Recipe

Compute a Laplace transform directly from the definition

Use when: you need \mathcal{L}\{f(t)\}(s) for an f(t) not yet in a lookup table, or you're deriving or verifying a table entry

Goal: a closed-form F(s), together with the values of s for which the improper integral actually converges

Taught in Definition and motivation

1. 1

   Write down \int_0^\infty f(t)e^{-st}\,dt with the given f(t) plugged in, and combine any exponentials sharing base e into a single exponent.

   Example

   For f(t)=e^{3t}: \int_0^\infty e^{3t}e^{-st}\,dt = \int_0^\infty e^{-(s-3)t}\,dt — combine the exponents by adding them, 3t-st=-(s-3)t.

   Why: this integral IS the transform, by definition — every later step is just mechanics for evaluating it.

2. 2

   Replace the upper bound \infty with a variable T and wrap the whole integral in \lim_{T\to\infty}, since \int_0^\infty is shorthand for that limit, not a bound you can plug a number into directly.

   Example

   \int_0^\infty e^{-(s-3)t}\,dt = \lim_{T\to\infty}\int_0^T e^{-(s-3)t}\,dt

   Why: skipping straight to "evaluate at \infty" hides exactly the question this step makes explicit — whether that limit even exists, which is where the transform's domain of validity comes from.

3. 3

   Find the antiderivative in t (treating s as a constant), apply the fundamental theorem of calculus over [0,T], then let T\to\infty and see what survives.

   Example

   \int_0^T e^{-(s-3)t}\,dt = \left[-\dfrac{1}{s-3}e^{-(s-3)t}\right]_0^T = -\dfrac{1}{s-3}e^{-(s-3)T}+\dfrac{1}{s-3}. Taking T\to\infty: if s>3, the exponent -(s-3)T\to-\infty, so e^{-(s-3)T}\to0, leaving \dfrac{1}{s-3}.

   • the coefficient of T in the exponent (after Step 1's combining) is negative — e^{(\cdots)T}\to0 as T\to\infty and the limit exists — this is the "s>" condition reported alongside the answer.
   • that coefficient is zero or positive — the exponential term stays at 1 or grows without bound, the limit fails to exist, and the transform simply doesn't exist for that s.

   Why: this is exactly where every transform's "s>\text{something}" restriction comes from — it isn't decoration, it's the actual convergence requirement uncovered by taking the limit honestly instead of skipping it.