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

[catcherintherye]

Homework Statement

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

Homework Equations

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?

 

[christianjb]

There's no singularity at 0 for sin(x)/x. (Use l'hopital's rule to give Lim x->0 =1)

 

[catcherintherye]

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, I've 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??

 

[Dick]

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 $$\int^a_{-a} e^{i k x} dk$$. Do you see the relation between your function and the Fourier transform of a step function?