Laplace transform of power function


In the defining integralDlmfPlanetmath (http://planetmath.org/ImproperIntegral)

{tr}=0e-sttr𝑑t

of the Laplace transformDlmfMathworldPlanetmath of the power functionDlmfDlmfPlanetmathttr,  we make the substitution (http://planetmath.org/SubstitutionForIntegration)  u:=st:

{tr}=0e-u(us)rdus=1sn+10e-uur+1-1𝑑u

Here we have assumed that  r>-1  and s>0.  According to the definition of the gamma functionDlmfDlmfMathworldPlanetmath, the last integral is equal to Γ(r+1).  Thus we obtain

{tr}=Γ(r+1)sr+1. (1)

The special case  r=-12  gives the result

{1t}=πs. (2)
Title Laplace transform of power function
Canonical name LaplaceTransformOfPowerFunction
Date of creation 2013-03-22 18:17:42
Last modified on 2013-03-22 18:17:42
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 6
Author pahio (2872)
Entry type Derivation
Classification msc 44A10
Related topic EvaluatingTheGammaFunctionAt12