# $\int\exp(-jnw_0t)\,dt$ integral calculus.

I seem to forgot these parts of integral calculus. I am trying to determine the Fourier coefficient in complex exponential form.

Here, $t$ is the variable being integrated and $n$ is for all integers $\mathbb{Z}$

$$\int e^{-jnw_0t}\,dt = \begin{cases} -\frac{1}{jnw_0}e^{-jnw_0t + c} &\text{for } n \neq 0 \\[2ex] t + c &\text{for }n = 0 \end{cases}$$

Why isn't the answer to the general integral the former for $n \neq 0$? Why is there a constant added inside the exponential and not outside? Could my textbook be wrong?

When $n=0$, the integrand is identically $1$. In other cases, its a complex exponential with non-zero frequency (which when integrated gives a complex exponential with the same frequency).
• thanks, of course, $\int(1)=t$ – RohanLatinSindhi Jun 21 '14 at 5:07
I suppose that there is a typo somewhere since $$\int e^{-jnw_ot}dt = -\frac{1}{jnw_o}e^{-jnw_ot}+c$$ instead