Time differentiation and integration

When all instantaneous quantities in an equation are written in terms of phasors such as in (A.54), then $\mathrm{Re}\{...\}$ is dropped from all terms of the equation and subsequently the factor $\exp(j\omega t)$, that is common to all terms of the equation and is never zero, is dropped from all terms, a phasor equation is obtained in which all terms are complex and dependent on position, but independent of time. It is important to understand what differentiation and integration with respect to time become in the phasor domain. The partial derivative with respect to time of the electric field $\underline{\mathcal{ E}}(\underline{r},t)$ of (A.51) or (A.56) yields:

$$\frac{\partial}{\partial t} \underline{\mathcal{E}}(\underline{r}... ...\mathrm{Re}\{ j \omega \underline{E}(\underline{r}) e^{j \omega t } \} \space ,$$ (2.58)

hence the phasor of $\partial \underline{\mathcal{E}}/\partial t$ is $j \omega$ times the phasor of $\underline{\mathcal{E}}$. Time differentiation becomes multiplication by $j \omega$ in phasor domain:

$$\boxed{ \frac{\partial}{\partial t} \rightarrow j \omega \cdot }\space .$$ (2.59)

Similarly, it is easily seen that integration with respect to time becomes multiplication by $1/(j\omega)$ in phasor domain:

$$\boxed{ \int^t \, dt \rightarrow \frac{1}{j \omega} \cdot }\space .$$ (2.60)

We conclude that the phasor of the time derivative of a time-domain quantity is $j \omega$ times the phasor of that quantity, and that the phasor of the indefinite integral with respect to time of a time-domain quantity is $1/(j\omega)$ times the phasor of that quantity.