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I Integral e^(ikx^3)

  1. Dec 5, 2017 #1
    Show that $$\int_0^\infty dx\exp(ikx^3) , k>0$$ may be written as integral from 0 to ##\infty## along the line ##arg(z) = \frac{\pi}{6}##.

    I'd appreciate it if you can help me how to approach this problem. My initial impression was to expand the integrand out
    $$\sum^{\infty}_{n=0}\frac{(ikx^3)^n}{n!}$$
    but did not how to obtain the ##arg(z)## condition. I plugged the integral in wolframalpha and gave me an expression with a Gamma function, which the lecture has covered but I'm not sure how to apply here.

    Thanks
     
  2. jcsd
  3. Dec 5, 2017 #2

    fresh_42

    Staff: Mentor

    Which expression in terms of the gamma-function do you get from the substitution ##t = -ikx^3\,##?
     
  4. Dec 5, 2017 #3
    Wolfram says $$\frac{\Gamma(\frac{4}{3})}{\sqrt[3]{-ik}}.$$
    I have been told that using ##i=\exp(i\frac{\pi}{2})## helps but I can only get as far as writing the integrand as ##\exp(\exp(i\frac{\pi}{2})kx^3))## and not sure what to do after.
     
  5. Dec 5, 2017 #4

    fresh_42

    Staff: Mentor

    That wasn't my question. I've asked what do you get, not WolframAlpha. The substitution of ##i## might help to properly calculate the powers of ##i## but I'd rather write ##k=ic## to begin. In any case one has to be cautious with the handling of the purely imaginary constants.

    The substitution ##t \mapsto -ikx^3## allows to write ##x## as power of ##t## as well as the integral in terms of the gamma function.
     
  6. Dec 5, 2017 #5

    Delta²

    User Avatar
    Gold Member

    I think the exercise asks not to compute the value of the integral but to express it as a contour integral with the contour being the line ##arg(z)=\frac{\pi}{6}##.
     
  7. Dec 5, 2017 #6
    Yes, the expression is what I am looking for.
    I am solving one of the past exam papers for my university and it is the first part of the question.
     

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  8. Dec 5, 2017 #7
    Okay, so I am looking for the expression of the contour integral as Delta said. I allowed ##i=\exp(i\frac{\pi}{2})## and extend the integral to z space. Then let ##z=r\exp(i\theta)##, yielding
    $$ \int _C dz e^{{kr^2e^{i(\pi/2+3\theta)}}}.$$

    Do you think this is correct so far? Maybe ##z=r\exp(i\theta)## is unnecessary.
     
    Last edited: Dec 5, 2017
  9. Dec 5, 2017 #8
    What I have now is:
    Choose ##z=re^{i\pi/6}## so ##dz=e^{i\pi/6}dr##. Plug in the integral ->
    $$e^{i\pi/6}\int^\infty_{-\infty} e^{ikr^3e^{i\pi/2}}$$ and note ##e^{i\pi/2}=i##
    Finally, the integral becomes $$e^{i\pi/6}\int^\infty_{-\infty} e^{-kr^3}.$$
    Does this seem right?
     
  10. Dec 6, 2017 #9

    Delta²

    User Avatar
    Gold Member

    The limits of integration should be from 0 to ##+\infty## (the contour is the positive part of the line ##z=re^{i\pi/6}##.

    However in order to be rigorous and clear you have to provide a function ##f:\mathbb{C}->\mathbb{C}## such that
    ##\int_0^{\infty}f(z(r))z'(r)dr=\int_0^{\infty}e^{ikr^3}dr##
     
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