\theta}}e^{i\theta}d\theta + \int_{0}^{A}e^{iw}e^{-Ate^{iw}}dt

    Now, I assume one of these three components goes to zero as A goes to infinity and indeed the second one (with \theta as variable) does. Then I'm stuck because what I find \int_{0}^{A = x = \infty}e^{-Ax}dx = 1/A doesn't lead me to anything remotely like the answer Wolfram's online calculator gave me if I use integration by parts


    Can anyone help me with this problem, what am I doing wrong?
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  2. #2
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    I'm not familiar with complex analysis, so this probably won't be of help, but both can be done by using by-parts (twice).
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  3. #3
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    I know, but I don't think I'd get away with that if this problem was in my exam.
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  4. #4
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    Let  f(z) = e^{-Az }~~,~~ A = \sqrt{a^2 + b^2 }



    Consider the path  C as you have drawn , since  f(z) is analytic everywhere , we have  \int_C e^{-Az}~dz = 0


     \int_{C_1} e^{-Az} ~dz +  \int_{C_2} e^{-Az} ~dz + \int_{C_3} e^{-Az} ~dz = 0


     C_1 is the straight line starting from  O to  R  ~( R + 0i )

     C_2 is the arc  R \to R e^{i \omega} ~~ \omega = \angle( a+ib)

     C_3 is a straight line  R e^{i \omega} \to O


    Then we have  \int_{C_1} e^{-Az} ~dz = \int_0^R e^{-Ax}~dx = \frac{1}{A}

     0 < |\int_{C_2} e^{-Az}~dz|  <  \int_{C_2} e^{-A|z|}~|dz| = \int_{C_2} e^{-AR} ~|dz| \to 0 ~~ \text{as}~ R \to \infty

     \int_{C_2} e^{-Az} ~dz  = 0


      \int_{C_3} e^{-Az} ~dz

    Sub  z = x e^{i \omega}

    we have

     -\int_0^R e^{-Ax  e^{i \omega} } ~  e^{i \omega} ~dx


    but   A e^{i \omega}  = a + bi ~ \implies  e^{i \omega} = \frac{a+bi}{\sqrt{a^2+b^2} }

    so
       \int_{C_3} e^{-Az} ~dz = -\frac{a+bi}{\sqrt{a^2+b^2} } \int_0^R e^{-ax - ibx} ~dx

    By joining the three integrals we obtain

      \frac{1}{A}  + 0   -\frac{a+bi}{\sqrt{a^2+b^2} } \int_0^R e^{-ax - ibx} ~dx = 0

     \frac{1}{a+bi} = \int_0^{\infty} e^{-ax} ( \cos(bx) - i \sin(bx) )~dx = \frac{a-bi}{a^2+b^2 }


    By comparing the real part and the Im part , we have

      \int_0^{\infty} e^{-ax} \cos(bx) ~dx = \frac{a}{a^2+b^2}

      \int_0^{\infty} e^{-ax} \sin(bx) ~dx = \frac{b}{a^2+b^2}
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  5. #5
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    Thank you very much, the A e^{i \omega} = a + bi ~ \implies e^{i \omega} = \frac{a+bi}{\sqrt{a^2+b^2} } part was the missing link. Every contour integral needs its own set of tricks it seems, very annoying, but at least I can have some peace of mind now.
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