Laplace transform

For \(s\) large enough the improper integral converges whenever \(f\) is piecewise continuous and of exponential order, that is, \(|f(t)|\le Me^{at}\). That is why \(\mathcal L\{e^{at}\}=1/(s-a)\) holds only for \(s>a\), while \(e^{t^2}\) has no transform. The operation is linear: \(\mathcal L\{\alpha f+\beta g\}=\alpha F+\beta G\), which is enough to transform sums of basic functions.

The basic transforms come straight from the integral: \(\mathcal L\{1\}=1/s\), \(\mathcal L\{t^n\}=n!/s^{n+1}\), \(\mathcal L\{\sin bt\}=b/(s^2+b^2)\) and \(\mathcal L\{\cos bt\}=s/(s^2+b^2)\). Multiplying by \(e^{at}\) shifts \(s\) to \(s-a\), so \(\mathcal L\{te^{t}\}=1/(s-1)^2\). A common mistake is flipping the sign of the shift.

The payoff comes when solving equations: transform the whole equation, solve for \(F(s)\) with algebra, and return to \(f(t)\) via the inverse transform, using tables and partial fractions. Initial conditions enter the process automatically, avoiding the search for a general solution followed by fitting constants. It is especially convenient when the forcing term \(g(t)\) is discontinuous or an impulse.