Find the inverse Laplace transformation f(t) that corresponds to $F(s)=\frac{s}{(s^2+a^2)^2}$

Question

I am currently completing a past exam for Complex Analysis (I have an upcoming exam), and cannot seem to get my head around the inverse Laplace transformation for the following:

$F(s)=\frac{s}{(s^2+a^2)^2}$

All notes for my course have been taken from Complex Variables 3rd Edition by A. David Wunsch - While there are similar questions contained within my text, I have found the explanations have left me at a loss of how to solve. Any help would be greatly appreciated.

Answer 1

There's a property about Laplace transformations that states

$$ L[t^n f(t)] = (-1)^n\frac{{\rm d}F(s)}{{\rm d}s} \tag{1} $$

In your case, it is easy to see that

$$ \frac{s}{(s^2 + a^2)^2}= -\frac{1}{2}\frac{{\rm d}}{{\rm d}s}\color{blue}{\frac{1}{a^2 + s^2}} = -\frac{1}{2}\frac{{\rm d}}{{\rm d}s}\color{blue}{\frac{L[\sin(at)]}{a}} \tag{2} $$

If you compare (1) and (2) you will that the function you're looking for is

$$ \frac{t \sin(at)}{2a} $$

Answer 2

You have at east two ways to find the inverse Laplace transform of $$F(s)=\frac{s}{(s^2+a^2)^2}$$

You may use the convolution formula, with $$\frac{s}{(s^2+a^2)^2}= \frac{s}{(s^2+a^2)} \frac{1}{(s^2+a^2)} $$

Thus $$f(t) = \cos t * ((1/a) \sin t)$$

You may also use the fact that $$F(s)=\frac{s}{(s^2+a^2)^2}= -\frac {1}{2} \frac {d}{ds} \frac{1}{(s^2+a^2)}$$

Thus $$f(t) = \frac {1}{2a} t\sin t$$

Answer 3

Use the convolution property which states

$$ f_1(t) * f_2(t) \longleftrightarrow F_1(s)F_2(s) $$

Here

$$ F_1(s) = \frac{s}{s^2+a^2}\longleftrightarrow \cos(a t) u(t)\\ F_2(s) = \frac{1}{s^2+a^2}\longleftrightarrow \frac{1}{a}\sin(a t)u(t) $$

Here $u(t)$ is the unit step function and the convolution is defined as

$$ (f_1*f_2)(t)=\int_0^t f_1(\tau)f_2(\tau-t)d\tau $$

Answer 4

As already noted you can figure this out by either of two properties of the Laplace transform: The fact that $L[f*g]=L[f]L[g]$ or the fact that if $L[f]=F(s)$ then $L[tf(t)]=-F'(s)$.

Seems to me students typically first meet the Laplace transform in a course on differential equations. Given that you say this is for a complex course you may not have the required background, but if you're familiar with the standard techniques for using Laplace transforms to solve differential equations it's interesting to note that you can also use differential equations to figure out things about Laplace transforms!

Consider the IVP $$y''+a^2y=\cos(at),\quad y(0)=y'(0)=0.$$If you try to solve this using the Laplace transform you run into exactly the inverse Laplace transform you ask about. Ok, so now solve the IVP by another method, say Undetermined Coefficients...