# Why Fourier series has summation and Fourier transform has integration symbol in their respective formulae?

Fourier transform for aperiodic signal is given by

$$X(\omega) = \int\limits_{t=-\infty}^{+\infty} x(t) e^{-j \omega t} dt. \quad (1)$$

Fourier series for periodic signal is given by

$$y(t) = \sum\limits_{m=0}^{+\infty} a_m \cos(w_m t) + \sum\limits_{m=0}^{+\infty}b_m \sin(w_m t). \quad (2)$$

I want to ask that

Why Fourier series has summation and Fourier transform has integration symbol in their respective formulae although both $x(t)$ and $y(t)$ are continuous signals only? Is it just because one is used for aperiodic and one is for periodic signal?

• The Fourier transform and Fourier series are different things. There are connections (in functional analysis you deal with something called Pontryagin duality where you learn exactly how they're connected), but what you're asking is like asking "why is there a summation in Taylor series but an integral to find areas?". They're two different things. – user223391 Jun 4 '15 at 17:19
• your question was asked before , i think you can find an explanation here : math.stackexchange.com/questions/221137/… – mounir ben salem Jun 4 '15 at 17:24
• @mvw sir,1.in case of Laplace transform ->we multiplied $exp(-p.t)$ term by Fourier transform formula to obtain Laplace transform formula which is useful when we can't find Fourier transform of function which is continuously increasing with time. 2.for z transform -> it is first of all used only in case of discrete signals which is derived from Fourier transform multiplying it with radius $r$ 3.I don't know about wavelets – pandu Jun 4 '15 at 17:56

Another way is to realize the Fourier transform as a limit of Fourier series. Specifically, you can take a function and cut it off outside some large interval. Then you can take the Fourier series of the new function on the interval. Note that there are some questions about how this Fourier series converges if the new function is not periodic. But we are at least guaranteed convergence in $L^2$ if the original function was $L^2$.