Evaluation of Integral using Laplace Transform

Theorem

Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of the real function $f$.

Then:

$\ds \int_0^{\to \infty} \map f t \rd t = \map F 0$

assuming the integral is convergent.

Proof

By definition of Laplace transform:

$\ds \int_0^{\to \infty} e^{-s t} \map f t \rd t = \map F s$

The result follows by taking the limit as $s \to 0$.

$\blacksquare$

Examples

Example 1

$\ds \int_0^\infty \dfrac {e^{-t} - e^{-3 t} } t \rd t = \ln 3$

Sources

• 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Evaluation of Integrals