# Preface

In this section, we discuss the inverse Laplace transform procedures. In particular, it presents the partial fraction decomposition method. Two other methods (convolution rule and the residue theorem) are presented in the following sections.

# Inverse Laplace transform

Let F(λ) be the Laplace transform of some function. This means that there exists a function f(t), defined on half-line, such that

$$\label{EqInverse.1} F(\lambda ) = f^L (\lambda ) = \left[ {\cal L} f \right] \left(\lambda \right) = \int_0^{\infty} f(t)\, e^{-\lambda\,t} \,{\text d}t .$$
For a given function F(λ) of a complex variable λ that is analytic in a half plane Reλ > s for some real s ∈ ℝ, determination of a function f(t) whose Laplace transform is F(λ) is called the inverse Laplace transform; it is usually denoted as $${\cal L}^{-1} \left[ F(\lambda ) \right] = f(t) ,$$ so F(λ) and f(t) are related via Eq.\eqref{EqInverse.1}. Actually, determination of the inverse Laplace transformation is equivalent to solving the integral equation of the first kind \eqref{EqInverse.1}. It is known that such integral equation is an ill-posed problem, and its solution usually requires a regularization. It was the English mathematician Thomas John I'Anson Bromwich (1875--1929) who gave a regularization for the inverse Laplace transform that can be expressed as the contour integral:
\begin{align} f(t) &= {\cal L}^{-1} \left[ F(\lambda ) \right] = \mbox{P.V. } \frac{1}{2\pi{\bf j}} \, \int_{s-{\bf j}\infty}^{s+{\bf j}\infty} \, F(\lambda )\, e^{\lambda \,t} \,{\text d} \lambda = \lim_{N\to \infty} \frac{1}{2\pi{\bf j}} \, \int_{s-{\bf j}N}^{s+{\bf j}N} \, F(\lambda )\, e^{\lambda \,t} \,{\text d} \lambda \label{EqInverse.2} \\ & \notag \\ &= \frac{f(t+0) + f(t-0)}{2} = \frac{e^{st}}{2\pi} \left[ \int_0^{\infty} e^{{\bf j}\xi t} \,F(s+ {\bf j}\xi )\,{\text d} \xi + \int_0^{\infty} e^{{\bf j}\xi t} \,F(s- {\bf j}\xi )\,{\text d} \xi \right] \qquad ({\bf j}^2 = -1), \notag \end{align}
where the abbreviation P.V. stands for the Cauchy principal value, which indicates that the symmetrical regularization of the improper integration is done along the vertical line Reλ = s in the complex plane such that s is greater than the real part of all singularities of F(λ) and F(λ) is bounded on the line of integration. This integral is usually referred to as the Bromwich integral. The real number s in the Bromwich integral should exceed the abscissa of convergence for the function f(t). Formula \eqref{EqInverse.2} allows us to make the following observations.

Theorem 1: Let f(t) be a function of bounded variation on interval [0, ∞) that satisfies the Dirichlet conditions on every finite subinterval. Suppose that f(t) is of exponential order then its Laplace transform $$F(\lambda ) = f^L (\lambda ) = \int_0^{\infty} e^{-\lambda t} f(t)\,{\text d}t$$ exists and has the abscissa of convergence σ∈ℝ. Then for any s > σ, we have
• The inverse Laplace transform is identically zero for negative t:
${\cal L}^{-1} \left[ F(\lambda ) \right] (t) \equiv 0 \quad \mbox{ for } \quad t < 0.$
• The Bromwich integral with s > σ restores the function in such a way that at the point of discontinuity t0 of the function f(t), the Bromwich integral assigns its mean value:
$$\label{EqInverse.3} {\cal L}^{-1} \left[ F(\lambda ) \right] \left( t_0 \right) = \frac{f(t_0 +0) + f(t_0 -0)}{2} = \frac{1}{2}\,\lim_{\epsilon \to 0, \ \epsilon > 0} \, \left[ f(t_0 + \epsilon ) + f(t_0 - \epsilon ) \right] .$$
• $\frac{1}{2}\, f(0^{+} ) = \frac{1}{2}\, f(0+) = \mbox{P.V.}\,\frac{1}{2\pi{\bf j}} \, \int_{s - {\bf j}\infty}^{s + {\bf j}\infty} f^L (\lambda )\,{\text d}\lambda = \lim_{N\to +\infty} \frac{1}{2\pi{\bf j}} \, \int_{s - {\bf j}N}^{s + {\bf j}N} f^L (\lambda )\,{\text d}\lambda \qquad ({\bf j}^2 =-1).$

Remark: There are known three possible notations for a right-sided limit: f(t+) or f(t+0) or f(t+); correspondingly, the left-sided limit is denoted as f(t-0) or f(t-) or f(t-).

The inverse Laplace transform, if it exists, should be multiplied by the Heaviside function.

Example 2: The inverse Laplace transform always assign the half of the limit value at the origin. For example, the Laplaced transform of constant unit function is
${\cal L} \left[ \, 1 \,\right] = \int_0^{\infty} 1 \cdot e^{-\lambda t}{\text d} t = \int_0^{\infty} e^{-\lambda t}{\text d} t = \frac{1}{\lambda} .$
However,when inverse Laplace transform is applied to 1/λ, it results in
${\cal L}^{-1}_{\lambda \to t} \left[ \, \frac{1}{\lambda} \,\right] = H(t) = \begin{cases} 1 , & \ \mbox{ for} \quad t > 0 , \\ ½ , & \ \mbox{ for} \quad t = 0, \\ 0, & \ \mbox{ for} \quad t < 0. \end{cases}$

As another example, we consider a familiar cosine function. Its Laplace transform is

${\cal L}_{t\to\lambda} \left[ \, \cos t \,\right] = \int_0^{\infty} \cos t \cdot e^{-\lambda t}{\text d} t = \int_0^{\infty} \cos t\,e^{-\lambda t}{\text d} t = \frac{\lambda}{\lambda^2 + 1} .$
Application of the Inverse Laplace transform yields another function
${\cal L}^{-1}_{\lambda \to t} \left[ \, \frac{\lambda}{\lambda^2 + 1} \,\right] = f(t) = \begin{cases} \cos t , & \ \mbox{ for} \quad t > 0 , \\ ½ , & \ \mbox{ for} \quad t = 0, \\ 0, & \ \mbox{ for} \quad t < 0. \end{cases}$

So we see that the inverse laplace transform actually multiplies the cosine function by the Heaviside function.

End of Example 1

Although the inverse Laplace transform formula via indefinite Bromwich integral may seem intimidating, we will be showing that there are much easier methods to obtain the result for the inverse transform of a function. We focus on the practical approaches of finding the inverse Laplace transformation.

We observe that applications of the Laplace transformation to ordinary differential equations lead to functions F(λ) of a complex variable λ that are represented either by the ratio of two polynomials or by such ratio times the exponential function:

$F(\lambda ) = \frac{P(\lambda )}{Q(\lambda )} \qquad\mbox{or} \qquad F(\lambda ) = \frac{P(\lambda )}{Q(\lambda )} \, e^{-a\lambda} ,$
where $$P(\lambda ) = p_n \lambda^n + p_{n-1} \lambda^{n-1} + \cdots + p_0 \quad\mbox{and} \quad Q(\lambda ) = q_m \lambda^m + q_{m-1} \lambda^{m-1} + \cdots + q_0 .$$ In applications, the degree m of the denominator always exceeds n, the degree the numerator. Later, we will consider a special case when m = n.

It should be noted that the latter case (when ratio of two polynomials is multiplied by the exponential term) can be reduced to the former one (without exponential multiple) with the aid of the shift rule:

$$\label{EqInverse.4} {\cal L}^{-1} \left[ \frac{P(\lambda )}{Q(\lambda )} \, e^{-a\lambda} \right] = f(t-a)\, H(t-a) , \qquad \mbox{where} \quad f(t) = {\cal L}^{-1} \left[ \frac{P(\lambda )}{Q(\lambda )} \right] .$$
Therefore, we concentrate our attention on determination of the inverse Laplace transform of the rational functions
$$\label{EqInverse.5} F(\lambda ) = \frac{P(\lambda )}{Q(\lambda )} .$$

Partial Fraction Decomposition

One of the methods to find the inverse Laplace transform of a given fracion \eqref{EqInverse.5} is to use partial fraction decomposition (or expansion), a familair topic from calculus. This allows one to represent a ratio of two polynomials, Eq.\eqref{EqInverse.5}, as a sum of simple fractions, with denominators having either a single zero or a pair of complex conjugate zeroes. Every term in the sum has a simple form because the degree of numerator is strictly less than degree of corresponding denominator. Since the inverse Laplace transform of each term in the decomposition sum is known and tabulated, the restoration of the inverse Laplace transform of the required fraction \eqref{EqInverse.5} becomes straight forward---it is expressed as a sum of known functions.

The advantage of this apporach is obvious---we avoid application of the Bromwich integral \eqref{EqInverse.2} and its regularization because finding the inverse Laplace transform is an ill-posed problem. Application of partial fraction decomposition has a drawback of performing tedious calculations and being supplied with a table of Laplace transforms of elementary functions. Fortunately, a computer algebra system such as Mathematica weakens the majority of these inconveniences.

Theorem 2: Suppose that the ratio of two relatively prime polynomials is
$F(\lambda ) = \frac{P(\lambda )}{\left( \lambda - \lambda_1 \right)\left( \lambda - \lambda_2 \right) \cdots \left( \lambda - \lambda_n \right)}$
where λ1, λ2, …, λn are distinct real numbers and P(λ) is a polynomial of degree less than n. Then
$F(\lambda ) = \frac{a_1}{\lambda - \lambda_1} + \frac{a_2}{\lambda - \lambda_2} + \cdots + \frac{a_n}{\lambda - \lambda_n} ,$
where 𝑎k, k = 1, 2, …, n, can be computed from F(λ) by ignoring the factor λ−λk and setting λ = λk elsewhere.
Example 2: Find the inverse Laplace transform of
$F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{\lambda^3 + 5\lambda^2 + 2 \lambda -8} . \tag{2.1}$
First, we factor the denominator
$\lambda^3 + 5\lambda^2 + 2 \lambda -8 = \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right) . \tag{2.2}$
Factor[s^3 + 5 s^2 + 2 s $Minus] 8] (-1 + s) (2 + s) (4 + s) Then using partial fraction decomposition, we represent the function (1.1) as a sum of simple terms \[ F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{\lambda^3 + 5\lambda^2 + 2 \lambda -8} = \frac{a}{\lambda +4} + \frac{b}{\lambda +2} + \frac{c}{\lambda -1} . \tag{2.3}$
Apart[1/(s^3 + 5 s^2 + 2 s $Minus] 8)] 1/(15 (-1 + s)) - 1/(6 (2 + s)) + 1/(10 (4 + s)) To find the values of coefficients 𝑎, b, and c, we have two options. First, we utilize Theorem 2. So we ignore the factor λ+4 in the denominator and set λ = −4. This yields \[ a = \frac{6 + 2(-4) -3(-4)^2}{(-4+2)(-4-1)} = \frac{-50}{10} =5.$
Similarly, the other coefficients are given by
$b = \frac{6 +2(-2) -3(-2)^2}{(4-2)(-1-2)} = \frac{-10}{-6} = \frac{5}{3},$
and
$c = \frac{6 +2(1) -3(1)^2}{(1+4)(1+2)} = \frac{5}{15} = \frac{1}{3}$
This gives
$F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{\lambda^3 + 5\lambda^2 + 2 \lambda -8} = \frac{-5}{\lambda +4} + \frac{5/3}{\lambda +2} + \frac{1/3}{\lambda -1} . \tag{1.4}$
We check with Mathematica
Apart[(6 + 2*s - 3*s^2)/(s + 4)/(s + 2)/(s - 1)]
1/(3 (-1 + s)) + 5/(3 (2 + s)) - 5/(4 + s)
Another option to find the values of these coefficients is to multiply Eq.(2.3) by the denominator:
$6 + 2\lambda -3 \lambda^2 = a\left( \lambda +2 \right)\left( \lambda -1 \right) + b \left( \lambda +4 \right)\left( \lambda -1 \right) + c \left( \lambda +2 \right)\left( \lambda +4 \right) .$
Upon setting in this equation λ = −4, we get
$6 + 2(-4) -3 (-4)^2 = a\left( -4 +2 \right)\left( -4 -1 \right) \qquad \Longrightarrow \qquad a=5.$
In a similar way, we determine the values of coefficients b and c. Using Mathematica, we find the inverse Laplace transform:
InverseLaplaceTransform[(6 + 2*s - 3*s^2)/(s + 4)/(s + 2)/(s - 1), s, t]
1/3 E^(-4 t) (-15 + 5 E^(2 t) + E^(5 t))
Collecting all terms, we find
${\cal L}^{-1} \left[ \frac{6 + 2\lambda -3 \lambda^2}{\lambda^3 + 5\lambda^2 + 2 \lambda -8} \right] = \left[ \frac{1}{3}\, e^t + \frac{5}{3}\,e^{-2t} - 5\, e^{-4t} \right] H(t) ,$
where H(t) is the Heaviside function.    ■
Theorem 3: Let Q(λ) be a polynomial of degree n ≥ 1, and P(λ) be a polynomial of degree < n. Then the number of arbitrary constants in the partial fraction expansion of P(λ)/Q(λ) equals degree n of the denominator.
Example 2: Find the inverse Laplace transform of
$F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^3} . \tag{2.1}$
We expand this function into the sum of simple terms:
$F(\lambda ) = \frac{a}{\lambda +4} + \frac{b}{\lambda +2} + \frac{c_1}{\lambda -1} + \frac{c_2}{(\lambda -1)^2} + \frac{c_3}{(\lambda -1)^3} , \tag{2.2}$
which has five constants. Using the common denominator, we get
$\frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^3} = \frac{a \left( \lambda +2 \right) \left( \lambda -1 \right)^3 + b \left( \lambda +4 \right) \left( \lambda -1 \right)^3 + c_1 \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^2 + c_2 \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right) + c_3 \left( \lambda +4 \right) \left( \lambda +2 \right)}{\left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^3} .$
Two sides have the same denominators, so to be equal they should have the same numerators:
$6 + 2\lambda -3 \lambda^2 = a \left( \lambda +2 \right) \left( \lambda -1 \right)^3 + b \left( \lambda +4 \right) \left( \lambda -1 \right)^3 + c_1 \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^2 + c_2 \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right) + c_3 \left( \lambda +4 \right) \left( \lambda +2 \right) .$
Since this is a polynomial equation, it should be valid for all real or complex λ. So both sides are the same whatever value of λ is chosen. The right-hand side suggests that if we take λ = −4, then −2, and finally λ = 1, we get
\begin{align*} 6 + 2(-4) -3(-4)^2 &= -50 = 10\,a \qquad \Longrightarrow \qquad a = -1/5, \\ 6 + 2(-2) -3(-2)^2 &= -10 = 2\,b \qquad \Longrightarrow \qquad b = 5/27, \\ 6 + 2(1) -3(1)^2 &= 5 = 15\,c_3 \qquad \Longrightarrow \qquad c_3 = 1/3. \end{align*}
The remaining coefficients (or all coefficients) may be obtained by equating like powers of λ:
Expand[(s + 2)*a*(s - 1)^3 + b*(s + 4)*(s - 1)^3 + c1*(s + 4)*(s + 2)*(s - 1)^2 + c2*(s + 4)*(s + 2)*(s - 1) + c3*(s + 4)*(s + 2)]
-2 a - 4 b + 8 c1 - 8 c2 + 8 c3 + 5 a s + 11 b s - 10 c1 s + 2 c2 s + 6 c3 s - 3 a s^2 - 9 b s^2 - 3 c1 s^2 + 5 c2 s^2 + c3 s^2 - a s^3 + b s^3 + 4 c1 s^3 + c2 s^3 + a s^4 + b s^4 + c1 s^4

In the following Mathematica codes, we use s instead of λ because I am too lazy to type six letters. Other people who are not like me can enjoy typing lambda.

Coefficient[(s + 2)*a*(s - 1)^3 + b*(s + 4)*(s - 1)^3 + c1*(s + 4)*(s + 2)*(s - 1)^2 + c2*(s + 4)*(s + 2)*(s - 1) + c3*(s + 4)*(s + 2),s,0]
-2 a - 4 b + 8 c1 - 8 c2 + 8 c3
Coefficient[(s + 2)*a*(s - 1)^3 + b*(s + 4)*(s - 1)^3 + c1*(s + 4)*(s + 2)*(s - 1)^2 + c2*(s + 4)*(s + 2)*(s - 1) + c3*(s + 4)*(s + 2),s,1]
5 a + 11 b - 10 c1 + 2 c2 + 6 c3
Coefficient[(s + 2)*a*(s - 1)^3 + b*(s + 4)*(s - 1)^3 + c1*(s + 4)*(s + 2)*(s - 1)^2 + c2*(s + 4)*(s + 2)*(s - 1) + c3*(s + 4)*(s + 2),s,2]
-3 a - 9 b - 3 c1 + 5 c2 + c3
Coefficient[(s + 2)*a*(s - 1)^3 + b*(s + 4)*(s - 1)^3 + c1*(s + 4)*(s + 2)*(s - 1)^2 + c2*(s + 4)*(s + 2)*(s - 1) + c3*(s + 4)*(s + 2),s,3]
-a + b + 4 c1 + c2
\begin{align*} &\lambda^0 : \qquad \phantom{-}6 = -2a -4b + 8 c_1 - 8 c_2 + 8 c_3 , \\ &\lambda^1 : \qquad \phantom{-}2 = 5a + 11 b - 10 c_1 +2 c_2 + 6 c_3 , \\ &\lambda^2 : \qquad -3 = -3 a - 9b -3 c_1 + 5 c_2 + c_3 , \\ &\lambda^3 : \qquad \phantom{-}0 = -a + b + 4 c_1 + c_2 , \\ &\lambda^4 : \qquad \phantom{-}0 = a + b + c_1 . \end{align*}
Solving the coresponding system of algebraic equations (of course, with the aid of Mathematica), we obtain
Solve[{-2 a - 4 b + 8 c1 - 8 c2 + 8 c3 == 6, 5 a + 11 b - 10 c1 + 2 c2 + 6 c3 == 2,-3 a - 9 b - 3 c1 + 5 c2 + c3 == -3, -a + b + 4 c1 + c2 == 0, a + b + c1 == 0}, {a, b, c1, c2, c3}]
{{a -> -(1/5), b -> 5/27, c1 -> 2/135, c2 -> -(4/9), c3 -> 1/3}}
$a = - \frac{1}{5} , \quad b = \frac{5}{27} , \quad c_1 = \frac{2}{135} , \quad c_2 = - \frac{4}{9} , \quad c_3 = \frac{1}{3} .$
Therefore,
$F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^3} = \frac{-1/5}{\lambda +4} + \frac{5/27}{\lambda +2} + \frac{2/135}{\lambda -1} + \frac{-4/9}{(\lambda -1)^2} + \frac{1/3}{(\lambda -1)^3} . \tag{2.3}$
Apart[(6 + 2*s - 3*s^2)/(s + 4)/(s + 2)/(s - 1)^3]
1/(3 (-1 + s)^3) - 4/(9 (-1 + s)^2) + 2/(135 (-1 + s)) + 5/( 27 (2 + s)) - 1/(5 (4 + s))
Then its inverse Laplace transform becomes
${\cal L}^{-1} \left[ F(\lambda ) \right] = \left[ -\frac{1}{5}\, e^{-4t} + \frac{5}{27} \, e^{-2t} + \frac{2}{135}\, e^{t} -\frac{4}{9}\,t\, e^t + \frac{t^2}{6}\, e^{t} \right] H(t) .$
InverseLaplaceTransform[(6 + 2*s - 3*s^2)/(s + 4)/(s + 2)/(s - 1)^3, s, t]
1/270 E^(-4 t) (-54 + 50 E^(2 t) + 4 E^(5 t) - 120 E^(5 t) t + 45 E^(5 t) t^2)
■
Example 4: Find the inverse Laplace transform of
$F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} . \tag{4.1}$
We expand this function into the sum of simple fractions:
$F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} = \frac{a}{\lambda +2} + \frac{b}{\lambda -1} + \frac{c_1 + c_2 \lambda}{\lambda^2 +6\lambda + 13} . \tag{3.2}$
Then we apply the inverse Laplace transform to each term and obtain the required inverse Laplace value:
${\cal L}^{-1} \left[ \frac{1}{\lambda +2} \right] = e^{-2t} H(t) , \qquad {\cal L}^{-1} \left[ \frac{1}{\lambda - 1} \right] = e^{t} H(t)$
InverseLaplaceTransform[1/(s + 2), s, t]
E^(-2 t)
InverseLaplaceTransform[1/(s - 1), s, t]
E^t
and
${\cal L}^{-1} \left[ \frac{c_1 + c_2 \lambda}{\lambda^2 +6\lambda + 13} \right] = \frac{1}{2} \,e^{-3t} \left[ c_1\, \sin 2t + c_2 \left( 2\,\cos 2t - 3\,\sin 2t \right) \right] H(t) , \tag{4.3}$
where H(t) is the Heaviside function.
InverseLaplaceTransform[(c1 + c2*s)/(s^2 + 6 s + 13), s, t]
1/4 E^((-3 - 2 I) t) (I c1 + (2 - 3 I) c2 - I c1 E^(4 I t) + (2 + 3 I) c2 E^(4 I t))

To justify formula (3.3), we make substitution s = λ + 3 and obtain

$\frac{c_1 + c_2 \lambda}{\lambda^2 +6\lambda + 13} = \frac{c_1 - 3\, c_2 + c_2 \left( \lambda + 3 \right)}{\left( \lambda + 3 \right)^2 + 4} = \frac{c_1 - 3\, c_2 + c_2 s}{s^2 + 2^2} .$
The inverse Laplace transforms of the following fractions are well-known
${\cal L}^{-1}_{s\to t} \left[ \frac{1}{s^2 + 2^2} \right] = \frac{1}{2}\,\sin 2t \,H(t) \qquad\mbox{and} \qquad {\cal L}^{-1}_{s\to t} \left[ \frac{s}{s^2 + 2^2} \right] = \cos 2t\,H(t) .$
Now the inverse Laplace transforms (3.3) follow from the attenuation rule:
${\cal L} \left[ e^{-at} f(t) \right] = f^L \left( \lambda + 3 \right) . \tag{4.4}$

So we need to determine the values of coefficients in Eq.(4.2). A usual way to acomplish this goal is to add all simple fractions and equate the result to the given function F(λ):

$\frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} = \frac{a\left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right) + b \left( \lambda +2 \right) \left( \lambda^2 +6\lambda + 13 \right) + c \left( \lambda +2 \right) \left( \lambda -1 \right) }{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} .$
Equating numerators, we get a system of algebraic equations
Expand[a*(s - 1)*(s^2 + 6*s + 13) + b*(s + 2)*(s^2 + 6*s + 13) + (c1 + c2*s)*(s^2 + s - 2)]
-13 a + 26 b - 2 c1 + 7 a s + 25 b s + c1 s - 2 c2 s + 5 a s^2 + 8 b s^2 + c1 s^2 + c2 s^2 + a s^3 + b s^3 + c2 s^3
\begin{align*} \lambda^3 : \ & a + b + c_2 &= 0 , \\ \lambda^2 : \ & 5a + 8b + c_1 + c_2 &= -3 , \\ \lambda^1 : \ & 7a + 25 b + c_1 -2 c_2 &= 2 , \\ \lambda^0 : \ & -13 a + 26 b - 2c_1 &= 6 . \end{align*}
Solving this system of algebraic equations, we obtain
Solve[{-13 a + 26 b - 2 c1 == 6, 7 a + 25 b + c1 - 2 c2 == 2, 5 a + 8 b + c1 + c2 == -3, a + b + c2 == 0}, {a, b, c1, c2}]
{{a -> 2/3, b -> 1/12, c1 -> -(25/4), c2 -> -(3/4)}}
$a = \frac{2}{3}, \quad b= \frac{1}{12} , \quad c_1 = - \frac{25}{4}, \quad c_2 = - \frac{3}{4} .$
So the expansion of F(λ) into simple fractions becomes
$F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} = \frac{2/3}{\lambda +2} + \frac{1/12}{\lambda -1} - \frac{1}{4} \cdot \frac{25 + 3 \lambda}{\lambda^2 +6\lambda + 13} .$
We check with Mathematica:
Apart[(6 + 2*s - 3 s^2)/(s + 2)/(s - 1)/(s^2 + 6*s + 13)]
1/(12 (-1 + s)) + 2/(3 (2 + s)) + (-25 - 3 s)/(4 (13 + 6 s + s^2))
Therefore, we get
${\cal L}^{-1} \left[ \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} \right] = \left[ \frac{1}{12}\, e^t + \frac{2}{3}\, e^{-2t} - e^{-3t} \left( 2\,\sin 2t + \frac{3}{4}\, \cos 2t \right) \right] H(t) .$
■
Example 5: Find the inverse Laplace transform of
$F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)^2} . \tag{5.1}$
We expand this fraction into the sum of simple terms:
$F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)^2} = \frac{a}{\lambda +2} + \frac{b}{\lambda -1} + \frac{c_1 + c_2 \lambda}{\lambda^2 +6\lambda + 13} + \frac{c_3 + c_4 \lambda}{\left( \lambda^2 +6\lambda + 13 \right)^2} . \tag{5.2}$
In order to find the values of coefficients in Eq.(5.2), we combine its right-hand side into one fraction and equate it to the given one. This yields the polynomial equation
$6 + 2\lambda -3 \lambda^2 = a \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)^2 + b \left( \lambda + 2 \right) \left( \lambda^2 +6\lambda + 13 \right)^2 + \left( c_1 + c_2 \lambda \right) \left( \lambda^2 + \lambda -2 \right) \left( \lambda^2 +6\lambda + 13 \right) + \left( c_3 + c_4 \right) \left( \lambda^2 + \lambda -2 \right) .$
Upon expansion and equating the coefficients of like powers of λ, we come to the system of algebraic equations
Expand[a*(s - 1)*(s^2 + 6*s + 13)^2 + b*(s + 2)* (s^2 + 6*s + 13)^2 + (c1 + c2*s)*(s^2 + s - 2)*(s^2 + 6*s + 13) + (c3 + c4*s)*(s^2 + s - 2)]
-169 a + 338 b - 26 c1 - 2 c3 + 13 a s + 481 b s + c1 s - 26 c2 s + c3 s - 2 c4 s + 94 a s^2 + 280 b s^2 + 17 c1 s^2 + c2 s^2 + c3 s^2 + c4 s^2 + 50 a s^3 + 86 b s^3 + 7 c1 s^3 + 17 c2 s^3 + c4 s^3 + 11 a s^4 + 14 b s^4 + c1 s^4 + 7 c2 s^4 + a s^5 + b s^5 + c2 s^5
\begin{align*} &\lambda^0 : \qquad \phantom{-}6 = -169 a + 338 b -26 c_1 -2 c_3 , \\ &\lambda^1 : \qquad \phantom{-}2 = 13 a + 481 b + c_1 -26 c_2 + c_3 -2 c_4 , \\ &\lambda^2 : \qquad -3 = 94 a + 280 b + 17 c_1 + c_2 + c_3 + c_4 , \\ &\lambda^3 : \qquad \phantom{-}0 = 50 a + 86 b+ 7 c_1 + 17 c_2 + c_4 , \\ &\lambda^4 : \qquad \phantom{-}0 = 11 a + 14 b + c_1 + 7c_2 , \\ &\lambda^5 : \qquad \phantom{-}0 = a + b + c_2 . \end{align*}

The right-hand side expressions are sorted by powers of λ (here, s). For convenience, these were determined with Mathematica:
Coefficient[ a*(s - 1)*(s^2 + 6*s + 13)^2 + b*(s + 2)* (s^2 + 6*s + 13)^2 + (c1 + c2*s)*(s^2 + s - 2)*(s^2 + 6*s + 13) + (c3 + c4*s)*(s^2 + s - 2), s, 0]
-169 a + 338 b - 26 c1 - 2 c3
Coefficient[ a*(s - 1)*(s^2 + 6*s + 13)^2 + b*(s + 2)* (s^2 + 6*s + 13)^2 + (c1 + c2*s)*(s^2 + s - 2)*(s^2 + 6*s + 13) + (c3 + c4*s)*(s^2 + s - 2), s]
13 a + 481 b + c1 - 26 c2 + c3 - 2 c4
Coefficient[ a*(s - 1)*(s^2 + 6*s + 13)^2 + b*(s + 2)* (s^2 + 6*s + 13)^2 + (c1 + c2*s)*(s^2 + s - 2)*(s^2 + 6*s + 13) + (c3 + c4*s)*(s^2 + s - 2), s, 2]
94 a + 280 b + 17 c1 + c2 + c3 + c4
Coefficient[ a*(s - 1)*(s^2 + 6*s + 13)^2 + b*(s + 2)* (s^2 + 6*s + 13)^2 + (c1 + c2*s)*(s^2 + s - 2)*(s^2 + 6*s + 13) + (c3 + c4*s)*(s^2 + s - 2), s, 3]
50 a + 86 b + 7 c1 + 17 c2 + c4
Coefficient[ a*(s - 1)*(s^2 + 6*s + 13)^2 + b*(s + 2)* (s^2 + 6*s + 13)^2 + (c1 + c2*s)*(s^2 + s - 2)*(s^2 + 6*s + 13) + (c3 + c4*s)*(s^2 + s - 2), s, 4]
11 a + 14 b + c1 + 7 c2
Coefficient[ a*(s - 1)*(s^2 + 6*s + 13)^2 + b*(s + 2)* (s^2 + 6*s + 13)^2 + (c1 + c2*s)*(s^2 + s - 2)*(s^2 + 6*s + 13) + (c3 + c4*s)*(s^2 + s - 2), s, 5]
a + b + c2

We solve this system of equations and obtain

Solve[{-169 a + 338 b - 26 c1 - 2 c3 == 6, 13 a + 481 b + c1 - 26 c2 + c3 - 2 c4 == 2, 94 a + 280 b + 17 c1 + c2 + c3 + c4 == -3, 50 a + 86 b + 7 c1 + 17 c2 + c4 == 0, 11 a + 14 b + c1 + 7 c2 == 0, a + b + c2 == 0}, {a, b, c1, c2, c3, c4}]
{{a -> 2/15, b -> 1/240, c1 -> -(9/16), c2 -> -(11/80), c3 -> -(25/4), c4 -> -(3/4)}}
$a = \frac{2}{15}, \quad b= \frac{1}{240} , \quad c_1 = - \frac{9}{16} , \quad c_2 = - \frac{11}{80} , \quad c_3 = - \frac{25}{4} , \quad c_4 = - \frac{3}{4} .$
This allows us to rewrite the given fraction as
$F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)^2} = \frac{2/15}{\lambda +2} + \frac{1/240}{\lambda -1} - \frac{45 + 11 \lambda}{80 \left( \lambda^2 +6\lambda + 13 \right)} - \frac{25 + 3 \lambda}{4 \left( \lambda^2 +6\lambda + 13 \right)^2} . \tag{5.3}$
We check with Mathematica:
Apart[(6 + 2*s - 3 s^2)/(s + 2)/(s - 1)/(s^2 + 6*s + 13)^2]
1/(240 (-1 + s)) + 2/(15 (2 + s)) + (-25 - 3 s)/( 4 (13 + 6 s + s^2)^2) + (-45 - 11 s)/(80 (13 + 6 s + s^2))
The inverse Laplace transforms of first two simple fractions are known from the previous example.
${\cal L}^{-1} \left[ \frac{1}{\lambda +2} \right] = e^{-2t} H(t) \qquad {\cal L}^{-1} \left[ \frac{1}{\lambda -1} \right] = e^t H(t) ,$
The third fraction can be written as
$\frac{45 + 11 \lambda}{80 \left( \lambda^2 +6\lambda + 13 \right)} = \frac{1}{80} \cdot \frac{12 + 11 \left( \lambda + 3 \right)}{\left( \lambda + 3 \right)^2 + 4} .$
We set s = λ + 3, and consider an auxiliary fraction
$F_1 (s) = \frac{12 + 11\,s}{s^2 + 4} .$
Its inverse Laplace transform is known
${\cal L}^{-1}_{s\to t} \left[ \frac{12 + 11\,s}{s^2 + 4} \right] = \left[ 11\,\cos 2t + 6 \sin 2t \right] H(t) .$
InverseLaplaceTransform[(12 + 11*s)/(s^2 + 4), s, t]
11 Cos[2 t] + 6 Sin[2 t]
Then using the attenuation rule, $${\cal L} \left[ e^{-at} f(t) \right] = f^L \left( \lambda + 3 \right) ,$$ we get
${\cal L}^{-1} \left[ \frac{45 + 11 \lambda}{80 \left( \lambda^2 +6\lambda + 13 \right)} \right] = \frac{1}{80} \cdot \left[ 11\,\cos 2t + 6\,\sin 2t \right] e^{-3t} H(t) .$
In order to find the inverse Laplace transform of the last term in right-hand side of Eq.(5.3), we rewrite it as
$\frac{25 + 3 \lambda}{4 \left( \lambda^2 +6\lambda + 13 \right)^2} = \frac{1}{4}\cdot \frac{16 + 3 \left( \lambda + 3 \right)}{\left[ \left( \lambda + 3 \right)^2 + 2^2 \right]^2}$
Again, setting s = λ + 3, we consider another auxiliary fraction
$F_2 (s) = \frac{16 + 3s}{\left( s^2 + 2^2 \right)^2} ,$
which has the inverse Laplace transform
${\cal L}^{-1}_{s\to t} \left[ F_2 (s) \right] = {\cal L}^{-1}_{s\to t} \left[ \frac{16 + 3s}{\left( s^2 + 2^2 \right)^2} \right] = \frac{1}{4} \left[ 3t \,\sin 2t -8t\,\cos 2t + 4\,\sin 2t \right] H(t) .$
InverseLaplaceTransform[(16 + 3*s)/(s^2 + 4)^2, s, t]
1/4 (-8 t Cos[2 t] + 4 Sin[2 t] + 3 t Sin[2 t])
Using the attenuation rule, we obtain
${\cal L}^{-1}_{\lambda\to t} \left[ \frac{25 + 3 \lambda}{4 \left( \lambda^2 +6\lambda + 13 \right)^2} \right] = \frac{1}{16} \left[ 3t ,\sin 2t -8t\,\cos 2t + 4\,\sin 2t \right] e^{-3t} H(t) .$
Collecting all inverse Laplace transformations of simple fractions into one expression, we obtain
${\cal L}^{-1} \left[ \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)^2} \right] = \left[ \frac{2}{15}\, e^{-2t} + \frac{1}{240}\, e^t - \frac{1}{16}\,e^{-3t} \left( 4\,\sin 2t -8t\,\cos 2t + 3t\,\sin 2t \right) - \frac{1}{80}\,e^{-3t} \left( 11\,\cos 2t + 6\,\sin 2t \right) \right] H(t) .$
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Example 6: Mathematica allows one to find the inverse Laplace transform in a straight forward way. For example, consider the rational function

$F(\lambda ) = \frac{4}{\lambda \left( 4 + \lambda^2 \right)} . \tag{6.1}$
Since Mathematica has a build-in command to determine the inverse Laplace transform, we apply it first and obtain

InverseLaplaceTransform[4/s/(4 + s^2), s, t]
Out[8]= 4 (1/4 - 1/4 Cos[2 t])
Simplify[%]
Out[9]= 2 Sin[t]^2
${\cal L}^{-1}_{\lambda\to t} \left[ \frac{4}{\lambda \left( 4 + \lambda^2 \right)} \right] = \left[ 1 - \cos 2t \right] H(t) = 2\,\sin^2 t \,H(t) . \tag{6.2}$
We check the answer by transforming the function F(λ) into the sum of simple terms:
Apart[4/s/(4 + s^2), s]
Out[10]= 1/s - s/(4 + s^2)
$F(\lambda ) = \frac{4}{\lambda \left( 4 + \lambda^2 \right)} = \frac{1}{\lambda} + \frac{\lambda}{4 + \lambda^2} .$
We recognize immediately that
${\cal L}^{-1} \left[ \frac{1}{\lambda} \right] = H(t) \qquad\mbox{and} \qquad {\cal L}^{-1} \left[ \frac{\lambda}{4 + \lambda^2} \right] = \cos 2t \,H(t) .$
Adding these expressions to come to the inverse Laplace transform of the required function. Indeed, we take the Laplace transform of function (6.2) and obtain F(λ).
LaplaceTransform[2 Sin[t]^2, t, lambda]
Out[11]= 4/(lambda (4 + lambda^2))
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1. Doetsch, G., Guide to the Applications of the Laplace and Z-Transforms, 1971, Van Nostrand Reinhold Company, 2nd Edition, London.
2. Doetsch, G., Introduction to the Theory and Application of the Laplace Transformation, 1974, Springer, Berlin.
3. Man, Y-K., An improved Heaviside approach to partial fraction expansion and its applications, International Journal of Mathematical Education in Science and Technology, Volume 40, 2009, Issue 7, pp. 808--814. https://doi.org/10.1080/00207390902825310
4. Man, Y-K., Introducing the improved Heaviside approach to partial fraction decomposition to undergraduate students: results and implications from a pilot study, International Journal of Mathematical Education in Science and Technology, Volume 43, 2012 - Issue 7, pp. 911--922. https://doi.org/10.1080/0020739X.2012.662292