Fourier transform of cos x the answer involves δ functions

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data

    This is an example provided by my lecturer in his notes. He puts practically zero working in.

    When i work the problem through i do not get the same answer as he does.

    In this section i have copied the exact text from the problem:

    Find the Fourier transform of cos(x). Your answer will include delta functions

    We shall use the definition [tex]\delta(k)=\int\limits_\infty^\infty dx \ e^{ikx}[/tex]



    he then simply writes:

    [tex]Fourier[cos(x)]=\frac{1}{\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{-ikx}\ \frac{e^{ix}+e^{-ix}}{2}=\sqrt{\frac{\pi}{2}}\left[\delta(k-1)+\delta(k+1)\right][/tex]

    As you can imagine this makes me very irritated so I start to solve it myself

    2. Relevant equations

    [tex]Fourier[f(x)]=\frac{1}{\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{-ikx}\ f(x)[/tex]

    [tex]\delta(k)=\int\limits_\infty^\infty dx \ e^{ikx}[/tex]

    3. The attempt at a solution

    So i work it through line by line:

    [tex]Fourier[cos(x)]=\frac{1}{\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{-ikx}\ \frac{e^{ix}+e^{-ix}}{2}[/tex]

    [tex]Fourier[cos(x)]=\frac{1}{2\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{-ikx}\left(\ e^{ix}+e^{-ix}\right)[/tex]

    [tex]Fourier[cos(x)]=\frac{1}{2\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{-ikx+ix}+e^{-ikx-ix}[/tex]

    [tex]Fourier[cos(x)]=\frac{1}{2\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{ix(-k+1)}+e^{ix(-k-1)}[/tex]

    [tex]Fourier[cos(x)]=\frac{1}{2\sqrt{2\pi}}\int\limits_\infty^\infty dx \ e^{ix(-k+1)}+\frac{1}{2\sqrt{2\pi}}\int\limits_\infty^\infty dx \e^{ix(-k-1)}[/tex]

    now recognise

    [tex]\delta(k)=\int\limits_\infty^\infty dx \ e^{ikx}[/tex]

    can be used. However the definition for the fourier transform shows that the constant should be:


    therefore we need a constant such that [tex]\frac{1}{2\sqrt{2\pi}}*x=\frac{1}{2\pi}[/tex]

    this implies x = [tex]\frac{2}{\sqrt{2\pi}}[/tex]

    this leads to:

    [tex]Fourier[cos(x)]=\frac{1}{2\sqrt{2\pi}}x\frac{2}{\sqrt{2\pi}}\ \delta(-k+1)+\frac{1}{2\sqrt{2\pi}}x\frac{2}{\sqrt{2\pi}}\ \delta(-k-1)[/tex]

    [tex]Fourier[cos(x)]=\frac{1}{2\pi}\ \delta(-k+1)+\frac{1}{2\pi}}\ \delta(-k-1)[/tex]

    then recognise [tex]\delta(k)=\delta(-k)[/tex]

    [tex]Fourier[cos(x)]=\frac{1}{2\pi}\ \delta(k-1)+\frac{1}{2\pi}}\ \delta(k+1)[/tex]

    This is of course not:


    The constant is incorrect.

    I am certain that he is correct and that I am wrong.

    Why have I got it wrong?

    Thanks very much for your time.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 8, 2010 #2


    Staff: Mentor

    I don't have time to look at your work, now, other than to notice that infinity is both the lower and upper limit of integration for all your integrals. I'll try to take a closer look later today.
  4. Apr 8, 2010 #3
    it looks at first like if your definition of dirac delta were true, then the coefficients would remain unaltered simply by direct substitution. please clarify why i'm seeing this incorrectly, since my fourier analysis isnt very good.

    you remain unstated what the actual definition of the fourier transform that says that the coefficient must be 1/2pi.

    presumably, if there is no normalizing factor, then you just multiply and divide by 2pi/2pi and the 1/2pi goes into the evaluating the integral with the remaining 2pi/1 as a coefficient. this with the original coefficient should give you sqrt(pi/2), right?
  5. Apr 8, 2010 #4


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    I think the definition of the delta function should be


    With the Fourier transform, you have different conventions for where the constants go, but I don't think you have that with the delta function. That factor in the front needs to be there.


    The Fourier transform of the delta function is:

    [tex]F[\delta(x)] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \delta(x)e^{-ikx}\,dx = \frac{1}{\sqrt{2\pi}}[/tex]

    so its inverse is

    [tex]\delta(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F[\delta(x)]e^{ikx}\,dk = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ikx}\,dk[/tex]
    Last edited: Apr 8, 2010
  6. Apr 8, 2010 #5
    Thanks for your replies everyone :)

    You are correct that is the definition of the delta function sorry about that.

    I still don't understand how he gets that though.

    @Xaos: I'm sorry I'm not sure i understand you :(
  7. Apr 8, 2010 #6
    sorry i don't know latex so well.

    if delta=1/2pi*int(exp(ikx))
    then int(exp(ikx))=2pi/2pi*int(exp(ikx))=2pi*delta.

    but if you have 1/2sqrt(2pi)*int(exp(ikx)), then
    and this is = sqrt(pi/2)delta which is what you want.
    Last edited: Apr 8, 2010
  8. Apr 9, 2010 #7
    legendary. Thanks Xaos :)
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