why does integral of exponent becomes 1 / something?

Question

Could anyone explain to me why at step 3, after integration it becomes $1/(a-jw)$?

Answer 1

Let $z=b+ic,b >0$ $$\int_{0}^{\infty} e^{-zt} dt=-\left .\frac{e^{-zt}}{z}\right|_{0}^{\infty}=\frac{1}{z}=\frac{1}{b+ic}.$$ As $e^{-zt}$ vanishes at $t=\infty$ if only $\Re(z)=b>0$$

Answer 2

Integral of exponent does not become 1/something. The basic rule is

$$ \int e^{ax} dx = \frac{e^{ax}}{a} +c ......->1$$

In the integrations you perform above,you integrate, for example, $e^{(a-j\omega)t}$ with respect to $dt$. Put $a-j\omega$ as $a$ above in 1. You just get the coefficient of the variable (wrt which you are performing integration) as the denominator. It happens in logarithmic and trigonometric integrations too. I think learning the basics of integration from any good book will help you.