Fourier transform, complex exponential and infinity

[fishingspree2]

I'm taking the Fourier transform of a signal. This integral has bounds from -∞ to ∞, but since the signal is 0 for negative t, the bounds become 0 to ∞

doing the integration, the antiderivative I get is e^{t*(-3-jω+2j)} where j is sqrt(-1)

Now I have to evaluate this at t=infinity (since it is a proper integral)...I don't really know how to do this since (-3-jω+2j) is a complex number.

infinity times a negative number is negative infinity
infinity times a positive number is positive infinity
infinity times a complex number?

any help will be appreciated,

thank you very much

 

[jackmell]

Looks like you have:

$$e^{-3t+it(2-w)}\biggr|_0^{\infty}$$

where I'm using i cus' I'm not an engineer but same dif. Now assume w is a real number, then we could write:

$$e^{-3t}e^{it(2-w)}=e^{-3t}\left(\cos(t(2-w))+i\sin(t(2-w)\right)\biggr|_0^{\infty}$$

so that because of the $e^{-3t}$ real part, at the upper limit, that is zero and at the lower limit it's one.

 

[fishingspree2]

thank you! got it!