Example. Restore the original $f(t)$ from the image

$$
F(p)=\frac{1}{p^{3}-8}
$$
## Solution.

1. We write $F(p)$ as a sum of elementary fractions:

$$
F(p)=\frac{1}{p^{3}-8}=\frac{1}{(p-2)\left(p^{2}+2 p+4\right)}=\frac{1}{12} \frac{1}{p-2}-\frac{1}{12} \frac{p+4}{p^{2}+2 p+4}
$$

2. For each fraction, we find its original.

First fraction

$$
F_{1}(p)=\frac{1}{p-2}
$$

is the image of the function $f_{1}(t)=e^{2 t}$.

In the denominator of the second fraction, we isolate a complete square and write it in a form that allows us to use the shift theorem:

$$
F_{2}(p)=\frac{p+4}{p^{2}+2 p+4}=\frac{(p+1)}{(p+1)^{2}+(\sqrt{3})^{2}}+\frac{3}{(p+1)^{2}+(\sqrt{3})^{2}}
$$

By the shift theorem

$$
e^{-t} \cos (\sqrt{3} t) \longleftrightarrow \frac{(p+1)}{(p+1)^{2}+(\sqrt{3})^{2}}
$$

and

$$
e^{-t} \sin (\sqrt{3} t) \longleftrightarrow \frac{\sqrt{3}}{(p+1)^{2}+(\sqrt{3})^{2}}
$$

3. Using the linearity of the Laplace transform and the decomposition (1), we find the desired original

$$
f(t)=\frac{1}{12} e^{2 t}-\frac{1}{12} e^{-t} \cos (\sqrt{3} t)-\frac{\sqrt{3}}{12} e^{-t} \sin (\sqrt{3} t)
$$

Answer. $f(t)=\frac{1}{12} e^{2 t}-\frac{1}{12} e^{-t} \cos (\sqrt{3} t)-\frac{\sqrt{3}}{12} e^{-t} \sin (\sqrt{3} t)$.

Conditions of the Problems. Restore the original $f(t)$ from the image $F(p)$.

1. $F(p)=\frac{3 p}{2 p^{2}-2 p-4}$.
2. $F(p)=\frac{1}{p^{4}-2 p^{3}+p^{2}}$.
3. $F(p)=\frac{3 p^{2}+3 p-13}{p^{3}+4 p^{2}+13 p}$.
4. $F(p)=\frac{p}{(p+2)^{2}(p-1)}$.
5. $F(p)=\frac{p^{2}+4}{(p-1)(p+2)(p-3)}$.
6. $F(p)=\frac{p^{2}+p-4}{p^{2}-p^{3}}$.
7. $F(p)=\frac{1}{(p-1)^{3}}$.
8. $F(p)=\frac{p}{p^{4}-1}$.
9. $F(p)=\frac{p}{(p+2)^{2}(p-1)^{3}}$.
10. $F(p)=\frac{p+1}{p(p-1)(p-2)(p-3)}$.

Answers. 1. $f(t)=e^{2 t}+e^{-t} / 2 . \quad 2 . \quad f(t)=t+2-2 e^{t}+t e^{t}$. 3. $f(t)=-1+4 e^{-2 t} \cos 3 t-e^{-2 t} \sin 3 t / 3$. 4. $f(t)=e^{t} / 9-e^{2 t} / 9+$ $+2 t e^{-2 t} / 3$. 5. $f(t)=-5 e^{t} / 6+8 e^{-2 t} / 15+13 e^{3 t} / 10$. 6. $f(t)=2 e^{t}-4 t-3$. 7. $f(t)=t^{2} e^{t} / 2$. 8. $f(t)=(\operatorname{ch} t-\cos t) / 2.9 . f(t)=\left(3 t^{2}+2 t-2\right) e^{t} / 54+$ $+(2 t+1) e^{-2 t} / 27$. $\quad 10 . f(t)=-1 / 6+e^{t}-3 e^{2 t} / 2+2 e^{3 t} / 3$.

## 2.9. Restoration of the Original Using the Decomposition Theorem

Problem Statement. Restore the original $f(t)$ from its image $F(p)$, using the decomposition theorem.

Solution Plan.

If the image $F(p)$ of the original $f(t)$ is a single-valued function of $p$ and has only a finite number of singular points $p_{1}, p_{2}, \ldots, p_{n}$, then by the 2nd decomposition theorem

$$
f(t)=\sum_{k=1}^{n} \operatorname{res}_{p=p_{k}}\left[e^{p t} F(p)\right]
$$

1. Find the singular points of the function $F_{1}(p)=e^{p t} F(p)$ and determine their type.
2. Calculate the residues at these points.
3. Calculate the original $f(t)$ using formula (1) and write the answer.