# Laplace transform of derivative

Theorem. If the real function $t\mapsto f(t)$ and its derivative are Laplace-transformable and $f$ is continuous^{} for $t>0$, then

$\mathcal{L}\{{f}^{\prime}(t)\}=sF(s)-\underset{t\to 0+}{lim}f(t).$ | (1) |

Proof. By the definition of Laplace transform^{} and using integration by parts, the left hand side of (1) may be written

$${\int}_{0}^{\mathrm{\infty}}{e}^{-st}{f}^{\prime}(t)\mathit{d}t=\underset{t=0}{\overset{\mathrm{\infty}}{/}}{e}^{-st}f(t)+s{\int}_{0}^{\mathrm{\infty}}{e}^{-st}f(t)\mathit{d}t=\underset{t\to \mathrm{\infty}}{lim}{e}^{-st}f(t)-\underset{t\to 0}{lim}{e}^{-st}f(t)+sF(s).$$ |

The Laplace-transformability of $f$ implies that ${e}^{-st}f(t)$ tends to zero as $t$ increases boundlessly. Thus the last expression leads to the right hand side of (1).

Title | Laplace transform of derivative |
---|---|

Canonical name | LaplaceTransformOfDerivative |

Date of creation | 2013-03-22 18:24:54 |

Last modified on | 2013-03-22 18:24:54 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 5 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 44A10 |

Related topic | SubstitutionNotation |