INTEGRAL TRANSFORMS
(iii) Once again using the definition (13.53) we havef ̄n(s)=∫∞
0tne−stdt.Integrating by parts we findf ̄n(s)=[
−tne−st
s]∞
0+
n
s∫∞
0tn−^1 e−stdt=0+
n
sf ̄n− 1 (s), ifs> 0.We now have a recursion relation between successive transforms and by calculating
f ̄ 0 we can inferf ̄ 1 ,f ̄ 2 ,etc.Sincet^0 = 1, (i) above givesf ̄ 0 =^1
s, ifs> 0 , (13.55)andf ̄ 1 (s)=^1
s^2, f ̄ 2 (s)=2!
s^3,...,f ̄n(s)=n!
sn+1ifs> 0.Thus, in each case (i)–(iii), direct application of the definition of the Laplace transform
(13.53) yields the required result.
Unlike that for the Fourier transform, the inversion of the Laplace transformis not an easy operation to perform, since an explicit formula forf(t), givenf ̄(s),
is not straightforwardly obtained from (13.53). The general method for obtaining
an inverse Laplace transform makes use of complex variable theory and is not
discussed until chapter 25. However, progress can be made without having to find
anexplicitinverse, since we can prepare from (13.53) a ‘dictionary’ of the Laplace
transforms of common functions and, when faced with an inversion to carry out,
hope to find the given transform (together with its parent function) in the listing.
Such a list is given in table 13.1.
When finding inverse Laplace transforms using table 13.1, it is useful to notethat for all practical purposes the inverse Laplace transform is unique§and linear
so that
L−^1[
af ̄ 1 (s)+bf ̄ 2 (s)]
=af 1 (t)+bf 2 (t). (13.56)In many practical problems the method of partial fractions can be useful in
producing an expression from which the inverse Laplace transform can be found.
Using table 13.1 findf(t)iff ̄(s)= s+3
s(s+1).
Using partial fractionsf ̄(s) may be written
f ̄(s)=^3
s−
2
s+1.
§This is not strictly true, since two functions can differ from one another at a finite number of
isolated points but have thesameLaplace transform.