# laplace inverse

• Feb 21st 2013, 09:09 AM
prasum
laplace inverse
find laplace inverse of log(s+a)/(s+b)

do we have to apply power series in it
• Feb 21st 2013, 10:18 AM
hollywood
Re: laplace inverse
If you just want the answer, you can get that here:

Inverse Laplace transform

- Hollywood
• Feb 22nd 2013, 05:51 PM
TheEmptySet
Re: laplace inverse
Quote:

Originally Posted by prasum
find laplace inverse of log(s+a)/(s+b)

do we have to apply power series in it

You can use the fact that

$\mathcal{L}(tf(t))=\int_{0}^{\infty}e^{-st}[tf(t)]dt=\int_{0}^{\infty}-\frac{d}{ds}e^{-st}f(t)dt=-\frac{d}{ds}F(s)$

If you take the inverse transform of the above you get

$-tf(t)=\mathcal{L}^{-1}\left(\frac{d}{ds}F(s)\right) \implies f(t)=-\frac{1}{t}\mathcal{L}^{-1}\left(\frac{d}{ds}F(s)\right)$

Using this gives

$F(s)=\ln\left(\frac{s+a}{s+b}\right) \implies F'(s)=\frac{1}{s+a}-\frac{1}{s+b}$

$f(t)=-\frac{1}{t}\left(e^{-at}-e^{-bt} \right)=\frac{e^{-bt}-e^{-at}}{t}$