Inverse Laplace Calculator

The inverse Laplace transform takes a function of $s$ back to the time domain:

$\mathcal{L}^{-1}\{F(s)\} = f(t) \quad \Longleftrightarrow \quad \mathcal{L}\{f(t)\} = F(s)$

There is a contour-integral formula, but in practice nobody uses it. The working method is algebraic: reshape $F(s)$ until every piece matches a known transform, then read the table backwards. The result is unique wherever $f$ is continuous, so the answer you find is the answer.

The inverse is linear:

$\mathcal{L}^{-1}\{aF(s) + bG(s)\} = a\,\mathcal{L}^{-1}\{F(s)\} + b\,\mathcal{L}^{-1}\{G(s)\}$

A proper rational $F(s)$ — numerator degree strictly below the denominator's — always has an inverse built from exponentials, sines, cosines and powers of $t$.

The reverse table

$\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1, \quad \mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\} = e^{at}, \quad \mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\} = t^n$

$\mathcal{L}^{-1}\left\{\frac{a}{s^2+a^2}\right\} = \sin at, \quad \mathcal{L}^{-1}\left\{\frac{s}{s^2+a^2}\right\} = \cos at$

The procedure

1. Factor the denominator.
2. Split with partial fractions — distinct linear factors give $\frac{A}{s-a}$ terms, repeated factors need $\frac{A}{(s-a)^2}$ as well, and an irreducible quadratic needs a linear numerator $\frac{As+B}{s^2+bs+c}$.
3. Complete the square on any irreducible quadratic, turning $s^2+2s+5$ into $(s+1)^2+4$.
4. Apply the shift theorem: $\mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t)$. A shifted $s$ becomes an exponential factor in $t$.
5. Match the numerator to the table form — a sine needs the constant $a$ on top, so scale rather than guess.

Initial-value problems

Transform the equation using $\mathcal{L}\{y'\} = sY(s) - y(0)$, solve the resulting algebra for $Y(s)$, then invert. The initial conditions enter at the transform stage, so no arbitrary constants ever appear.

• Splitting a quadratic denominator into linear pieces: $\frac{1}{s^2+4}$ does not decompose over the reals. Leave irreducible quadratics intact.
• Forgetting to scale for sine: $\mathcal{L}^{-1}\left\{\frac{6}{s^2+9}\right\} = 2\sin 3t$, because the table entry needs a $3$ in the numerator.
• Shifting only part of the expression: after completing the square, every $s$ in the numerator must be rewritten in terms of $s + a$ too.
• Applying partial fractions to an improper fraction: divide first if the numerator's degree is at least the denominator's.
• Adding a constant of integration: the inverse transform yields one specific function; there is no $+C$.