Calculating Integral: Solving \int_{0.5}^{\infty }e^{-jwt}dt

[khdani]

hello,
how do i calculate this integral:
$$I = \int_{0.5}^{\infty }e^{-jwt}dt$$

i though it should be
$$I=0 - \frac{e^{-jw/2}}{-jw} = \frac{e^{-jw/2}}{jw}$$

but it seems that the first part isn't zero...why ?

 

[n!kofeyn]

$$\begin{align*} I &= \int_{0.5}^\infty e^{-jwt} \,dt \\ &= \lim_{a\to\infty} \int_{0.5}^a e^{-jwt} \,dt \\ &= \lim_{a\to\infty} \frac{e^{-jwt}}{-jw} \,\bigg|_{0.5}^a \\ &= \lim_{a\to\infty} \frac{e^{-jwa}}{-jw} - \frac{e^{-jw/2}}{-jw} \\ &= \left( \lim_{a\to\infty} \frac{e^{-jwa}}{-jw} \right) + \frac{e^{-jw/2}}{jw} \end{align*}$$

You handle improper integrals by using limits, but the reason why you don't have zero is because the limit on the left doesn't converge for a general w. The complex function e^-jwa is a periodic function.

 

[HallsofIvy]

$e^{-i\omega t}= cos(\omega t)- i sin(\omega t)$ for real t. That does NOT go to 0 as t goes to infinity.

 

[khdani]

i see.
how do i calculate the final result?

 

[Hermit]

i see.
how do i calculate the final result?

I think you can use : $$\lim_{x \to + \infty} \frac{\cos(x)}{x}= \lim_{x \to + \infty} \frac{\sin(x)}{x}=0$$

 

[cipher42]

I think you can use : $$\lim_{x \to + \infty} \frac{\cos(x)}{x}= \lim_{x \to + \infty} \frac{\sin(x)}{x}=0$$

I don't think that helps in this case, since his limit is with respect to a, and there is no a in the denominator of the limited term