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In the continuous Fourier series properties for a periodic continuous-time signal, we have time-integration property:

$$ \int_{-\infty}^t x(\alpha)d\alpha \leftrightarrow \frac{a_k}{jk\omega_0} $$

where $a_k$ is the Fourier series coefficients of the signal $x(t)$. Now I am confused. Is the above integral equivalent to $\int x(t)dt$ ?

For instance, the Fourier series coefficients of $\cos(\omega_0 t)$ are: $a_1=a_{-1}=\frac{1}{2}$, other $a_{k}=0$.

and the Fourier series coefficients of $\sin(\omega_0 t)$ are: $b_1=\frac{1}{2j}, b_{-1}=\frac{-1}{2j}$, other $b_{k}=0$.

and we have: $\int \cos(\omega_0t)dt=\frac{1}{\omega_0}\sin(\omega_0t)$

Then:

$$ \textrm{F.S.}\left\{\int \cos(\omega_0t)dt\right\} = \frac{a_k}{jk\omega_0} = \textrm{F.S.}\left\{\frac{1}{\omega_0}\sin(\omega_0t)\right\}= \frac{1}{\omega_0}b_k $$

Thus:

$$ b_k=\frac{a_k}{jk} $$

as we expected.

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  • $\begingroup$ I'm not sure what you're asking. Is it just the difference between $\int x(t) dt$ and $\int_{-\infty}^tx(\alpha)d\alpha$? $\endgroup$
    – Matt L.
    Aug 22, 2016 at 8:35
  • $\begingroup$ @Matt L.: Yes, my main question is this difference. But according to the Fourier series examples that I have mentioned in the question, it seems these two integral have the same meaning! $\endgroup$
    – AllEs
    Aug 22, 2016 at 9:54
  • $\begingroup$ @AllEs One integral is indefinite and the other is definite. How can they have the same meaning? $\endgroup$
    – MBaz
    Aug 22, 2016 at 14:04
  • $\begingroup$ @MBaz: I know, but the example demonstrates the time-integration property of the Fourier series on the indefinite integral! In addition, the definite integral cannot be calculated for the sine or the cosine signals directly. $\endgroup$
    – AllEs
    Aug 22, 2016 at 15:37
  • $\begingroup$ The fist integral would be solved as $\int_{-\infty}^{t}= x(\alpha)d\alpha = X(t) - \lim_{p\rightarrow -\infty}X(p)$, so the cosine function would not have a solution to that as it does not have a convergence limit towards $-\infty$. The indefinite integral just gives us the primitive function $X$ I have used in the last part to solve the integral, but the solution of the indefinite integral is $X + C$, where C is a constant of integration that is undefined as you do not know the limits of the integral. $\endgroup$ Aug 22, 2016 at 16:02

1 Answer 1

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Let's consider the complex form of the Fourier series of a $T$-periodic function $x(t)$:

$$x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}\tag{1}$$

with $\omega_0=2\pi/T$. The integration property says that the Fourier series of

$$y(t)=\int x(t)dt\tag{2}$$

is given by

$$y(t)=\sum_{k=-\infty}^{\infty}b_ke^{jk\omega_0t}\tag{3}$$

with

$$b_k=\frac{a_k}{jk\omega_0},\quad k\neq 0\tag{4}$$

where we require that $a_0=0$.

Note that in $(2)$ I used the indefinite integral, and I chose the integration constant to be zero, such that $b_0=0$. In some texts you may find the definite integral instead of the indefinite integral:

$$y(t)=\int_{-\infty}^{t} x(t)dt\tag{5}$$

The problem with $(5)$ is that for certain functions $x(t)$ the definite integral doesn't exist with $-\infty$ as the lower bound. Note that the lower bound in $(5)$ corresponds to the integration constant of the indefinite integral $(2)$. This becomes obvious if we consider the relation between the indefinite and the definite integral:

$$\int x(t)dt=X(t)+C\tag{6}$$

$$\int_{t_0}^tx(\tau)d\tau=X(t)-X(t_0)\tag{7}$$

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  • $\begingroup$ In other words, if $a_k=0$, then the RHS of equation $(3)$ is the FS of $X(t)$ ? $\endgroup$
    – MBaz
    Aug 22, 2016 at 18:23
  • $\begingroup$ @MBaz: If you mean $a_0=0$ then yes. If we define $X(t)$ to be zero-mean then $b_0=0$, otherwise we just have to adjust $b_0$. $\endgroup$
    – Matt L.
    Aug 22, 2016 at 18:27
  • $\begingroup$ Agh, yes, I mean $a_0=0$, sorry for the typo. $\endgroup$
    – MBaz
    Aug 22, 2016 at 18:29
  • $\begingroup$ @Matt L.: Thanks again Matt. Your response was great and I have accepted it. $\endgroup$
    – AllEs
    Aug 22, 2016 at 19:32

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