If f ( t ) = e b t − e a t t how to compute L { f ( t ) }I have a problem trying to solve∫+∞0e(b−s)ttI know that I can use the Ei (exponential integral function) but after that I don't know what's exactly this means.I begin with the definition L { f ( t ) } := ∫ 0 + ∞ e − s t f ( t ) d tIn my laplace transform table says that L { f ( t ) } = ln ⁡ s − a s − b How I know that is true?

Question

If   f  (  t  )  =                    e                  b          t                    −              e                  a          t                      t   how to compute       L    {  f  (  t  )  }I have a problem trying to solve∫+∞0e(b−s)ttI know that I can use the Ei (exponential integral function) but after that I don&#039;t know what&#039;s exactly this means.I begin with the definition       L    {  f  (  t  )  }  :=      ∫    0          +      ∞            e          −      s      t        f  (  t  )    d  tIn my laplace transform table says that       L    {  f  (  t  )  }  =  ln  ⁡            s      −      a              s      −      b      How I know that is true?

Lorenzo Aguilar · Accepted Answer

Your integral can instead be converted to a double integral first by noting that      ∫          s      −      b              s      −      a            e          −      x      t          d  x  =                    e                  −          (          s          −          b          )          t                    −              e                  −          (          s          −          a          )          t                      t    .Then                              L                {        f        (        t        )        }                            =                  ∫          0          ∞                                                    e                              −                (                s                −                b                )                t                                      −                          e                              −                (                s                −                a                )                t                                              t                        d        t                                          =                  ∫          0          ∞                          ∫                      s            −            b                                s            −            a                                    e                      −            x            t                                  d        x                d        t                                          =                  ∫                      s            −            b                                s            −            a                                    ∫          0          ∞                          e                      −            x            t                                  d        t                d        x                                          =                  ∫                      s            −            b                                s            −            a                                                d            x                    x                                                  =        ln        ⁡                  (                                    s              −              a                                      s              −              b                                )                .            Note when   a  ≠  b, for convergence we require   s  &amp;gt;  max  {  a  ,  b  }

puntdald8 · Accepted Answer

Here, we will use Feynman&#039;s Trick for differentiating under the integral.Let F(s) be defined by the integral                    (1)                    F        (        s        )                            =                  ∫          0          ∞                                                    e                              −                (                s                −                b                )                t                                      −                          e                              −                (                s                −                a                )                t                                              t                        d        t            Differentiating under the integral in (1) reveals                              F          ′                (        s        )                            =                  ∫          0          ∞                          (                      e                          −              (              s              −              a              )              t                                −                      e                          −              (              s              −              b              )              t                                )                        d        t                                          (2)                                  =                  1                      s            −            a                          −                  1                      s            −            b                              Next, integrating both sides of (2) we obtain      ∫          s        ∞        F    ′    (  u  )    d  u  =      ∫    s    ∞        (          1              u        −        a              −          1              u        −        b              )      d  ufrom which we see that  F  (  s  )  =  log  ⁡      (                  s        −        a                    s        −        b              )

Try to solve int_0^(oo) (e^((b−s)t))/(t)

Answered question

Answer & Explanation

New Questions in Differential Equations