Fourier transform of ##e^{-a |t|}\cos{(bt)}##

schniefen
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Homework Statement
Find the Fourier transform of ##f(t)=e^{-a |t|}\cos{(bt)}##, where ##a## and ##b## are positive constants.
Relevant Equations
I use the convention ##\tilde{f}(\omega)= \int_{-\infty}^{\infty}f(t)e^{-i\omega t} \mathrm{d}t##. Also, I can use the fact that ##f(t)=-i\Theta(t)e^{-i\omega_0 t-\gamma t}## has Fourier transform ##\tilde{f}(\omega)=\frac{1}{\omega-\omega_0+i\gamma}##, although this is not explicitly hinted at.
First,

##\tilde{f}(\omega)=\int_{-\infty}^{\infty}e^{a|t|}\cos(bt)e^{-i\omega t} \mathrm{d}t##​

We can get rid of the absolute value by splitting the integral up

##\int_{-\infty}^{0}e^{at}\cos(bt)e^{-i\omega t} \mathrm{d}t+ \int_{0}^{\infty}e^{-at}\cos(bt)e^{-i\omega t} \mathrm{d}t##
Using ##\cos(x)=\frac12(e^{ix}+e^{-ix})##, the first integral reduces to

##\frac12\int_{-\infty}^{0}e^{at}(e^{ibt}+e^{-ibt})e^{-i\omega t}\mathrm{d}t=\frac12\int_{-\infty}^{0}e^{t(a+i(b-\omega))}\mathrm{d}t+\frac12\int_{-\infty}^{0}e^{t(a+i(-b-\omega))}\mathrm{d}t ##​

My strategy then is to rewrite the two integrands on the form ##-i\Theta(t)e^{-i\omega_0 t-\gamma t}## and use the known formula, as given under Relevant equations. Is this an approach you would have taken?
 
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I think you are completely justified in carrying out the integration just as you would if you were to integrate ## \int e^{at} \ dt ##, without resorting to a formula for the F.T. Consider the Euler formula ## e^{ix}=\cos{x}+i \sin{x} ## for the details for making the integration valid when complex numbers are involved. Edit: e.g. You can use the complex numbers to do the integration, and then verify that it gives the correct result for ## \int e^{-at} \cos(bt) \, dt ## by taking the derivative of the real part of the result you get for ## \int e^{-at} e^{ibt} \, dt ##, and see that you do indeed get ## e^{-at} \cos(bt) ##.

See also https://www.physicsforums.com/threa...le-of-emitted-radiation.1048791/#post-6839518
for a recent homework posting that is somewhat related.
 
Last edited:
schniefen said:
##\frac12\int_{-\infty}^{0}e^{at}(e^{ibt}+e^{-ibt})e^{-i\omega t}\mathrm{d}t=\frac12\int_{-\infty}^{0}e^{t(a+i(b-\omega))}\mathrm{d}t+\frac12\int_{-\infty}^{0}e^{t(a+i(-b-\omega))}\mathrm{d}t ##

My strategy then is to rewrite the two integrands on the form ##-i\Theta(t)e^{-i\omega_0 t-\gamma t}## and use the known formula, as given under Relevant equations. Is this an approach you would have taken?
Starting from two integrals in the last line, change intgral variables from ##t## to ##-t## so that integrand become ##[0,+\infty)##.
Then you may write it
\int_0^{+\infty} g_k(t)dt = \int_{-\infty}^{+\infty} g_k(t)\theta (t) dt
where
g_1(t)=e^{-ta-i(b-\omega)t}
g_2(t)=e^{-ta-i(-b-\omega)t}
so that you may make use of what you prepared.
 
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