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I Imaginary sinusoidal exponential

  1. May 1, 2016 #1
    I need to convince myself that

    ##e^{ikr\text{cos}\ \theta}=\frac{\text{sin}\ \theta}{kr}##.

    Could you please help me with it?
     
  2. jcsd
  3. May 1, 2016 #2

    blue_leaf77

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    Cannot be true, the left side is complex whereas the right side is real.
     
  4. May 1, 2016 #3
    Sorry, there was a typo.

    How about this one:

    ##\text{exp}({ikr\text{cos}\ \theta}) = \frac{\text{exp}(ikr)-\text{exp}(-ikr)}{2ikr}##?
     
  5. May 1, 2016 #4

    blue_leaf77

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    What do you know about the Euler formula for a complex exponential ##e^{i\phi}##?
     
  6. May 1, 2016 #5
    The Euler formula is ##e^{i\phi}= \text{cos}\ \phi + \text{sin}\ \phi##.

    I don't see how ##\text{exp}(ikr\text{cos}\ \theta) = \frac{\text{sin}(kr)}{kr}##.
     
  7. May 1, 2016 #6

    blue_leaf77

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    It's ##e^{i\phi}= \text{cos}\ \phi + i\text{sin}\ \phi##.
    Indeed, I too, do not see how that can be true.
     
  8. May 1, 2016 #7
    The following is taken from page 39 of Matthew Schwartz's 'Introduction to Quantum Field Theory.' My above posts refer to the third and fourth lines of the following calculation:

    Capture.jpg a

    Would you please have a look and let me know what I'm missing in the previous posts?
     
  9. May 1, 2016 #8

    blue_leaf77

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    $$\int_{-1}^1 d(\cos \theta) e^{ikr\cos \theta} = \frac{1}{ikr}e^{ikr\cos \theta}\Big|_{-1}^1 = \frac{e^{ikr}-e^{-ikr}}{ikr}$$
    Then in the third line you have ##\frac{\sin kr}{k}## as the integrand to be integrated from 0 to infiinity. The function ##\frac{\sin kr}{k}## is an even function, thus
    $$
    \int_0^\infty \frac{\sin kr}{k} \, dk = \frac{1}{2} \int_{-\infty}^\infty \frac{\sin kr}{k} \, dk
    $$
     
  10. May 1, 2016 #9
    Thanks! I feel like a total idiot now! :frown:
     
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