Simple Fourier transformation calculation

[schniefen]

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?

 

[Office_Shredder]

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?