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P. 47

Properties of Laplace Transform


                  If we define g(t) = 0 for t < 0, then g(t − τ) = 0 for t < τ and we can write (2.21)
                  as
                                                         ∫     ∫
                                                            ∞    ∞
                                 L(f(t)) · L(g(t)) =                e −st f(τ)g(t − τ)dtdτ.
                                                           0    0
                  Due to the hypotheses on f and g, the Laplace integrals of f and g converge absolutely
                  and hence, in view of the preceding calculation,

                                             ∫    ∫
                                                ∞    ∞
                                                        |e −st f(τ)g(t − τ)|dtdτ
                                              0     0
                  converges. This fact allows us to reverse the order of integration, ∗ so that

                                                        ∫    ∫
                                                           ∞    ∞
                                L(f(t)) · L(g(t)) =                e −st f(τ)g(t − τ)dτdt =
                                                          0    0
                                            ∫    (∫                           )
                                               ∞      t
                                         =              e −st f(τ)g(t − τ)dτ     dt =
                                             0       0
                                   ∫          (∫                      )
                                      ∞            t
                                =        e −st      f(τ)g(t − τ)dτ       dt = L[(f ∗ g)(t)].               2
                                     0            0
               . . . .
                .


                   Example 2.38,
                                                                       1
                                                   at
                                                        bt
                                              L(e ∗ e ) =                       .
                                                               (s − a)(s − b)
                   Moreover,

                          (                   )                 ∫  t
                                                                                       at
                                    1                                                 e − e   bt
                       −1                            at    bt         aτ b(t−τ)
                     L       (s − a)(s − b)      = e ∗ e =          e e        dτ =     a − b   , a ̸= b.
                .                                                 0
               . . . . .
                   Example 2.39, Find
                                                          (            )
                                                                 1
                                                       −1
                                                     L                    .
                                                             2
                                                            s (s − 1)
                .
               . . . . .
                   Solution. Previously we applied a partial fraction decomposition. But we
                   can also write
                                                       1          1      1
                                                              =     ·        ,
                                                   2
                                                  s (s − 1)      s 2  s − 1
                                        2
                                               t
                   where L(t) = 1/s , L(e ) = 1/(s − 1). By the convolution theorem,
                                                   1      1               t
                                                      ·        = L(t ∗ e ),
                                                   s 2  s − 1
                   and thus
                                             (             )
                                                     1
                                          −1                          t     t
                                        L                     = t ∗ e = e − t − 1.
               . . . .                          s (s − 1)
                                                 2
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