Integral $\int_{-\infty}^0 e^{(-3i+\omega)t} $

Question

Let's say I am integrating this function: $e^{(-3i+\omega)t}$ from $t=-\infty $ to $t=0$ [Note: $\omega$ is just a constant]

The same function could be rewritten in this form(i believe?) : $e^{-(3i-\omega)t}$

Knowing that $e^{-\infty}$ will give me $0$, I am confused by the fact integrating the same function over the same period give me two differnt answers.

Case 1:

$\int_{-\infty}^0 e^{(-3i+\omega)t} = \dfrac{e^{(-3i+\omega)t}}{(-3i+\omega)}\bigg|_{-\infty}^0 = 0$ $(e^{-\infty}=0)$

Case 2:

$\int_{-\infty}^0 e^{-(3i-\omega)t} = -\dfrac{e^{-(3i-\omega)t}}{(3i-\omega)}\bigg|_{-\infty}^0 = \infty$ $(e^{\infty}=\infty)$

I believe I am doing something wrong? What am i missing?

Answer 1

The second case also converges to zero, check your signs.

$$ \int exp(-(3i-w)t) \mathrm{dt}= - \frac{exp(-(3i-w)t)}{3i-w} = - \frac{exp((w-3i)t)}{3i-w}$$

That should convince you.

Answer 2

Actually neither of your results are correct, since $e^{(\omega-3i)t}=1$ when $t=0$. So your integral should be (and the negative infinity term does go to $0$ as mentioned in the other answers): $$\int_{-\infty}^0e^{(\omega-3i)t}\,dt=\frac{1}{\omega-3i}$$