Laplace Transform Table
A searchable table of Laplace transforms, with the properties and theorems that do the work.
Showing 31 of 31. Click any formula to copy it.
Every entry assumes f(t) = 0 for t < 0 and holds where the integral converges — for the exponential that means Re(s) > a. The derivative rules are where initial conditions enter a differential equation, which is what makes the transform worth using: an ODE in t becomes an algebraic equation in s.
The Laplace transform turns a function of time into a function of s, converting calculus into algebra. Differentiation becomes multiplication by s, so a differential equation becomes an equation you can solve by rearranging: f′(t) → sF(s) − f(0).
Why anyone bothers
Solving a linear differential equation directly means finding a complementary function, guessing a particular integral, and fitting the constants to the initial conditions. The Laplace transform replaces all of that with one mechanical route: transform the equation, solve the resulting algebra for F(s), and transform back.
The step that makes it work is the derivative rule. Because f′(t) transforms to sF(s) − f(0), the initial conditions enter the algebra automatically rather than being fitted at the end. A second-order equation becomes a quadratic in s with f(0) and f′(0) already built in, which is why the method suits initial value problems so particularly well.
The hard part is coming back
Going forward is a lookup. Coming back is where the work sits, because the F(s) you end up with is rarely in the table as it stands. Partial fractions is the standard move — split the expression into pieces that each match a known pair — followed by completing the square when the denominator has complex roots, which turns it into the damped sine and cosine forms.
defined for Re(s) large enough that the integral converges; f(t) is taken to be zero for t < 0
- 1 Transform every term. Apply the table to each side, using f′(t) → sF(s) − f(0) and f″(t) → s²F(s) − sf(0) − f′(0).
- 2 Substitute the initial conditions. They are already in the transformed derivatives, so they enter as numbers rather than unknown constants.
- 3 Solve algebraically for F(s). Everything is now polynomial in s, so this is rearranging rather than calculus.
- 4 Break F(s) into recognisable pieces. Partial fractions, then completing the square where the denominator has complex roots.
- 5 Read each piece back through the table. The inverse transform of the sum is the sum of the inverse transforms, which gives f(t).
The pairs used most
Each holds where the defining integral converges — for the exponential, Re(s) > a.
| f(t) | F(s) |
|---|---|
| 1 | 1 ÷ s |
| t | 1 ÷ s² |
| tⁿ | n! ÷ s^(n+1) |
| e^(at) | 1 ÷ (s − a) |
| sin(at) | a ÷ (s² + a²) |
| cos(at) | s ÷ (s² + a²) |
| e^(at)sin(bt) | b ÷ ((s − a)² + b²) |
| u(t − a) | e^(−as) ÷ s |
| δ(t) | 1 |
Details that catch people out
The sign in the hyperbolic pairs is the commonest slip: sin(at) gives a ÷ (s² + a²) but sinh(at) gives a ÷ (s² − a²). Reading the wrong one turns an oscillation into exponential growth. The second is forgetting that the transform only sees t ≥ 0 — every f(t) here is implicitly multiplied by the unit step, which is why a delayed function needs the second shifting theorem rather than a straight substitution.
The final value theorem carries a condition worth remembering: lim(t→∞) f(t) = lim(s→0) sF(s) only holds when that limit actually exists. Apply it to an undamped oscillation and it returns a number for something that never settles, so check the poles are in the left half-plane before trusting it.