Fourier Transform of e-x*e-(x*coswt) w.r.t Frequency

[rem]

Homework Statement

it's related with SHM.its a trivial question.and it's definitely got an ans.i need to do Fourier transformation for
e(iwt) e-(q**2/bk(2sin**2(wt)/2))dt.
btw the limits -inf to +inf.whereb=kt,k=boltzmannconst.w=(k/m)**.5(k here is spring const.)do Fourier transform and get it in terms of frequency

Homework Equations

that"s the only eqnbut can be modified to e-x*e-(x*coswt) where x=const.
("though some friends say it'll result in error function")

The Attempt at a Solution

e-(q**2/bk(2sin**2(wt)/2))
=e-(q**2/bk(1-coswt)) (taking q**2/bk=const.say x)
=e-(x(1-coswt))
=e-x*e-(x*coswt)

 

[cepheid]

It's really hard to read the function you're trying to calculate the FT for. Are you saying that the function is

f(t) = \exp{\left(-\frac{q^2}{bk} \frac{2\sin^2(wt)}{2}\right)}

and you want to calculate the Fourier transform?

\hat{F}(\omega) = \int_{-\infty}^{\infty} \exp{\left(-\frac{q^2}{bk} \frac{2\sin^2(wt)}{2}\right)} \exp{(-i \omega t)} dt

 

[rem]

It's really hard to read the function you're trying to calculate the FT for. Are you saying that the function is

f(t) = \exp{\left(-\frac{q^2}{bk} \frac{2\sin^2(wt)}{2}\right)}

and you want to calculate the Fourier transform?

\hat{F}(\omega) = \int_{-\infty}^{\infty} \exp{\left(-\frac{q^2}{bk} \frac{2\sin^2(wt)}{2}\right)} \exp{(-i \omega t)} dt

"yes".
the given eqn is
"e-(2q^2/bk(sin^2(wt/2)))

where b=kT,T=absolute temp,k=bolzmann const. and w=(k/m)^(1/2) t=time period.

the above fn. depends on time.i have to apply Fourier transform to convert it into "w" frequency.

the alternate method i came up with is by treating "q,b,w" as const. say"x"
and converting sin^2(wt/2) as "1-coswt"

so the final eqn became e-(x(1-coswt))
((e-x)(e(xcoswt))
other wise have to Fourier transform for this eqn.