Mathematical Methods for Physics and Engineering - Matematica.NET

INTEGRAL TRANSFORMS(iii) Once again using the definition (13.53) we have∫ ∞¯f n (s) = t n e −st dt.0Integrating by parts we find[ ] −t¯f n e −st ∞n (s) =+ n ∫ ∞t n−1 e −st dts0s 0=0+ n s ¯f n−1 (s), if s>0.We now have a recursion relation between successive transforms and by calculating¯f 0 we can infer ¯f 1 , ¯f2 ,etc.Sincet 0 = 1, (i) above gives¯f 0 = 1 , if s>0, (13.55)sand¯f 1 (s) = 1 s , 2 ¯f2 (s) = 2!s , ..., 3 ¯fn (s) = n! if s>0.s n+1Thus, 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 for f(t), given ¯f(s),is not straightforwardly obtained from (13.53). The general method for obtainingan inverse Laplace transform makes use of complex variable theory and is notdiscussed until chapter 25. However, progress can be made without having to findan explicit inverse, since we can prepare from (13.53) a ‘dictionary’ of the Laplacetransforms 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 linearso thatL −1[ a ¯f 1 (s)+b ¯f 2 (s) ] = af 1 (t)+bf 2 (t). (13.56)In many practical problems the method of partial fractions can be useful inproducing an expression from which the inverse Laplace transform can be found.◮Using table 13.1 find f(t) if¯f(s) = s +3s(s +1) .Using partial fractions ¯f(s) may be written¯f(s) = 3 s − 2s +1 .§ This is not strictly true, since two functions can differ from one another at a finite number ofisolated points but have the same Laplace transform.454