laplace inverse

**prasum** · February 21st 2013, 09:09 AM

find laplace inverse of log(s+a)/(s+b)

do we have to apply power series in it

**hollywood** · February 21st 2013, 10:18 AM

If you just want the answer, you can get that here:

Inverse Laplace transform

- Hollywood

**TheEmptySet** · February 22nd 2013, 05:51 PM

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

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

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

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