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

I Can't find a solution to this integral

  1. Mar 23, 2016 #1

    I try to solve this integral, but I failed... I have tried Euler's formula but it only got more and more complex. Can anyone help me?

    \int_{0}^{T}\int_{0}^{L}\frac{e^{2i(\omega t+\alpha x)}}{\sqrt{1-a e^{2i(\omega t+\alpha x)}}}dx dt
  2. jcsd
  3. Mar 23, 2016 #2


    User Avatar
    Gold Member

    Hi, you must decide respect what variable you want to start ..., after I suggest a substitution of the exponential ...
  4. Mar 23, 2016 #3
    Thanks for your post. I start with dx. I can write

    [tex]e^{2i(\omega t + \alpha x)} = \cos{(2\omega t + 2\alpha x)} + i \sin{(2\omega t + 2\alpha x)} [/tex]
    and then for the first term
    [tex]\cos{(2\omega t + 2\alpha x)} = \cos{(2\omega t)}\cos{(2\alpha x)} - \sin{(2\omega t)}\sin{(2\alpha x)} [/tex]
    Thus, the first integral to solve would still be:
    [tex] \cos{(2\omega t)} \int_{0}^{L} \frac{\cos{(2\alpha x)}}{\sqrt{1-a e^{2i (\omega t + \alpha x)}}} [/tex]

    I can of course use Euler's formula in the root as well, but it does not help yet.

    Or I substitute with
    [tex]k = 2 i (\alpha x + \omega t) [/tex]
    [tex]dx = \frac{1}{2i\alpha} dk [/tex]
    and come up with an integral like
    [tex]\frac{1}{2i\alpha} \int \frac{e^k}{\sqrt{1-ae^k}} dk[/tex]
    Last edited: Mar 23, 2016
  5. Mar 23, 2016 #4


    User Avatar
    Gold Member

    exactly! :wink: Now substitute ##1-ae^{k}=w## so ##e^{k}dk=-\frac{dw}{a}## and the integral become elementary ...
  6. Mar 30, 2016 #5
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted