# Transforming piecewise continuous functions

• djeitnstine

#### djeitnstine

Gold Member
I was just reflecting upon my math courses and wondered why can we transform any piecewise continuous functions by using transforms such as laplace transforms or converting to Fourier series by simply adding the required integrals on the respective bounds?

I was just reflecting upon my math courses and wondered why can we transform any piecewise continuous functions by using transforms such as laplace transforms or converting to Fourier series by simply adding the required integrals on the respective bounds?

Your statement is a little confusing. In any case, continuity is not necessary for Laplace transforms, which are integrals from 0 to oo, while Fourier series are obtained by integrals over a finite interval. The function has to meet certain conditions related to integrability, but continuity is not one of them.

Yes but why can the integrals simply be added in both cases?

Example:

$$f(t)= \left\{^{5, t<1}_{sin(t), t>1}$$ So the laplace transform of this would be

$$L\left\{ f(t) \right\} = \int_0^1 5e^{-st}dt + \int_1^{\infty} sin(t)e^{-st}dt$$

Why can they just be added?

Last edited:
Yes but why can the integrals simply be added in both cases?

Example:

$$f(t)= \left\{^{5, t<1}_{sin(t), t>1}$$ So the laplace transform of this would be

$$L\left\{ f(t) \right\} = \int_0^1 5e^{-st}dt + \int_1^{\infty} sin(t)e^{-st}dt$$

Why can they just be added?

Well.

$$L\left\{ f(t) \right\} = \int_0^{\infty} e^{-st} f(t) dt$$

$$L\left\{ f(t) \right\} = \int_0^1 f(t)e^{-st}dt + \int_1^{\infty} f(t)e^{-st}dt$$

[because the integration operator is LINEAR]

which then becomes

$$L\left\{ f(t) \right\} = \int_0^1 5e^{-st}dt + \int_1^{\infty} sin(t)e^{-st}dt$$

Hmm. I should be hitting myself in the head.