# Convolution Integral

1. Mar 18, 2013

### mateomy

Boas Ch. 8, Sec. 10 #3

Use the convolution integral to find the inverse transforms of:
$$\frac{p}{(p^{2}-1)^{2}} = \frac{p}{p^{2}-1} \frac{1}{p^{2}-1}$$

I'm completely confused with these things. Are we supposed to figure out the inverse Laplace transform and then use that within our convolution integral? I am completely lost. Just looking for some advice, thanks.

I know the convolution integral takes the form of:
$$\int_{-\infty}^\infty\,f(x)g(z-x)dx$$

2. Mar 18, 2013

### Ray Vickson

For functions defined on [0,∞) the convolution is, instead:
$$\int_0^z f(x) g(z-x) \, dx \text{ for }z \geq 0.$$
And yes, you are supposed to figure out the inverse Laplace of p/(p^2-1), then do a convolution, exactly as it says.

3. Mar 18, 2013

### mateomy

Just to clarify, you mean '..inverse Laplace of
$$\frac{p}{(p^{2}-1)^{2}}$$
right?

4. Mar 18, 2013

### mateomy

So far...
$$\int \frac{p}{(p^{2}-1)^{2}}dp$$
getting...
$$\frac{-1}{2(p^{2}-1)}$$

Now do I look for the inverse transform? Because If I do that my answer slightly varies from the book's. Boas shows:
$$\frac{tsinht}{2}$$
whereas I'm getting
$$\frac{sinht}{2}$$

..hmmm?

(thanks for the help)

5. Mar 18, 2013

### mateomy

However, if I actually paid attention to what you had said maybe I'd be doing something different. I'll try the 'real' method now...

6. Mar 18, 2013

### mateomy

After finding my inverse transform, being $cosh(at)sinh(at)$ I used their exponential forms to integrate.
$$\int_0^\infty \left(\frac{e^{at}+e^{-at}}{2}\right)\frac{e^{at}-e^{-at}}{2}dt$$
...which I end up getting
$$\frac{1}{2}\left(cosh(at)-t\right)$$
which still isn't right.

Clearly, I'm confused.

7. Mar 18, 2013

### vela

Staff Emeritus
That's not a convolution integral. You want to convolve sinh and cosh.

8. Mar 18, 2013

### mateomy

Then I'm even more lost...

9. Mar 18, 2013

### vela

Staff Emeritus
Look at Ray's integral in post #2 and yours. Yours doesn't have a $z$. I'm not sure how you came up with your integral.