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How to integrate the delta function of complex variable?

  1. Dec 11, 2011 #1
    It is easy to integrate the delta function of real variable. But when the argument of the delta function is complex, I get stuck. For example, how to calculate the integral below, where u is a complex constant:
    \int_{ - \infty }^{ + \infty } {f\left( x \right)\delta \left( {ux} \right){\text{d}}x} = ?
  2. jcsd
  3. Dec 11, 2011 #2


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    Try y = ux.
  4. Dec 11, 2011 #3
    This method is only valid in the case when u is real.
  5. Dec 12, 2011 #4


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    In the present example, one could require that the complex delta function shares the following property with the usual one:

    [tex]\delta(ux) = \frac{1}{|u|}\delta(x),[/tex]
    where |u| is now interpreted as the complex modulus.

    Otherwise, if you had an example such as

    [tex]\int dx~f(x) \delta(x-w),[/tex]
    where w is a complex number, then I would think one would need to define the delta function as

    [tex]\oint_C dz~f(z)\delta(z-w),[/tex]
    where z is complex, and the integral is equal to f(w) if w is a point enclosed by C, and zero otherwise. This is basically like requiring that w be a simple pole and then evaluating the integral via the residue theorem. In fact, recall that

    [tex]\frac{1}{2\pi i}\oint_C dz~\frac{f(z)}{z-w} = f(w).[/tex]

    So, one might define a contour delta function by

    [tex]\delta(z-w) = \frac{1}{2\pi i} \frac{1}{z-w},[/tex]
    where the integral is to be taken around some contour.
  6. Dec 13, 2011 #5
    Thanks for Mute's answer. But I think there are two flaws in the definition of the contour delta function. First, this definition is only valid for the analytic function, namely it cannot apply to function such as f(z)=Re(z). Second, the way it selects the marked point differs from the usual definition of the delta function, namely the point selected by the contour delta function is enclosed by the path C rather than lying in C.

    After a few days of thinking, I come up with an idea to generalize the delta function to the complex plane. Recall that the delta function can be expressed as
    \delta \left( z \right) = \frac{1}{2}\frac{{{\text{dsgn}}\left( z \right)}}{{{\text{d}}z}} = \frac{1}{2}\frac{{\text{d}}}{{{\text{d}}z}}\left( {\frac{z}{{\left| z \right|}}} \right)
    where z is real and sgn(z) is the sign function. We may generalize this expression to the complex plane to obtain the complex delta function. However, this delta function generally cannot select the marked point even the point lies in the path C, unless the path C is parallel to the x axis, viz
    \int_{w \in L}^{} {f\left( z \right)\delta \left( {z - w} \right){\text{d}}z} = f\left( w \right)
    where L is a path parallel to the x axis.

    With the definition of delta function I give above, the integral in my first post can be calculated like this:
    \int_{ - \infty }^{ + \infty } {f\left( x \right)\delta \left( {ux} \right){\text{d}}x} = \frac{1}{2}\int_{ - \infty }^{ + \infty } {\left[ {f\left( x \right)\frac{{\text{d}}}{{{\text{d}}\left( {ux} \right)}}\left( {\frac{{ux}}{{\left| {ux} \right|}}} \right)} \right]{\text{d}}x} = \frac{1}{{\left| u \right|}}\int_{ - \infty }^{ + \infty } {f\left( x \right)\delta \left( x \right){\text{d}}x} = \frac{f(0)}{{\left| u \right|}}
  7. Dec 13, 2011 #6


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    The contour could be chosen to skirt around the point w in a semi-circle, which is equivalent to the contour running through the point. In this case the delta function would have to be modified slightly because only half of the residue would be acquired by having the contour run through the point. Although the definition wouldn't work for non-analytic functions, it should be otherwise fine for meromorphic functions, so long as there are no poles at the point where the "delta function" pings.

    Why can you generalize the delta function to complex numbers in this way? The sgn(x) function is not analytic. As a result it seems like you end up being able to work only with points for which you can formally manipulate the delta function of real argument as though some numbers were complex.

    Which is the result I suggested at the start of my earlier post! ;) However, again, this is a result of formally manipulating the delta function of real argument and then assuming the result holds for u complex, with |u| suitably re-interpreted as the modulus of u. I don't think you've really defined a complex delta function.

    Since the delta function isn't really a function to begin with, one could always just try defining it by its desired properties:

    [tex]\oint_C dz f(z) \delta(z-w) = f(w),[/tex]
    where C is a contour that goes through the point w. I would guess one could show that most of the properties of the regular delta function continue to hold, though definitions such as derivatives of other generalized functions like the sgn function may not hold, or would need to be suitably formalized. This is probably what I would try if f(z) is not analytic.
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