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Contour integration: why is this integral non-zero

  1. Apr 24, 2012 #1
    Just out of curiosity, why doesn't

    [tex]\int_{1}^{\infty} dx \sqrt{x-1} \mbox{ }e^{-x^4} [/tex]

    vanish by contour integration?

    The contour starts just above the real x-axis at x=1, goes rightward to x=infinity, then counter-clockwise from 0 to 2pi, then it goes leftward back to x=1.

    Since the integrand has no poles, the entire integral should be zero. Along the real axis, the leftward and rightward contributions are equal. At infinity the contributions are zero. So it seems that the entire integral should be zero.

    This is absurd of course, but I can't figure out where I went wrong in my logic.
     
  2. jcsd
  3. Apr 24, 2012 #2
    ....
     
  4. Apr 24, 2012 #3
    No. I get:

    [tex]\int_1^{\infty} f+\int_R f+\int_{\infty}^1 (-f)+\int_{\rho}=0[/tex]

    It's zero around the small indentation around the origin. so:

    [tex]2\int_{1}^{\infty} f dz=-\int_{R} f dz[/tex]

    And since the integral is a positive real number, the limit of the integral over the large circle must be the negative of twice that number.
     
  5. Apr 24, 2012 #4


    What "large circle" and what "little indentation around the origin"? I can't see any in the OP's original post...

    In fact, the integral is defined only for [itex] x\geq 1[/itex] , not even close to the origin.

    DonAntonio
     
  6. Apr 24, 2012 #5
    I've attached a picture of the contour that I'm talking about.

    Apologies for not being clear.

    Shouldn't the integral along the contour of the large circle give a zero value, since the exponential term makes the integrand zero around the large circle?
     

    Attached Files:

  7. Apr 24, 2012 #6


    Oh, I see now! Well, you know the square root is a double-branched function, and when you go around on that circle, its value

    jumps, so it is not that "obvious" what goes on there.

    Of course, there's a simple reason why your integral can't be zero: the function in the integral is non zero in

    a whole non-trivial interval in [itex][1,\infty)[/itex], since the function is always positive in [itex](1,\infty)[/itex] ...

    DonAntonio
     
  8. Apr 25, 2012 #7
    I think that there is a mistake in the next equation :
    It should be :

    [tex]\int_1^{\infty} f+\int_R f+\int_{\infty}^1 f+\int_{\rho} f = 0[/tex]
    or
    [tex]\int_1^{\infty} f+\int_R f-\int_1^{\infty} f+\int_{\rho} f = 0[/tex]
    [tex]\int_R f-\int_{\rho} f = 0[/tex]
    So, the contour integration leads to zero and it is not in contradiction with [tex]\int_1^{\infty} f > 0[/tex]
    In fact, the chosen contour is not convenient to compute the integral.
     
  9. Apr 25, 2012 #8
    I don't think so Jacquelin. It's negative along the contour beneath the real axis and goes from infinity to 1 so it's correct as I've written it and although we cannot check that numerically because of the large exponent, we can check the formula with an equivalent one with smaller exponent:
    [tex]\int_{c} \sqrt{z-1} e^{-z}dz[/tex]
    which agrees with the formula I stated.

    Also, sorry in my confusion of my first post. The small contour was going around the point 1 and not zero.
     
    Last edited: Apr 25, 2012
  10. Apr 25, 2012 #9
    Thanks everyone. It turns out, as some have noted, that the integral over the large circle is not zero.

    I chose the argument to be -z^4 because I thought no matter if you insert +i∞ or -i∞, it'll become -∞^4, which makes the exponential vanish. But there are other directions between those two directions where the integral does not vanish (and even blows up).
     
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