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Integrating sine where argument goes to infinity.

  1. Sep 25, 2014 #1
    After some integration, i am getting a form [itex]e^{i \alpha\phi+i\beta\phi\sin(\phi-\phi')-i\gamma\sin\phi} [/itex], where ##\alpha, \beta, \gamma## are constants. Now i want to apply the limit where ##\phi ## ranges from 0 to ##\infty ## (ya, in the argument of sine we will encounter ##\infty ## which is exactly my problem). How can i do that?
     
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  3. Sep 25, 2014 #2

    dextercioby

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    What's the original problem? We need to see a context. Please, post it.
     
  4. Sep 26, 2014 #3
    Okay. i am trying to derive a dispersion relation in magnetized plasma with kinetic approach. By the by i am taking only two terms, just forget about the last term in the above equation. We will encounter a term##I=
    \int_0^\infty cos(\phi-\phi')e^{g_1(\phi')}\,d\phi'\\## where ##g_1(\phi')= i \alpha \phi'+i\beta sin(\phi-\phi')##. Differentiating ##g_1(\phi')## with respect to ##\phi'## and then substituting back into ##I_1## will give
    ##I_1=\alpha\int_0^\infty e^{g_1(\phi') d\phi'}+ \gamma\int_0^\infty \frac{dg_1(\phi')}{d\phi'}e^{g_1(\phi')}d\phi'##. It is the second term that causes trouble. From second term, ##\frac{dg_1(\phi')}{d\phi'}e^{g_1(\phi')}d\phi= d(e^{g_1(\phi')})##. So ##I_1= firstterm + \gamma\int_0^\infty d(e^{g_1(\phi')})##. Now, i think, you can make sense out of it.
    Thanks in advance...
     
  5. Sep 26, 2014 #4

    HallsofIvy

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    The trig function do NOT have a limit as the argument goes to infinity.
     
  6. Sep 26, 2014 #5
    i understand that sir. But, when you do the maths it comes into picture. And this is why i posted the question here.
    Thank you...
     
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