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Fourier transform f(x)=sinax/x, a>0

  1. Mar 12, 2007 #1
    1. The problem statement, all variables and given/known data
    I am trying to show given f(x)=(sinax)/x, a>0

    that the transform is 0, |k|>a
    (pi/2)^1/2, |k|<a
    2. Relevant equations



    3. The attempt at a solution

    so far i have f transform =1/(2pi)^1/2.[integral from -inf to +inf]exp[-ikx](sinax)/x.dk, i am concerned about the singularity at x =0, does this compel me to use contour integration?
     
  2. jcsd
  3. Mar 12, 2007 #2
    There's no singularity at 0 for sin(x)/x. (Use l'hopital's rule to give Lim x->0 =1)
     
  4. Mar 13, 2007 #3
    okay so i've used l'hopital to evaluate the quotient, so this tells me my rational function q->1 as x-> 0 right ? but i don't see how this helps me in the evaluation of the transform? I suppose I have to findintegral (-inf, +inf) of exp[-ikx](sinax)/x. First I use the fact that my integrand is an even function, ive ended up with lim R->inf of the integral(0,R) of (sinaxcoskx)/xdx. not really sure if this right and the form of the answer seems to suggest that contour integration was used??
     
    Last edited: Mar 13, 2007
  5. Mar 13, 2007 #4

    Dick

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    You could do contour integration, but since you know the answer, it's easier to show it's right. Write sin(ax)/x as (exp(iax)-exp(-iax))/2ix. Now compare this expression with [tex]\int^a_{-a} e^{i k x} dk[/tex]. Do you see the relation between your function and the fourier transform of a step function?
     
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