Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Exponential integral

  1. Jul 31, 2010 #1
    Hi i was wondering if anyone could help me with a fairly simple integration.

    [tex]\int exp(iab/c)exp(-iaz) da[/tex]

    Where the integral is from 0 to infinity

    Im not sure how to proceed with this and am basically teaching myself
    Thanks in advance for any help.
     
  2. jcsd
  3. Jul 31, 2010 #2

    Pengwuino

    User Avatar
    Gold Member

    Do you have any conditions on b,c, and z? This integral is not going to be solvable as far as I can tell. Remember, [tex] e^{ix} = cos(x) + isin(x)[/tex] and you'll end up trying to take a limit at infinity which does not exist for cosines and sines.
     
    Last edited: Jul 31, 2010
  4. Jul 31, 2010 #3
    I say the answer is 0
    First consolidate the integrand to exp(iaK) where K = (b/c) - z, then expand the exponential to cos(aK) +i*sin(aK)
    then [tex]
    \int cos(aK) da + i\int sin(aK) da = 0
    [/tex]
    since integrating a sine or cosine gives zero, since there are equal positive and negative areas for every cycle.
     
  5. Jul 31, 2010 #4

    Pengwuino

    User Avatar
    Gold Member

    unless the limit is at infinity, which means the limit is not defined.
     
  6. Aug 1, 2010 #5
    In the wording of the problem, Skullmonkee forgot to specify if the coefficients b, c, z are real or complex.
    Suppose b, c real and z complex. If z=x+i y , with x , y real and y<0
    then, the integral (from 0 to infinity) is convergent = 1/(-y+i(x-b))
     
  7. Aug 1, 2010 #6
    The correct answer follows:

    Let [itex]k = i(b/c-z)[/itex]. Then the integral is the same as
    [tex]\int_0^\infty \exp(ka) \;da[/tex]
    which converges precisely when [itex]\operatorname{Re}(k) < 0[/itex]. In this case, the value of the integral is
    [tex]\frac{1}{k} \exp(ka) \biggr|_{a=0}^\infty = -\frac{1}{k} = \frac{i}{b/c-z}.[/tex]

    Pengwuino and ACPower assume that [itex]b/c - z[/itex] is real, but if that is not the case, the integrand may not be purely sinusoidal (that is, [itex]k[/itex] above may have nonzero real part). JJacquelin gives an incorrect value for the integral (there is a missing c).
     
    Last edited: Aug 1, 2010
  8. Aug 1, 2010 #7
    Adriank is right, I forgot c in my equation. Thanks for bringing the mistake to our attention.
    If b and c are real and z complex ( i.e.: z=x+i y , with x and y real), the condition of convergence is y<0 as already said.
    Of course, if all b, c, z are complex a more general condition has to be derived from :
    [Real part of i((b/c)-z)] < 0
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook