What am I doing wrong when I try to deduce the Laplace transform formula? The Laplace transform of a function $f(t)$ is the projection of $f(t)$ vector (indexed with $t$) onto the linearly independent set of vectors $e^{st}$. The projection of a vector $\vec{v}$ onto another vector $\vec{w}$ is$$\vec{v}_{\vec{w}} = {{\vec{v} \cdot \vec{w}}\over{\vec{w} \cdot \vec{w}}}.$$Therefore,$$f(t)_{e^{st}} = {{f(t) \cdot e^{st}}\over{e^{st} \cdot e^{st}}} = \lim_{dt \to \infty} {{\sum_{k = 0}^\infty f(dt \cdot k) \cdot e^{e \cdot dt \cdot k}}\over{\sum_{k = 0}^\infty e^{2s \cdot dt \cdot k}}} = {{\int_0^\infty f(t) \cdot e^{st} \cdot dt}\over{\int_0^\infty e^{2st}\cdot dt}},\tag*{$(1)$}$$and this should be equal to$$\int_0^\infty f(t) \cdot e^{-st} \cdot dt.\tag*{$(2)$}$$If I am right up to this point, how can I prove that $(1)$ equals $(2)$? And if I did make a mistake somewhere, where?
 A: Actually, your heart is in the right place; you just have the wrong inner product in mind. Instead of taking the inner product of $f$ and $g$ to be$$\int_0^\infty f(t)g(t)\,dt,$$you should take it to be the average value of $f(t)g(-t)$, in an appropriate sense; this will be proportional to$$\int_{-\infty}^\infty f(t)g(-t)\,dt,$$and thus you can use this in the numerator instead of the inner product itself, as though everything were just uniformly rescaled.
You should think of the restriction to positive values as part of the definition of $f$, but not part of the basis vectors $t \mapsto e^{st}$ or part of the definition of the relevant inner product.
It will be obvious that these vectors are of unit magnitude on this definition. However, asserting their orthogonality may make you uncomfortable, as the relevant integrals will not be convergent in the ordinary sense. But there are natural senses in which the average value of $t \mapsto e^{(s_1 - s_2)t}$ should be considered zero whenever $s_1 - s_2$ is nonzero; for one thing, this average value should obviously be unchanged when shifting $t$, and thus when multiplying $e^{s_1 - s_2}$, which only zero would be. If that is unsatisfactory for you, you might instead interpret our averages as taken along the imaginary axis rather than the real one; this amounts to just rotating into the Fourier transform.
