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Calculate the integrals

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data
    1) Calculate the integral:
    [tex]$\int_{-\infty }^{\infty }{{{\left( \frac{\sin \left( x \right)}{x} \right)}^{2}}dx}$[/tex]

    2) Calculate the integral:
    [tex]$\int_{-\infty }^{\infty }{{{\left( \frac{1-\cos \left( \lambda \pi \right)}{{{\lambda }^{2}}} \right)}^{2}}d\lambda }$[/tex]

    2. Relevant equations

    I think it has to be solved with finding the Fourier transform or something. I'm not quite sure...


    3. The attempt at a solution
    ???
    Well, I really don't know where to start.

    Help ?
     
  2. jcsd
  3. Oct 1, 2009 #2
    You need to use contour integration in the complex plane:

    Choose a closed contour part of which runs from negitive infinity to possitive infinity on the real axis. Then use either the 'Cauchy Integral Formula' or the 'Residue Theorm' to integrate the function over your chosen contour. Details of these methods given here: http://en.wikipedia.org/wiki/Methods_of_contour_integration.
     
    Last edited: Oct 1, 2009
  4. Oct 1, 2009 #3
    Hmmm, these questions are from a Fourier analysis course. I have never heard of what you have described above :)
     
  5. Oct 1, 2009 #4

    Dick

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    You'll need to learn it. You need the residue theorem to calculate a lot of fourier integrals. For the first one you want to replace the z^2 in the denominator with (z-ie)^2 where e>0 to move the pole off the real axis. Then take the limit e->0 at the end. Similar trick for the second one.
     
  6. Oct 1, 2009 #5
    Hmmm... It just seems wierd that my teacher wants me to figure out a new integrating trick on my own ?

    Is that really the other way to solve it ?
    I've been told, as a hint, to look at Plancherels Formula. And then I could neglect the squared paranthes around the whole function, because the inner product would give the same.

    'Cause I don't know about you're way you do it - even though it might be smart :)
     
  7. Oct 1, 2009 #6

    Dick

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    That is a good hint to avoid the residue stuff. Did you look at look at Plancherel's theorem? Any thoughts on how to use it?
     
  8. Oct 1, 2009 #7
    Not much really...

    It states:

    Suppose f and g are square integrable. Then

    [tex]$<F\left[ f \right],F\left[ g \right]{{>}_{{{L}^{2}}}}=\,<f,g{{>}_{{{L}^{2}}}}$[/tex]

    [tex]${{\left\langle {{F}^{-1}}\left[ f \right],{{F}^{-1}}\left[ g \right] \right\rangle }_{{{L}^{2}}}}={{\left\langle f,g \right\rangle }_{{{L}^{2}}}}$[/tex]

    (The symbol in the start is suppose to be <. Don't know why it wouldnt write it out.)

    In particular,

    [tex]\[{{\left\| F\left[ f \right] \right\|}_{{{L}^{2}}}}={{\left\| f \right\|}_{{{L}^{2}}}}\][/tex]

    So as I said, since we take the inner product, we can neglect the squared paranthes, and then do <f,f> and <g,g> and get the same as if they were squared.

    But besides that, I'm pretty much clueless tbh...
     
    Last edited: Oct 1, 2009
  9. Oct 1, 2009 #8

    Dick

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    I'm not sure. But your first integral is the L^2 norm of the function f(x)=sin(x)/x. If you could figure out what the fourier transform of f(x) is, it might be easier to find the L^2 norm of that instead of the original function.
     
  10. Oct 1, 2009 #9
    Hmmm, the integral of f(x) is sqrt(pi/2) ?
     
  11. Oct 1, 2009 #10

    Dick

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    Why would you say that? Are you guessing?
     
  12. Oct 1, 2009 #11
    Well, I kinda looked it up :) But I'm not even sure that's right...

    I'm not really quite sure how to do it. As far as I can see the indefinite integral of just sin(x)/x = x - x^3/(3*3!) + x^5/(5*5!)...
    And then when I have to involve e*(-i*n*x) as well, I really get confused :S
    Is integration by parts still an option when there is 3 "parts" ?
     
  13. Oct 1, 2009 #12

    Dick

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    You are missing the Plancherel's clue. What's the fourier transform of sin(x)/x?? This is a fourier course, right? Hunt around a little in the course material.
     
  14. Oct 1, 2009 #13
    Hmmm, the clue being that the Fourier transform F[f] is equal the function f ?
    If not, how am I suppose to find the Fourier transform, when it's actually what I'm asking help to find ? :)
     
  15. Oct 1, 2009 #14

    Dick

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    No, no, no. F(f) is not equal to f. The L^2 norm of the two is equal. Define f(x)=1 if -1<=x<=1 and f(x)=0 otherwise. Can you find the fourier transform of f(x)?
     
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