# Inverse Laplace Transform(s/((s^2)+1)^2

1. Feb 4, 2011

### fysiikka111

1. The problem statement, all variables and given/known data
Find inverse Laplace transform of:
s/((s^2)+1)^2

2. Relevant equations

3. The attempt at a solution
(t/2)sin t
First shifting theorem or partial fractions don't work, I think. From a Laplace transforms table, sin at = a/((s^2)+(a^2)), which is almost the form that the problem is in.
Thanks

2. Feb 4, 2011

### hunt_mat

Here is something that may help, take the sin transform and differentiate it w.r.t s and what do you get?

Mat

3. Feb 4, 2011

### fysiikka111

d/ds of sin transform = -2sa/(s^2+a^2)^2
This would give an inverse of the problem of:
(d/ds)-sin(t)/2
I'm not sure I did that right.

4. Feb 4, 2011

### hunt_mat

Set a=-1 and what do you get?

5. Feb 4, 2011

### hunt_mat

Not quite, we have:
$$\frac{1}{s^{2}+a^{2}}=\int_{0}^{\infty}e^{-st}\sin atdt$$
Differentiate this w.r.t and tell me what you get?

6. Feb 4, 2011

### fysiikka111

I'm getting:
1/(a-s)
with integration-by-parts twice.

7. Feb 4, 2011

### Staff: Mentor

No, hunt_mat said to differentiate, not integrate.

8. Feb 4, 2011

### fysiikka111

Oh, right, sorry.
(d/ds)1/(s^2+a^2)=-2s/(s^2+a^2)^2

9. Feb 4, 2011

### hunt_mat

and on the RHS?

10. Feb 4, 2011

### fysiikka111

RHS:
integral from 0 to infinity(-s(e^-st)sin atdt

11. Feb 4, 2011

### hunt_mat

set a=-1 and what have you got?

12. Feb 4, 2011

### fysiikka111

If a=-1:
-2s/(s^2+a^2)^2 = -s/(-s-1)

13. Feb 4, 2011

### fysiikka111

Sorry:
-2s/(s^2+1)^2=-s/(-s-1)

14. Feb 4, 2011

### hunt_mat

No, if a=-1:
$$\frac{s}{(s^{2}+1)^{2}}=\int_{0}^{\infty}\frac{t}{2}e^{-st}\sin tdt$$
What does that tell you?

15. Feb 4, 2011

### fysiikka111

I know then that:
L{tsin(at)/2}=s/(s^2+a^2)?
I'm not sure.

16. Feb 4, 2011

### hunt_mat

Let me re-arrange the RHS:
$$\int_{0}^{\infty}\frac{t\sin t}{2}e^{-st}dt$$
Does it make sense now?

17. Feb 4, 2011

### fysiikka111

No, I'm sorry, I'm just not seeing it. Thanks for all your help, though.

18. Feb 4, 2011

### hunt_mat

Think of it like this, you know that:
$$\frac{1}{1-x}=1+x+x^{2}+x^{3}+x^{4}+\cdots$$
Now differentiate both sides to show that:
$$\frac{1}{(1-x)^{2}}=1+2x+3x^{2}+4x^{3}+\cdots$$
and now you know the power series for $$(1-x)^{2}$$ is just by differentiating the series.

19. Feb 4, 2011

### fysiikka111

So I know that differentiating both sides of the sin(at) LT will give me the square that I need in the denominator, but then I'm stuck with the d/ds on the sin side. I'm not sure what to do after this.
Thanks

20. Feb 4, 2011

### hunt_mat

when you differentiate the integral you take in differentiation under the integral sign:
$$\frac{d}{ds}\int_{0}^{\infty}e^{-st}\sin tdt=\int_{0}^{\infty}\frac{d}{ds}\left( e^{-st}\sin t\right) dt$$
Differentiating the integrand:
$$\frac{d}{ds}e^{-st}\sin t=-te^{-st}\sin t$$
As now we regard s as variable and t is fixed. So the above can be written as:
$$(-t\sin t)e^{-st}$$
and so you're taking the Laplace transform of something different now. That something is the answer to your question. Is this making sense now?