In terms of linear algebra, functions can be represented as vector with infinite number of components (Hilbert's space as they say in physics). The set of function $e^{-st}$ forms a vector space, and Fourier and Laplace transform X(s) = $\int_{a}^{ \infty} x(t) e^{-st} $ is a projection of $x(t)$ to $e^{-st}$, i.e. an inner product with basis function $e^{-st}$. Here $a= -\infty$ for Fourier and $0$ for Laplace.

Now in the case of Fourier, your basis function is $e^{-iwt}$ with real $w$ (sine and cosine of frequency $w$). Thus Fourier transform is inner product of $x(t)$ with the basis function of a single frequency $w$.

In the case of Laplace, basis functions are $e^{-(\sigma + iw)t}$, i.e. $e^{-\sigma t} (cos(wt) + i sin(wt))$, Laplace transform is an inner product of $x(t)$ with the basis function of frequency $w$ and decays at the rate specified by $\sigma$.

The transform gives you a weight at  $s$, and the inverse transform is a linear combination of all the basis function $e^{st}$ in which forward transforms are weights, gives you back $x(t)$.

In terms of linear algebra, functions can be represented as vector with infinite number of components (in the Hilbert's space as they say in physics or $L2$ space in mathematics). The set of function $e^{-st}$ with complex $s$ forms a vector space, and Fourier and Laplace transform X(s) = $\int_{a}^{ \infty} x(t) e^{-st} $ is a projection of $x(t)$ to $e^{-st}$, i.e. an inner product with basis function $e^{-st}$. Here $a= -\infty$ for Fourier and $0$ for Laplace.

Now in the case of Fourier, your basis function is $e^{-iwt}$ with real $w$ (sine and cosine of frequency $w$). Thus Fourier transform is inner product of $x(t)$ with the basis function of a single frequency $w$.

In the case of Laplace, basis functions are $e^{-(\sigma + iw)t}$, with real $\sigma$ and $w$, i.e. $e^{-\sigma t} (cos(wt) + i sin(wt))$, Laplace transform is like an inner product of $x(t)$ with the basis function of frequency $w$ and decays at the rate specified by $\sigma$.

The transform gives you a weight at $s$, and the inverse transform is a linear combination of all the basis function $e^{st}$ in which forward transforms are weights, gives you back $x(t)$.

In terms of linear algebra, functions can be represented as vector with infinite number of components (Hilbert's space as they say in physics). The set of function $e^{-st}$ forms a vector space, and Fourier and Laplace transform X(s) = $\int_{a}^{ \infty} x(t) e^{-st} $ is a projection of $x(t)$ to $e^{-st}$, i.e. an inner product with basis function $e^{-st}$. Here $a= -\infty$ for Fourier and $0$ for Laplace.

Now in the case of Fourier, your basis function is $e^{-iwt}$ with real $w$ (sines and cosines of frequency $w$). Thus Fourier transform is inner product of $x(t)$ with the basis function of a single frequency $w$.

In the case of Laplace, basis functions are $e^{-(\sigma + iw)t}$, i.e. $e^{-\sigma t} (cos(wt) + i sin(wt))$, Laplace transform is an inner product of $x(t)$ with the basis function of frequency $w$ and decays at the rate specified by $\sigma$.

The inner product gives you a weight, and the inverse tranform is a linear combination of the basis vectors using forward transform as weights, gives you back $x(t)$.

In terms of linear algebra, functions can be represented as vector with infinite number of components (Hilbert's space as they say in physics). The set of function $e^{-st}$ forms a vector space, and Fourier and Laplace transform X(s) = $\int_{a}^{ \infty} x(t) e^{-st} $ is a projection of $x(t)$ to $e^{-st}$, i.e. an inner product with basis function $e^{-st}$. Here $a= -\infty$ for Fourier and $0$ for Laplace.

Now in the case of Fourier, your basis function is $e^{-iwt}$ with real $w$ (sine and cosine of frequency $w$). Thus Fourier transform is inner product of $x(t)$ with the basis function of a single frequency $w$.

In the case of Laplace, basis functions are $e^{-(\sigma + iw)t}$, i.e. $e^{-\sigma t} (cos(wt) + i sin(wt))$, Laplace transform is an inner product of $x(t)$ with the basis function of frequency $w$ and decays at the rate specified by $\sigma$.

The transform gives you a weight at $s$, and the inverse transform is a linear combination of all the basis function $e^{st}$ in which forward transforms are weights, gives you back $x(t)$.