Evaluating $\int \limits_{-\infty}^{0} xe^{x(1-i\omega)} dx$
I did integration by parts and got
$$I = \frac{xe^{x(1 - i\omega)}}{1-i\omega} \bigg \rvert_{-\infty}^{0} - \int_{-\infty}^{0} \frac{e^{x(1-i\omega)}}{1- i\omega} dx$$
The problem is I don't know what happens when I plug in $-\infty$ in the first term.
I know that $\lim \limits_{x \to-\infty} xe^{x} = 0$, but here this $1 - i\omega$ factor in exponent really confuses me.
$$\lim_{x \rightarrow -\infty} xe^{x(1 - i\omega)} = z \big(\lim_{x \rightarrow -\infty}xe^x\big) $$ where $z = e^{-i \omega}$