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

[schniefen]

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?

 

[Charles Link]

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.

 

[anuttarasammyak]

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

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.