# The Laplace Transform

### Derivation

We all know power series (* _{0}Σ^{∞}a(n)x^{n}*). We can use them to define functions (say

*A(x)*). But what if instead of only accepting things from the Whole Numbers (or the Natural Numbers including zero), we accepted all Real Numbers greater than 0? How would this turn out? Let’s figure this out!

We know that the integral sums up infinitely many rectangles over all of the real numbers to determine the area under a curve. This idea can be translated to summing up infinitely many things in general, instead of just rectangles. So, if we can all agree that integrals are a way to add up infinitely many things continuously over the real numbers, let’s integrate our function!

In our case, the *n* in our power series equation is only for Whole Number values, so we can’t use that variable. Say instead we use *t*, which is a continuous variable over the real numbers, instead of the discrete *n* we were using. This is the same issue with *a(n)*, so we can replace it with *f(t)*. Now, we get the equation * _{0}∫^{∞}f(t)x^{t}dt = F(x)*. Well, we don’t want to have that dirty

*x*there with an exponent, we want

*e*there, since it is much easier to work with. Let’s do that.

We know that *x ^{t} = e^{ln(x)t}*, knowing that

*0 < x < 1*, since negative values of

*x*returns something imaginary and

*x > 1*usually doesn’t converge (just for the sake of this argument, take this with a light heart). This means that

*ln(x) < 0*(since

*x < 1, ln(x) < 0*). In order to make the exponent positive, let’s make the substitution

*s = -ln(x)*, or

*-s = ln(x)*. We then get our final product:

_{0}∫^{∞}f(t)e^{-st}dt = F(s)

We call this the Laplace Transform, since it transforms one function (*f(t)*) into another (*F(s)*) with a different variable. We denote it with a cursive L, but I can’t find HTML for that symbol, so let’s just use *L*. We say that *L{f(t)}* is the integral above. We insert our *f(t)*, looking like *L{f(t)} = F(s)*.

For example, *L{1} = _{0}∫^{∞}e^{-st}dt = [-(1/s)*

*e*. This means that

^{-st}]_{0}^{∞}= (0 - (-1/s)) = 1/s*L{1} = 1/s = F(s)*. Now that you have the gist of what to do, here are some problem questions:

This is an article of which I will make a corresponding video for. If anyone sees this, please give some criticisms. Also, I will NOT do a derivation from a Fourier series, since I will introduce that in a different article.

That one guy that's always on