# Integral help

1. Jul 20, 2009

### 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 ?

2. Jul 20, 2009

### 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.

3. Jul 20, 2009

### HallsofIvy

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

4. Jul 21, 2009

### khdani

i see.
how do i calculate the final result?

5. Jul 21, 2009

### Hermit

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

6. Jul 21, 2009

### cipher42

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