Solving Inverse Laplace Transform w/ Convolution Integral: Ch 8, Sec 10 #3

[mateomy]

Boas Ch. 8, Sec. 10 #3

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

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:
<br /> \int_{-\infty}^\infty\,f(x)g(z-x)dx<br />

 

[Ray Vickson]

Boas Ch. 8, Sec. 10 #3

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

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:
<br /> \int_{-\infty}^\infty\,f(x)g(z-x)dx<br />

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.

 

[mateomy]

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

 

[mateomy]

So far...
<br /> \int \frac{p}{(p^{2}-1)^{2}}dp
getting...
<br /> \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:
<br /> \frac{tsinht}{2}
whereas I'm getting
\frac{sinht}{2}

..hmmm?

(thanks for the help)

 

[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...

 

[mateomy]

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

Clearly, I'm confused.

 

[vela]

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

 

[mateomy]

Then I'm even more lost...

 

[vela]

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.