Inverse Laplace transform

The inverse Laplace transform recovers $f(t)$ from its Laplace transform $F(s)$:

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

If $F(s)=\mathcal{L}\{f(t)\}(s)$, then ${\mathcal{L}}^{-1}\{F(s)\}=f(t)$. The two operations are inverses.

In practice, you don’t compute the inverse from a definition — you recognize the $F(s)$ as one of the standard forms (or a combination via partial fractions, shifting, etc.) and read off $f(t)$ from a table.

Linearity

Like the forward transform, the inverse is linear:

${\mathcal{L}}^{-1}\{c_{1}F(s)+c_{2}G(s)\}=c_{1}f(t)+c_{2}g(t)$

So you can decompose $F(s)$ into pieces, invert each piece, and combine.

Standard inverse pairs

$F(s)$	${\mathcal{L}}^{-1}\{F(s)\}(t)$
$1/s$	$1$
$1/s^n$	$t^{n-1}/(n-1)!$
$1/(s-a)$	$e^{at}$
$1/(s-a{)}^k$	$t^{k-1}{e}^{at}/(k-1)!$
$\omega /(s^2+{\omega }^2)$	$\sin (\omega t)$
$s/(s^2+{\omega }^2)$	$\cos (\omega t)$
$\omega /((s-a{)}^2+{\omega }^2)$	$e^{at}\sin (\omega t)$
$(s-a)/((s-a{)}^2+{\omega }^2)$	$e^{at}\cos (\omega t)$
$e^{-cs}F(s)$	$u(t-c)f(t-c)$
$e^{-cs}/s$	$u(t-c)$ (Heaviside step function)
$e^{-cs}$	$\delta (t-c)$ (Dirac delta function)

Partial fractions

The standard technique for inverting rational $F(s)$. Decompose into a sum of simpler fractions you can invert directly.

For $F(s)=\frac{N(s)}{(s-a)(s-b)\cdots }$ (distinct linear factors):

$F(s)=\frac{A}{s-a}+\frac{B}{s-b}+\cdots$

Solve for $A,B,\dots$ by clearing denominators and matching coefficients (or using the cover-up method).

For repeated factors $(s-a{)}^k$, you get terms $\frac{A_1}{s-a}+\frac{A_2}{(s-a{)}^2}+\cdots +\frac{A_k}{(s-a{)}^k}$.

For irreducible quadratic factors $s^2+ps+q$, complete the square to write as $(s-\alpha {)}^2+{\beta }^2$, then use the shifted sine/cosine forms above.

Worked example 1: simple partial fractions

Invert $\frac{2}{s^2+3s-4}$.

Factor: $s^2+3s-4=(s+4)(s-1)$. Decompose:

$\frac{2}{(s+4)(s-1)}=\frac{A}{s+4}+\frac{B}{s-1}$

Multiply through: $2=A(s-1)+B(s+4)$. Cover-up at $s=-4$: $2=-5A$, $A=-2/5$. At $s=1$: $2=5B$, $B=2/5$.

So:

${\mathcal{L}}^{-1}\left\{\frac{2}{(s+4)(s-1)}\right\}=-\frac{2}{5}{e}^{-4t}+\frac{2}{5}{e}^t$

Worked example 2: shifted denominator

Invert $\frac{1}{(s-1{)}^2}$.

Recognize as $F(s-1)$ where $F(s)=1/s^2$. By the first shifting theorem:

${\mathcal{L}}^{-1}\{F(s-a)\}=e^{at}{\mathcal{L}}^{-1}\{F(s)\}$

${\mathcal{L}}^{-1}\{1/s^2\}=t$. So ${\mathcal{L}}^{-1}\{1/(s-1{)}^2\}=e^t\cdot t=t{e}^t$.

Worked example 3: complete the square

Invert $\frac{6s^2+28}{(s^2-2s+5)(s+2)}$.

Complete the square in the quadratic factor: $s^2-2s+5=(s-1{)}^2+4=(s-1{)}^2+2^2$.

Decompose into partial fractions matching the structure:

$\frac{6s^2+28}{((s-1{)}^2+4)(s+2)}=\frac{A(s-1)+2B}{(s-1{)}^2+4}+\frac{C}{s+2}$

The form $A(s-1)+2B$ is chosen because it matches the inverse forms: $(s-1)/((s-1{)}^2+4)$ inverts to $e^t\cos (2t)$, and $2/((s-1{)}^2+4)$ inverts to $e^t\sin (2t)$.

Solve for $A,B,C$ by clearing denominators and matching coefficients. Multiply through by the full denominator:

$6s^2+28=(A(s-1)+2B)(s+2)+C((s-1{)}^2+4)$

Substitute $s=-2$ (the root of the second denominator) to isolate $C$: $6(4)+28=0+C(9+4)$, so $52=13C$ and $C=4$.

Expand the remaining terms and match coefficients of $s^2,s^1,s^0$. The $s^2$ coefficients give $A+C=6$, so $A=2$. The $s^0$ coefficients give $-2A+4B+4C\cdot 1-2\cdot 0\cdot 1+4C\cdot 1\dots$ — easier to substitute $s=1$ (root of the quadratic’s “shifted form”) and read off: $6+28=0+C\cdot 4$, which redundantly confirms $C$. Substituting $s=0$ gives $28=(-A+2B)(2)+C(5)=-2A+4B+5C$, so $28=-4+4B+20$, giving $B=3$.

So $A=2$, $B=3$, $C=4$.

So:

$F(s)=\frac{2(s-1)}{(s-1{)}^2+4}+\frac{6}{(s-1{)}^2+4}+\frac{4}{s+2}$

For the first piece: ${\mathcal{L}}^{-1}\{2(s-1)/((s-1{)}^2+4)\}=2e^t\cos (2t)$.

For the second: ${\mathcal{L}}^{-1}\{6/((s-1{)}^2+4)\}={\mathcal{L}}^{-1}\{3\cdot 2/((s-1{)}^2+4)\}=3e^t\sin (2t)$.

For the third: ${\mathcal{L}}^{-1}\{4/(s+2)\}=4e^{-2t}$.

Total: $f(t)=2e^t\cos (2t)+3e^t\sin (2t)+4e^{-2t}$.

Bromwich integral (formal definition)

The inverse Laplace transform has a formal definition as a contour integral in the complex plane:

$f(t)=\frac{1}{2\pi i}{\int }_{\gamma -i\infty }^{\gamma +i\infty }{e}^{st}F(s)ds$

where $\gamma$ is chosen larger than the real part of any singularity of $F(s)$. This is the Bromwich integral.

In practice, nobody computes inverse Laplace transforms this way. Tables and partial fractions are vastly more practical.

When you can’t invert

If $F(s)$ doesn’t decompose into recognizable forms, you may need:

• More aggressive partial fraction work.
• Series expansion of $F(s)$ and inverting term by term.
• Numerical inversion (advanced, beyond this course).

For the forward transform context, see Laplace transform. For its use in solving ODEs end-to-end, see Method of Laplace transform. For the full partial-fraction technique (including repeated and complex-conjugate poles), see Partial fraction decomposition. For improper rational functions that require polynomial division first, see Polynomial division for improper rational functions.