How to Find the Inverse Laplace Transform of 1/[s²(s² + ω²)]

For the usual Laplace-transform convention with time variable $t\ge 0$, the inverse transform can be found directly by partial fractions. The apparent difficulty is that the denominator contains both a repeated factor, $s^2$, and a quadratic factor, $s^2+\omega^2$.

Assume first that $\omega\ne 0$. The result is

$$\mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+\omega^2)}\right\} =\frac{t}{\omega^2}-\frac{\sin(\omega t)}{\omega^3}.$$

The case $\omega=0$ must be handled separately: then the transform is $1/s^4$, whose inverse is $t^3/6$.

Partial fractions

Set the rational function equal to a sum of terms with the same denominator factors:

$$\frac{1}{s^2(s^2+\omega^2)} =\frac{A}{s^2}+\frac{B}{s^2+\omega^2}.$$

Multiplying through by $s^2(s^2+\omega^2)$ gives

$$1=A(s^2+\omega^2)+Bs^2 =A\omega^2+(A+B)s^2.$$

Matching the constant term and the coefficient of $s^2$ gives $A\omega^2=1$ and $A+B=0$. Therefore $A=1/\omega^2$ and $B=-1/\omega^2$, so

$$\frac{1}{s^2(s^2+\omega^2)} =\frac{1}{\omega^2}\left(\frac{1}{s^2}-\frac{1}{s^2+\omega^2}\right).$$

This decomposition is valid when $\omega\ne0$. It is already arranged in terms of standard Laplace-transform pairs.

Apply the transform pairs

The relevant pairs are

$$\mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\}=t, \qquad \mathcal{L}^{-1}\left\{\frac{1}{s^2+\omega^2}\right\} =\frac{\sin(\omega t)}{\omega}.$$

Using linearity and retaining the factor $1/\omega^2$ gives

$$\begin{aligned} f(t) &=\frac{1}{\omega^2}\left(t-\frac{\sin(\omega t)}{\omega}\right)\\ &=\frac{t}{\omega^2}-\frac{\sin(\omega t)}{\omega^3}. \end{aligned}$$

The extra factor of $1/\omega$ in the sine term comes from its transform pair. It is easy to miss when applying the pair to $1/(s^2+\omega^2)$.

Worked example: $\omega=3$

Substitute $3$ for $\omega$ in the decomposition and inverse-transform each term:

$$\begin{aligned} \frac{1}{s^2(s^2+9)} &=\frac{1}{9}\left(\frac{1}{s^2}-\frac{1}{s^2+9}\right),\\ \mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+9)}\right\} &=\frac{1}{9}\left(t-\frac{\sin(3t)}{3}\right)\\ &=\frac{t}{9}-\frac{\sin(3t)}{27}. \end{aligned}$$

To check, transform the two time-domain terms separately. The transform of $t/9$ is $1/(9s^2)$, and the transform of $\sin(3t)/27$ is $1/[9(s^2+9)]$. Their difference is

$$\frac{1}{9s^2}-\frac{1}{9(s^2+9)} =\frac{1}{s^2(s^2+9)},$$

as required.

An independent check using convolution

The original expression is a product of $1/s^2$ and $1/(s^2+\omega^2)$. Their inverse transforms are $t$ and $\sin(\omega t)/\omega$, respectively. The product rule for Laplace transforms says that the inverse of the product is their convolution:

$$f(t)=\int_0^t (t-u)\frac{\sin(\omega u)}{\omega}\,du.$$

Evaluating the integral gives $t/\omega^2-\sin(\omega t)/\omega^3$, agreeing with the partial-fraction result. This provides another route if you are studying the convolution theorem, though partial fractions are shorter here.

The special case $\omega=0$

When $\omega=0$, do not substitute into the partial-fraction coefficients: they contain division by $\omega^2$. Instead simplify the original expression first:

$$\frac{1}{s^2(s^2+0)}=\frac{1}{s^4}, \qquad \mathcal{L}^{-1}\left\{\frac{1}{s^4}\right\}=\frac{t^3}{3!}=\frac{t^3}{6}.$$

This is also the limit of the nonzero-$\omega$ answer. Expanding the sine near zero gives $\sin(\omega t)=\omega t-(\omega t)^3/6+\cdots$, so the leading terms cancel and the expression approaches $t^3/6$.

Common mistakes

• Omitting the sine-pair factor: $1/(s^2+\omega^2)$ transforms to $\sin(\omega t)/\omega$, not just $\sin(\omega t)$.
• Using the wrong sign: the coefficient of $1/(s^2+\omega^2)$ is negative. The decomposition must reproduce the original fraction when recombined.
• Applying the formula at $\omega=0$ without a limit: the displayed coefficients are undefined there, even though the original transform has a well-defined value.
• Confusing $\omega$ with $\omega^2$ in the sine argument: the standard pair uses $s^2+\omega^2$ in the denominator and $\sin(\omega t)$ in time.