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?


  • $\begingroup$ Could you explain why the proof-verification tag is on this question? $\endgroup$
    – Varun Iyer
    Commented Jan 27, 2015 at 22:39
  • $\begingroup$ sorry, I removed it, I thought that tag is for proof or verification. $\endgroup$
    – eric
    Commented Jan 27, 2015 at 22:43

1 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.

  • $\begingroup$ @eric: You are very welcome! $\endgroup$ Commented Jan 27, 2015 at 22:53

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