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


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


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

as we expected.

  • $\begingroup$ 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 '16 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 '16 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 '16 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 '16 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. $\endgroup$ – Josu Etxezarreta Martinez Aug 22 '16 at 16:02

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


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



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

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:

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


  • $\begingroup$ In other words, if $a_k=0$, then the RHS of equation $(3)$ is the FS of $X(t)$ ? $\endgroup$ – MBaz Aug 22 '16 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 '16 at 18:27
  • $\begingroup$ Agh, yes, I mean $a_0=0$, sorry for the typo. $\endgroup$ – MBaz Aug 22 '16 at 18:29
  • $\begingroup$ @Matt L.: Thanks again Matt. Your response was great and I have accepted it. $\endgroup$ – AllEs Aug 22 '16 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.