Find inverse Laplace Transform of : $\frac{1}{(s^2+a^2)^2}$

Question

Question :

Find inverse Laplace of :

$$\dfrac{1}{(s^2+a^2)^2}$$

My try :

$$\dfrac{1}{(s^2+a^2)^2}=-\frac{1}{2s}\frac{\mathrm d}{\mathrm ds}\left( \frac{1}{s^2+a^2}\right)$$

I need use this identity :

$tf(t)\stackrel{\mathcal{L}}\longleftrightarrow-\frac{\mathrm d}{\mathrm ds}F(s)$

And :

$\int_{0}^{t}g(\tau)d\tau\stackrel{\mathcal{L}}\longleftrightarrow\frac{1}{s}G(s)$

but I don't understand how I applied ?

Answer 1

The second rule says$$\mathcal L^{-1}\left[\frac1sG(s)\right]=\int_0^t g(\tau)\mathrm d\tau$$where $g(\tau)=\mathcal L^{-1}[G](\tau)$. In this example, our $g(t)$ is $$g(t)=\mathcal L^{-1}\left[-\frac d{ds}\left(\frac1{s^2+a^2}\right)\right]=tf(t)$$ where $$f(t)=\mathcal L^{-1}\left[\frac{1}{s^2+a^2}\right]$$So combining these together: $$\mathcal L^{-1}\left[\frac1{(s^2+a^2)^2}\right]=\int_0^t\tau\mathcal L^{-1}\left[\frac1{s^2+a^2}\right](\tau)\mathrm d\tau$$

From here you can integrate this to get the final result $$\frac1{2a^3}(\sin at - at \cos at)$$

Answer 2

Using the fact that $(s^2+a^2)^2=(s-ai)^2(s+ai)^2$, we can write the given expression under the form :

$$\dfrac{1}{4ai}\dfrac{1}{s}\left(\dfrac{1}{(s-ai)^2}-\dfrac{1}{(s+ai)^2}\right)$$

which gives, reading backwards the classical LT table :

$$\dfrac{1}{4ai}\text{Prim}\left(te^{ait}-te^{-ait}\right)=\dfrac{1}{4ai}\text{Prim}\left(t(e^{ait}-e^{-ait})\right)$$

where "Prim" means "take the primitive function" of what follows.

Otherwise said :

$$\dfrac{1}{2a}\dfrac{1}{\require{cancel}\cancel{2i}}\text{Prim}\left(t \cancel{2i} \sin(at)\right)$$

Let us integrate by parts :

$$\dfrac{1}{2a}\left( [-t \tfrac1{a} \cos(at)]- \text{Prim} ((-\tfrac1{a})\cos(at))\right)$$

$$\dfrac{1}{2a^2}\left( -t \cos(at)+\tfrac1{a}\sin(at)\right)$$

$$\dfrac{1}{2a^3}\left(-at \cos(at)+\sin(at)\right)$$

Answer 3

As

$$ F(s) = \int_0^{\infty}e^{-st}f(t)dt $$

we have

$$ \frac{d}{ds}F(s) = \frac{d}{ds}\int_0^{\infty}e^{-st}f(t)dt = -\int_0^{\infty}e^{-st}t f(t)dt $$

then knowing that $\mathcal{L}^{-1}\left(\frac{1}{s^2+a^2}\right) = \frac 1a\sin(a t)$

$$ -\frac{d}{ds}\left(\frac{1}{s^2+a^2}\right) = \int_0^{\infty}e^{-st} \frac ta\sin(a t)dt $$

then

$$ \frac{2s}{(s^2+a^2)^2}\Leftrightarrow \frac ta\sin(a t) $$

and

$$ \frac{1}{2s}\frac{2s}{(s^2+a^2)^2}\Leftrightarrow \frac 12\int_0^t\frac {\tau}{a}\sin(a \tau)d\tau = \frac{\sin (a t)-a t \cos (a t)}{2 a^3} $$

Answer 4

Hint: Other method, which isn't the last one is using convolution $${\cal L}^{-1}\left(\dfrac{1}{(s^2+a^2)^2}\right)=\dfrac{1}{a^2}\int_0^x\sin a(x-t)\sin at\ dt$$

Answer 5

$$ \frac{1}{(s^2+a^2)^2}= \frac{1}{(s+ia)^2(s-ia)^2} \\ = -\frac{1}{4a^2}\left[\frac{1}{s+ia}-\frac{1}{s-ia}\right]^2 \\ = -\frac{1}{4a^2}\left[\frac{1}{(s+ia)^2}-2\frac{1}{(s+ia)(s-ia)}+\frac{1}{(s-ia)^2}\right] \\ = -\frac{1}{4a^2}\left[\frac{1}{(s+ia)^2}-\frac{1}{ia}\left(\frac{1}{s-ia}-\frac{1}{s+ia}\right)+\frac{1}{(s-ia)^2}\right] \\ = \frac{1}{4a^2}\frac{d}{ds}\left[\frac{1}{s+ia}+\frac{1}{s-ia}\right]+\frac{1}{4ia^3}\left(\frac{1}{s-ia}-\frac{1}{s+ia}\right) \\ = \frac{1}{4a^2}\frac{d}{ds}\int_0^{\infty}e^{-(s+ia)t}+e^{-(s-ia)t}dt \\ +\frac{1}{4ia^3}\int_0^{\infty}e^{-(s-ia)t}-e^{-(s+ia)t}dt \\ = \frac{1}{4a^2}\int_0^{\infty}-t(e^{-(s+ia)t}+e^{-(s-ia)t})dt \\ +\frac{1}{4ia^3}\int_0^{\infty}e^{-(s-ia)t}-e^{-(s+ia)t}dt \\ = \int_0^{\infty}e^{-st}\left[\frac{1}{4a^2}\left(-te^{-iat}-te^{iat}\right)+\frac{1}{4ia^3}(e^{iat}-e^{-iat})\right]dt \\ = \int_0^{\infty}e^{-st}\left[-\frac{1}{2a^2}t\cos(at)+\frac{1}{2a^3}\sin(at)\right]dt $$ So the inverse Laplace transform is $$ -\frac{1}{2a^2}t\cos(at)+\frac{1}{2a^3}\sin(at). $$

Answer 6

Other method might be useful is complex method, the function $f(s)=\dfrac{e^{st}}{(s^2+a^2)^2}$ has two poles $s=\pm ia$ of order $2$, with residues $$\operatorname{Res}_{s=ia}f(s)= \lim_{s\to ia}\dfrac{d}{ds}\left((s-ia)^2f(s)\right)=e^{ iat}\dfrac{iat-1}{-4ia^3}$$ $$\operatorname{Res}_{s=-ia}f(s)= \lim_{s\to-ia}\dfrac{d}{ds}\left((s+ia)^2f(s)\right)=e^{-iat}\dfrac{iat+1}{-4ia^3}$$ thus the sum of residues is $$\dfrac{at\cos at-\sin at}{-2a^3}$$