Fourier Transform of product of heaviside step function and another function

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The discussion focuses on finding the response of a damped harmonic oscillator driven by a force involving a Heaviside step function and an exponential term. The user initially struggles with taking the Fourier transform of the function f(t) and encounters convergence issues when using integration by parts. A suggestion is made to simplify the integration by recognizing that t^2 e^{iωt} can be expressed as the second derivative of e^{iωt} with respect to ω. This approach leads to a more manageable integral and ultimately resolves the user's difficulties. The conversation highlights effective techniques for handling Fourier transforms involving step functions.
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Homework Statement


A damped harmonic oscillator is driven by a force of the form f(t)=h(t) t^2 Exp(-t), where h(t) is a Heaviside step function. The Oscillator satisfies the equation x''+2x'+4x=f(t). Use pencil-and-paper methods involving Fourier transforms and inverse transforms to find the
response of the oscillator, x(t), assuming that x(0)=0 and x'(0)=1.

Homework Equations


The Fourier Transform F[f(t)]
The Inverse Fourier Transform F^(-1)[f(ω)]
Integration by parts

The Attempt at a Solution


First thing I did was take the Fourier transform of the left hand side, which I'm sure I got correct. The part I'm stuck on is taking the Fourier transform of f(t). I used integration by parts using
u=t^2/e^t dv=h(t)(Exp[i ω t])
du=(2t-t^2)/e^t v=πδ(ω)+i/ω

but now when I try to complete the integration, i get an integral that does not converge because of the i/ω. There have been multiple heaviside problems that I have been working on that all end up the same, and I'm not sure where to go from here.
 
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In general actually, how would one go about solving the Fourier transform of a function with the form f(t)H(t)

for example, how could I do the Fourier transform of tH(t)? Again, I've tried doing it by parts, but nothing good comes of it
 
The way you split the integrand up is unusual. The step function allows you to write the integral as
$$\int_{-\infty}^\infty t^2e^{-t}h(t) e^{i\omega t}\,dt = \int_0^\infty t^2e^{(i\omega-1)t}\,dt,$$ which you can integrate by parts. A trick you could use to avoid integration by parts is to note that ##t^2 e^{i\omega t} = -\frac{\partial^2}{\partial \omega^2}e^{i\omega t}##, so you can say
$$\int_0^\infty t^2e^{(i\omega-1)t}\,dt = -\frac{\partial^2}{\partial \omega^2}\int_0^\infty e^{(i\omega-1)t}\,dt$$
 
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Thank you so much you are a life saver. My homework became super easy after your post!
 

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