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Fourier transform problem

  1. Mar 26, 2014 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    Suppose [itex]f(x)[/itex], [itex]-\infty<x<\infty[/itex], is continuous and piecewise smooth on every finite interval, and both [itex]\int_{-\infty}^\infty |f(x)|dx[/itex] and [itex]\int_{-\infty}^\infty |f'(x)|dx[/itex] are absolutely convergent. Show the fourier transform of [itex]f'(x)[/itex] is [itex]i\mu F(\mu)[/itex].


    2. Relevant equations [tex]F(\mu)=\frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\mu x}dx[/tex] where [itex]F(\mu)[/itex] is the fourier transform of [itex]f(x)[/itex]



    3. The attempt at a solution
    transforming [itex]f'(x)[/itex] and integrating by parts yields [tex]\frac{f(x)e^{-i\mu x}}{2\pi}\Biggr|_{-\infty}^\infty+i\mu \underbrace{\frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\mu x}dx}_{F(\mu)}[/tex]

    my question here is, evidently [itex]\frac{f(x)e^{-i\mu x}}{2\pi}\Biggr|_{-\infty}^\infty=0[/itex] but how? any insight is greatly appreciated!

    also, where have i used the absolutely convergent material? did it show up in the integration by parts when i had to integrate [itex]f'(x)[/itex] and then again in order to show that [itex]\frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\mu x}dx[/itex] exists and may in fact equal [itex]F(\mu)[/itex]?

    thanks all!
     
    Last edited: Mar 26, 2014
  2. jcsd
  3. Mar 26, 2014 #2

    LCKurtz

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    Aren't you missing the minus sign I added in red?
     
  4. Mar 26, 2014 #3

    joshmccraney

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    ahh shoot, totally forgot to put this! (although when writing it i remembered, hence the positive integral. at any rate, positive or negative doesn't really affect this limit right (since we evaluate from negative to positive infinity)?

    either way, any insight?
     
  5. Mar 26, 2014 #4

    joshmccraney

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    i just changed it back to the way it should be (with the negative). thanks
     
  6. Mar 26, 2014 #5

    jbunniii

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    Note that
    $$|F(\mu)| = \left|\int_{-\infty}^{\infty} f(x) e^{-i\mu x} dx\right| \leq \int_{-\infty}^{\infty}|f(x) e^{-i\mu x}| dx = \int_{-\infty}^{\infty}|f(x)| dx$$
    So the absolute integrability of ##f## ensures that the Fourier transform is ##F(\mu)## is well defined for all ##\mu##.

    Similarly, the absolute integrability will allow you to compute the Fourier transform of ##f'## using the usual integral definition.
     
  7. Mar 26, 2014 #6

    jbunniii

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    [edit] deleted, see next post instead.
     
    Last edited: Mar 26, 2014
  8. Mar 26, 2014 #7

    jbunniii

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    Regarding why
    $$\frac{f(x)e^{-i\mu x}}{2\pi}\Biggr|_{-\infty}^\infty=0$$
    this is a bit subtle. Note that absolute integrability and smoothness of ##f## are not enough to guarantee this: consider a function ##f## which is zero in most places but has a smooth "bump" centered at each integer ##n##, such that as ##|n|## increases, the bumps maintain the same height but become narrower, so that the area of the bump centered at ##n## or ##-n## is ##1/2^{|n|}##. Then the integral of ##|f|## is finite, but ##\lim_{x \rightarrow \infty} f(x)## and ##\lim_{x \rightarrow -\infty} f(x)## do not exist.

    So you must make use of the fact that the derivative ##f'## is also absolutely integrable.
     
  9. Mar 26, 2014 #8

    micromass

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    Hint, prove that

    [tex]\lim_{x\rightarrow +\infty} f(x) = f(0) + \int_0^{+\infty} f^\prime(x)dx.[/tex]

    Thus, in particular, the limit exists. Now you can deduce that the limit is ##0##. (as jbun noticed, it might happen that the limit does not exist, which might have been a problem).
     
  10. Mar 27, 2014 #9

    joshmccraney

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    i'm lost on what you're saying. i realize you're beginning to explain why the given value is zero, but i am not sure what is going on. could you further your explanation?

    thanks a ton for helping!
     
  11. Mar 27, 2014 #10

    joshmccraney

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    [tex]f(0) + \int_0^{+\infty} f^\prime(x)dx = f(0) + \underbrace{\lim_{x\rightarrow +\infty} f(x) - f(0)}_{\text{fundamental theorem}} = \lim_{x\rightarrow +\infty} f(x)[/tex] but now i'm confused? i guess i'm not seeing how this relates? i'm guessing you're suggesting i substitute this integral into the above [itex]f(x)[/itex] but since the [itex]e^{-i\mu x}[/itex] is not in the integral, i'm missing how this helps?
     
  12. Mar 27, 2014 #11

    jbunniii

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    You are given that ##f'## is absolutely integrable. Therefore,
    $$\left|\int_{0}^{\infty} f'(x) dx\right| \leq \int_{0}^{\infty}|f'(x)| dx < \infty$$
    This means that ##\int_{0}^{\infty} f'(x) dx## is finite, so (by the equation you indicated above), ##\lim_{x \rightarrow \infty} f(x)## exists; let's call this limit ##L##. By a similar argument, ##\lim_{x \rightarrow -\infty} f(x)## also exists; let's call this limit ##M##.

    But you also know that ##\int_{-\infty}^{\infty} |f(x)| dx## is finite. I claim that this implies that ##L## and ##M## must both be zero. See if you can prove this claim.

    Assuming the claim is true, we have ##\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow -\infty} f(x) = 0##. Do you see why this immediately implies that
    $$\frac{f(x)e^{-i\mu x}}{2\pi}\Biggr|_{-\infty}^\infty=0$$
     
  13. Mar 27, 2014 #12

    joshmccraney

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    assuming that we can prove ##L=M=0##, yes, I definitely agree. i just took a quick glance so i'll try and work out the proof!

    honestly though, how on earth did you manage to think of using ##f(0) + \int_0^\infty f(x) dx = \lim_{x \rightarrow \infty} f(x)##?? this seems so random to just think of, although it is brilliant
     
    Last edited: Mar 27, 2014
  14. Mar 27, 2014 #13

    joshmccraney

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    okay i got you!

    so, to summarize, when showing $$\frac{f(x)e^{-i\mu x}}{2\pi}\Biggr|_{-\infty}^\infty=0$$ we consider ##\lim_{x \rightarrow \pm \infty} f(x)=0##. using the fact that [itex]\int_{-\infty}^\infty |f(x)|dx[/itex] converges ALMOST guarantees that in the limit ##f(x)## converges to zero (almost because of the counter example that you've shown). evidently we need to show that this limit exists.

    to show existence we make use of the equality [tex]\lim_{x\rightarrow +\infty} f(x) = f(0) + \int_0^{+\infty} f^\prime(x)dx[/tex] and couple this with the absolute convergence of ##f'(x)##. this guarantees that the above integral is finite. we know ##f(0)## exists, as implied by the piecewise continuity of ##f(x)## on ##-\infty < x < \infty##. therefore, the limit in question not only exists but converges to zero.

    does this sound right?

    thanks a ton you guys!!!
     
  15. Mar 27, 2014 #14

    jbunniii

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    I didn't follow your argument. How does this show that ##L = M = 0##?

    Try arguing as follows. We have established that ##L = \lim_{x \rightarrow \infty} f(x)## exists. Suppose ##L \neq 0##. Then by the definition of a limit, there exists some ##X > 0## such that ##|f(x)| > |L|/2## for all ##x > X##. Now what does this imply about ##\int_{X}^{\infty} |f(x)|dx##? You can argue similarly for ##M##.
     
  16. Mar 27, 2014 #15

    joshmccraney

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    my "proof" is predicated on the fact that, if ##\int_0^\infty f(x) dx## converges and ##\lim_{x \rightarrow \infty} f(x)## exists, then ##\lim_{x \rightarrow \infty} f(x)=0##

    is this not correct?
     
  17. Mar 27, 2014 #16

    joshmccraney

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    evidently, ##\int_{X}^{\infty} |f(x)|dx## does not converge, which violates the given conditions.

    please respond to my other post, though. i want you're feedback to make sure what i've written is correct.

    thanks a ton!
     
  18. Mar 27, 2014 #17

    jbunniii

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    It's correct, I just wanted to make sure you saw why it was true. If the function converges on some nonzero value, then that "tail" has infinite area.

    The rest of your summary looks fine.
     
  19. Mar 27, 2014 #18

    joshmccraney

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    cool just making sure! thanks again for the help (and the clever counter example mentioned above)
     
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