inverse laplace transform of F(s)

Question

Let $f(x)$ be some arbitrary function , $F(s)$ is laplace transform of it

I think

inverse laplace transform of $\frac {F(s)}{s+r}$

where, r is constant

may be $\int_0^t e^{-rt'}F(t') dt'$

however, the answer is $\int_0^t e^{-r(t-t')}F(t') dt'$

why it is like that?

Thanks

Answer 1

You need to use the property:

the Laplace of the convolution equals the product of the Laplace, i.e; $\mathcal{L}(f*g)=\mathcal{ L }f \mathcal {L}g $.

See here.