# Simple Fourier transformation calculation

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Homework Statement
Calculate the Fourier transform of ##p(t)=Ae^{-\gamma t}e^{-i\omega_0t}\Theta(t)##, where ##\Theta(t)## is the step function, i.e. ##1## for ##t\geq 0## and ##0## otherwise, and ##A##, ##\gamma## and ##\omega_0## are unspecified constants. Do not use the "known" formula. The answer given is ##A/(\gamma+i(\omega-\omega_0))##.
Relevant Equations
I guess the "known" formula alludes to the Laplace transform, which the Fourier transform reduces to, i.e. I use the notation ##\hat{p}(\omega)=\int_{-\infty}^{\infty} p(t)e^{-i\omega t}\mathrm{d}t## for the Fourier transform of ##p(t)##.
So,
##\hat{p}(\omega)=\int_{-\infty}^{\infty} p(t)e^{-i\omega t}\mathrm{d}t=A\int_{0}^{\infty}e^{-t(\gamma+i(\omega+\omega_0))}=A\left[-\frac{e^{-t(\gamma+i(\omega+\omega_0))}}{\gamma+i(\omega+\omega_0)}\right]_0^\infty,##​
provided ##\gamma+ i(\omega+\omega_0)\neq 0## for the last equality. Now, considering the answer given, why does the numerator go to ##0## at ##\infty##? The minus sign in front of the exponent could also be absorbed by the constants and then one could argue the numerator diverges at ##\infty##, or?

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It's definitely being assumed that##\gamma > 0## here. This is pretty common - otherwise why would they bother putting the minus sign in the first place?

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