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Fourier transform integration using well-known result

  1. Nov 11, 2016 #1
    Problem

    F denotes a forward fourier transform, the variables I'm transforming between are x and k
    - See attachment

    Relevant equations

    So first of all I note I am given a result for a forward fourier transform and need to use it for the inverse one.

    The result I am given to use, written out is :

    ## \int e^{-ikx} e^{-\alpha x^{2}} dx = \frac{1}{2\alpha} e^{-k^{2}/4\alpha} ##

    I note that ## F(k) C(t) ## gives me a function of k, so I apply ## F^{-1} ## on this I get a function of x.

    Attempt:

    My thoughts are I'm looking to change signs in my exponential terms so that it is , effectively a forward transform and then use the result, so just thinking of it as a integration result rather than a particular fourier transform.


    However if I do this my exponential terms are:


    ## e^{-(-ikx+\alpha k^{2})} ##

    , I don't know how I can then apply the result, completing the square looks like the only candidate to me , but this seems like it will be too scrappy, in particular with 'i' terms, how do I deal with this ## e^{ikx} ## term, if I'm on the right lines, is completion of the square necessary or is there some other approach to being able to use this result ?


    Many thanks in advance.
     

    Attached Files:

  2. jcsd
  3. Nov 11, 2016 #2

    blue_leaf77

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    Not too scrappy, it will only give you an extra constant phase factor.
    Anyway, you can just use the forward transform as a "template" with which you do the back transform. Note that in the RHS of the equation you are given, the expression is real so the sign of i should not matter.
     
  4. Nov 12, 2016 #3
    No sorry I'm still totally stuck, I think I see the logic in your arguement that the sign of i shouldn't matter but I don't see how I can apply the result given if the sign is wrong.

    Any more hints anyone?

    Cheers
     
  5. Nov 12, 2016 #4

    vela

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    How about making the change of variables ##k' = -k##?
     
  6. Nov 12, 2016 #5

    lurflurf

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    From your given transform you should see at one something reminiscent of the quadratic formula for Re(a)>0
    Which as you say relates to completing the square
    $$\int_{-\infty}^\infty\! \exp\left(-a x^2+2b x+c\right) \,\mathrm{d}x=\\
    \exp\left(\frac{b^2}{a}+c\right)\int_{-\infty}^\infty\! \exp\left(-a (x-b/a)^2\right) \,\mathrm{d}x=\\
    \exp\left(\frac{b^2}{a}+c\right)\sqrt{\frac{\pi}{a}}$$
    Notice the first integral cares what b is, but the second does not since the effect of b (and c) has been moved outside.
    See that since b is squared in the result -(-i)^2= -i^2=1 and the sign does not matter
     
    Last edited: Nov 12, 2016
  7. Nov 12, 2016 #6

    blue_leaf77

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    Even if you were to change the sign of ##i## or equivalently the sign of ##k## as suggested by vela above, you will get
    $$\int e^{ikx} e^{-\alpha x^{2}} dx = \frac{1}{2\alpha} e^{-k^{2}/4\alpha}$$
    If you are still wondering why this is true, try expanding ##e^{ikx}## into Cartesian form. The imaginary part of the integral that contains ##\sin kx## vanishes because the product with ##e^{-\alpha x^{2}}## results in an odd function and you integrate it over entire axis.
     
  8. Nov 12, 2016 #7
    yeh that's all fine.

    I don't know what result you've used in the last equality though, it wasn't using the result quoted in the OP which I'm trying to get at.
    Anyway so completing the square I get:

    ##e^{(-1/D)\frac{x^{2}D^{2}}{4}} \int e^{-1/D(k-ixD/2)^{2}} dk ##

    In order to then use the result given I need to write it in a similar complete square form in order to make a comparison...pretty sure this isn't what my lecturer was hinting at...
     
  9. Nov 12, 2016 #8

    blue_leaf77

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    What about changing the integration variable?
     
  10. Nov 13, 2016 #9

    lurflurf

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    • Member warned about providing too much help
    I don't know what Fourier convention is in use looks like a typo
    I have
    $$F(k)\left[e^{-\alpha x^2}\right]=
    \int_{-\infty}^\infty\! \exp\left(-i k x-\alpha x^2\right) \,\mathrm{d}x=
    \sqrt{\frac{\pi}{\alpha}}\exp\left(\frac{-k^2}{4\alpha}\right)$$
    let 1/alpha=4D t
    to see
    $$\frac{1}{\sqrt{4D \pi t}}F(k)\left[e^{-\frac{ x^2}{4D t}}\right]=
    \frac{1}{\sqrt{4D \pi t}}\int_{-\infty}^\infty\! \exp\left(-i k x-\frac{ x^2}{4D t}\right) \,\mathrm{d}x=
    \exp\left(-D k^2t\right)$$
    substitute it in
    $$\frac{Q}{2\pi}\int_{-\infty}^\infty\! \exp\left(i k x-D k^2t\right) \,\mathrm{d}x=QF^{-1}(x)\left[e^{-D k^2t}\right]=QF^{-1}(x)\left[\frac{1}{\sqrt{4D \pi t}}F(k)\left[e^{-\frac{ x^2}{4D t}}\right]\right]$$
    cancel the forward and inverse transforms to obtain the result
     
  11. Nov 13, 2016 #10
    exactly what i was looking for,- thank you.
     
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