# What is the time-integration property in the Fourier series analysis?

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.

• I'm not sure what you're asking. Is it just the difference between $\int x(t) dt$ and $\int_{-\infty}^tx(\alpha)d\alpha$? Aug 22, 2016 at 8:35
• @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! Aug 22, 2016 at 9:54
• @AllEs One integral is indefinite and the other is definite. How can they have the same meaning?
– MBaz
Aug 22, 2016 at 14:04
• @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. Aug 22, 2016 at 15:37
• 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. Aug 22, 2016 at 16:02

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:

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

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

• In other words, if $a_k=0$, then the RHS of equation $(3)$ is the FS of $X(t)$ ?
– MBaz
Aug 22, 2016 at 18:23
• @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 $b_0$. Aug 22, 2016 at 18:27
• Agh, yes, I mean $a_0=0$, sorry for the typo.
– MBaz
Aug 22, 2016 at 18:29
• @Matt L.: Thanks again Matt. Your response was great and I have accepted it. Aug 22, 2016 at 19:32