I have the function $$ F(s)=\frac{1}{1+s^2}\frac{1}{1+4s^2} $$ and I would like to know if exists a non-decreasing function $f(t)$ such that $F(s)$ is the two-sided Laplace transform of $f(t)$.
Of course $F(s)$ is the two-sided Laplace transform of $f(t)=u(t)[2\sin(t/2)-\sin(t)]/3$ with $u(t)$ Heaviside function, but is not non-decreasing. Somebody knows how I can find out if such a $f(t)$ exists?
Thank you
Consider the decomposition
$$\frac{1}{1+s^2}\frac{1}{1+4s^2}=\frac{A}{1+s^2}+\frac{B}{1+4s^2},$$
with $A$ and $B$ coefficients to determine.
Then
$$\frac{A}{1+s^2}+\frac{B}{1+4s^2}=\frac{(A+B)+s^2(4A+B)}{(1+s^2)(1+4s^2)} $$
implies $A+B=1$ and $4A+B=0$, or $A=-\frac{1}{3}$ and $ B=\frac{4}{3}$. In summary
$$F(s)=\frac{1}{1+s^2}\frac{1}{1+4s^2}=-\frac{1}{3}\frac{1}{1+s^2}+\frac{4}{3}\frac{1}{1+4s^2},$$
i.e.
$$F(s)=-\frac{1}{3}\mathcal L(\sin(t))+\frac{4}{3}\frac{1}{2}\mathcal L(\sin(2t))= -\frac{1}{3}\mathcal L(\sin(t))+\frac{2}{3}\mathcal L(\sin(2t)),$$
as $\mathcal L(\sin(2t))=\frac{\frac{1}{2}}{s^2+\frac{1}{4}}=2\frac{1}{4s^2+1}$.
EDIT: The OP asks for the 2-sided Laplace transform.
Using the above computations, and remembering that the bilateral Laplace transform $\mathcal B(f)$ of a function $f$ is defined as the improper integral $$\mathcal B(f)(s)=\int_{-\infty}^{\infty}e^{-st}f(t)dt, $$
then the causal functions $f$ and $g$ s.t.
$$f(t)=-\frac{1}{3}\sin(t),$$
$$g(t)=\frac{2}{3}\sin(2t),$$
for $t\geq 0$ and $f(t)=g(t)=0$ for $t<0$ satisfy $\mathcal B(f+g)=F(s)$.