By definition, the Laplace transform of a function $f$ is given by,

$$ L(f)(\lambda) = \int_0^\infty e^{-\lambda s}f(s) ds .$$

My question is two fold. I need help in findding the derivative of $L(f)(\lambda)$. Is there a relationship between $L(f)(\lambda)$ and its derivative $\dfrac{dL(f)(\lambda)}{d\lambda}$?

  • $\begingroup$ Derivative with respect to what? $\endgroup$
    – user170231
    Commented Jan 14, 2015 at 17:04
  • $\begingroup$ derivative with respect to $\lambda$ $\endgroup$
    – seth
    Commented Jan 14, 2015 at 17:11

1 Answer 1


Since $\lambda$ does not depend on $s$, we can differentiate under the integral sign to find

$$ \frac{d}{d \lambda} L(f)(\lambda) = \frac{d}{d \lambda} \int_0^\infty e^{-\lambda s}f(s) ds .$$

$$ = \int_0^\infty \frac{d}{d \lambda} (e^{-\lambda s})f(s) ds = \int_0^\infty (-s)e^{-\lambda s}f(s) ds = - L(sf)(\lambda).$$

This shows that differentiation in the Laplace (frequency) domain corresponds to multiplication in the original (time) domain.


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