# differentiation of Laplace transform with respect to parameter

If

 $f(t,x)\;\,\curvearrowleft\;\,F(s,x),$

then one can differentiate both functions with respect to the parametre $x$:

 $\displaystyle f^{\prime}_{x}(t,x)\;\,\curvearrowleft\;\,F^{\prime}_{x}(s,x)$ (1)

(1) may be written also as

 ${\displaystyle\mathcal{L}\{\frac{\partial}{\partial x}f(t,x)}\}\;=\;\frac{% \partial}{\partial x}\mathcal{L}\{f(t,x)\}.$ (2)

Proof.  We differentiate partially both sides of the defining equation

 $F(s,x)\;:=\int_{0}^{\infty}e^{-st}f(t,x)\,dt,$

on the right hand side under the integration sign (http://planetmath.org/differentiationundertheintegralsign), getting

 $\displaystyle F^{\prime}_{x}(s,x)\;=\;\int_{0}^{\infty}e^{-st}f^{\prime}_{x}(t% ,x)\,dx,$ (3)

which means same as (1).  Q.E.D.

Example.  If the rule

 $\frac{s}{s^{2}\!-\!a^{2}}\;\,\curvearrowright\;\,\cosh{at}$

is differentiated with respect to $a$, the result is

 $\frac{2as}{(s^{2}\!-\!a^{2})^{2}}\;\,\curvearrowright\;\,t\,\sinh{at}.$

## References

• 1 K. Väisälä: Laplace-muunnos.  Handout Nr. 163. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1968).
Title differentiation of Laplace transform with respect to parameter DifferentiationOfLaplaceTransformWithRespectToParameter 2014-03-09 12:52:11 2014-03-09 12:52:11 pahio (2872) pahio (2872) 2 pahio (2872) Theorem msc 44A10