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Imaginary components of real integrals

  1. Dec 29, 2012 #1
    Why does the incomplete gamma function have an imaginary component, when the exponential integral does not?
    [tex]\Gamma(0,z,\infty)\equiv\int^\infty_z \frac{e^{-t}}t dt[/tex]
    [tex]Ei(z)\equiv-\int^\infty_{-z} \frac{e^{-t}}t dt[/tex]
    Looking at how these integrals are usually defined I would have expected them to be nearly identical, [itex]\Gamma(0,z,\infty)=-Ei(-z)[/itex], for all z. Instead, while that is satisified for positive real z, the general relation is more complicated and particularly for negative real values [itex]\Gamma(0,x,\infty)=-Ei(-x)-i\pi[/itex] (e.g., according to MathWorld). Why is this?

    I understand that the integrands have a singularity at the origin of the complex plane. I think I can see how integrating a contour around this will contribute ±2πi, or half that if only skipping from the positive to the negative side of the real axis. But I can't see why this would only apply when we label it the gamma function and not the exponential integral?

    Plotting with Alpha/Mathematica (for real arguments) seems to confirm that the real components of [itex]\Gamma(0,x,\infty),\Gamma(0,x),E_1(x)[/itex] and Ei(x) all behave symmetrically, but that the imaginary component of Ei lacks the step-function which the others all exhibit.

    Curiously, I was able to obtain the step-function in the imaginary component of Ei by subtracting a tiny imaginary component from its argument. Should I conclude that, by some quirk, Ei alone is just being defined in a completely separate manner along the real axis (perhaps, defaulting to Cauchy principal values rather than complex analysis as the method to bypass the singularity)?

    More generally, for an arbitrary integrand with (multiple) singularities, is there a convention which determines which is the principal branch (e.g., why does [itex]\Gamma(0,-2)[/itex] involve negative rather than positive iπ)?
  2. jcsd
  3. Dec 30, 2012 #2


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    gamma(0,x,infinity)+Ei(-x)=1/2 (log(-x)-log(-1/x))-log(x)

    So that explains your observation in terms of branch cuts and what not. Normally the branch cut is taken to be the negative real axis. The fact that we have introduced a negative into the argument explains the two function regard opposite sides as negative.
  4. Dec 30, 2012 #3
    Thanks, lurflurf!
    So.. two questions..

    If the branch cut is customarily at the negative real axis, which side does that axis itself fall on? (For example, if we were to trace a contour from -1 to +1, do we pass the origin on the south side of the complex plane? And if there wasn't just one singularity on the real line but say multiple singularities arranged as the vertices of a pentagon in the complex plane, would there still be any convention or would each author place the cuts where they please?)

    And secondly, how do you obtain the formula with the logs?
  5. Dec 30, 2012 #4


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    That is just a fancy way to write a function that is
    0 when Re(x)>0
    -pi i when Re(x)<0

    it arises because in your two functions one path passes the singularity on the right and the other passes on the left so they differ by pi i. The same thing happens for 1/t if we pass t=0 on different sides. For a pentagon we would connect the five points with 4 lines. Which of the five to leave could very. If your want to write out a function in terms of standard function for use with tables or software great care is needed. Conventions do not always agree for different tables and software.
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