Solve Inverse Laplace Transform Using Input Integral Theorem

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Problem



Using the input integral principle below



$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s), s > c $$



Find $ mathscrL^-1 left[ frac1s(s^2+1) right](t) $ without using partial fractions.



Attempt



Letting $ f(t) = mathscrL^-1 left[ frac1s(s^2+1) right](t) $,



$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ mathscrL^-1 left[ frac1s(s^2+1) right](t) right] (s), s > c $$



$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s left[ frac1s(s^2+1) right], s > c $$



$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s^2(s^2+1) , s > c $$



Notes



Perhaps I've approached this problem incorrectly, but I'm confused as how to proceed with it. All I'm looking for is a hint or correct first step in solving this problem, with a little bit of explanation as to what the correct method toward solving this problem entails.



Also, I searched up what the input integral principle is and I'm not finding anything on it. Did my professor invent this name or is it an alias for something else?



That being said, any help is appreciated. Thanks!






integration differential-equations definite-integrals laplace-transform







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    up vote
    2
    down vote

    favorite












    Problem



    Using the input integral principle below



    $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s), s > c $$



    Find $ mathscrL^-1 left[ frac1s(s^2+1) right](t) $ without using partial fractions.



    Attempt



    Letting $ f(t) = mathscrL^-1 left[ frac1s(s^2+1) right](t) $,



    $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ mathscrL^-1 left[ frac1s(s^2+1) right](t) right] (s), s > c $$



    $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s left[ frac1s(s^2+1) right], s > c $$



    $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s^2(s^2+1) , s > c $$



    Notes



    Perhaps I've approached this problem incorrectly, but I'm confused as how to proceed with it. All I'm looking for is a hint or correct first step in solving this problem, with a little bit of explanation as to what the correct method toward solving this problem entails.



    Also, I searched up what the input integral principle is and I'm not finding anything on it. Did my professor invent this name or is it an alias for something else?



    That being said, any help is appreciated. Thanks!






    integration differential-equations definite-integrals laplace-transform







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Problem



      Using the input integral principle below



      $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s), s > c $$



      Find $ mathscrL^-1 left[ frac1s(s^2+1) right](t) $ without using partial fractions.



      Attempt



      Letting $ f(t) = mathscrL^-1 left[ frac1s(s^2+1) right](t) $,



      $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ mathscrL^-1 left[ frac1s(s^2+1) right](t) right] (s), s > c $$



      $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s left[ frac1s(s^2+1) right], s > c $$



      $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s^2(s^2+1) , s > c $$



      Notes



      Perhaps I've approached this problem incorrectly, but I'm confused as how to proceed with it. All I'm looking for is a hint or correct first step in solving this problem, with a little bit of explanation as to what the correct method toward solving this problem entails.



      Also, I searched up what the input integral principle is and I'm not finding anything on it. Did my professor invent this name or is it an alias for something else?



      That being said, any help is appreciated. Thanks!






      integration differential-equations definite-integrals laplace-transform







      share|cite|improve this question













      Problem



      Using the input integral principle below



      $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s), s > c $$



      Find $ mathscrL^-1 left[ frac1s(s^2+1) right](t) $ without using partial fractions.



      Attempt



      Letting $ f(t) = mathscrL^-1 left[ frac1s(s^2+1) right](t) $,



      $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ mathscrL^-1 left[ frac1s(s^2+1) right](t) right] (s), s > c $$



      $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s left[ frac1s(s^2+1) right], s > c $$



      $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s^2(s^2+1) , s > c $$



      Notes



      Perhaps I've approached this problem incorrectly, but I'm confused as how to proceed with it. All I'm looking for is a hint or correct first step in solving this problem, with a little bit of explanation as to what the correct method toward solving this problem entails.



      Also, I searched up what the input integral principle is and I'm not finding anything on it. Did my professor invent this name or is it an alias for something else?



      That being said, any help is appreciated. Thanks!






      integration differential-equations definite-integrals laplace-transform




      integration differential-equations definite-integrals laplace-transform






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      asked Sep 25 at 2:42









      Anthony Krivonos

      1929




      1929




















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          4
          down vote













          Compare these
          $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) $$
          $$ mathscrL left[ int_0^t colorredsin u du right] (s) = frac1s colorredfrac1(s^2+1) $$






          share|cite|improve this answer




















          • +1 color text makes things easy to understand
            – Isham
            Sep 25 at 3:27










          • You really saved my as(ymptote) here––super descriptive, and I was able to just take the inverse Laplace transform of both sides to quickly get the solution described by the other answerers. Also, thanks for just providing a hint, as I was not looking for the full solution. Lastly, do you mind answering the last part of the question, at least to the best of your ability and with accuracy, before I mark this as the accepted answer? Thanks!
            – Anthony Krivonos
            Sep 25 at 3:55










          • You are welcome and did you google "input integral principle" ?
            – Nosrati
            Sep 25 at 4:00










          • Actually I haven't ever heard this name, but look at merounak.files.wordpress.com/2016/02/… page 190.
            – Nosrati
            Sep 25 at 4:08


















          up vote
          1
          down vote













          Hint: The implied way is
          $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) = frac1sfrac1s^2+1 implies mathscrL left[ f(t) right] (s) = frac1s^2+1.$$






          share|cite|improve this answer



























            up vote
            1
            down vote













            Note that your $$f(t)= sin t $$ so $$ mathscrL^-1 left[ frac1s(s^2+1) right](t)=int _0^t sin u du = 1- cos t$$






            share|cite|improve this answer




















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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              4
              down vote













              Compare these
              $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) $$
              $$ mathscrL left[ int_0^t colorredsin u du right] (s) = frac1s colorredfrac1(s^2+1) $$






              share|cite|improve this answer




















              • +1 color text makes things easy to understand
                – Isham
                Sep 25 at 3:27










              • You really saved my as(ymptote) here––super descriptive, and I was able to just take the inverse Laplace transform of both sides to quickly get the solution described by the other answerers. Also, thanks for just providing a hint, as I was not looking for the full solution. Lastly, do you mind answering the last part of the question, at least to the best of your ability and with accuracy, before I mark this as the accepted answer? Thanks!
                – Anthony Krivonos
                Sep 25 at 3:55










              • You are welcome and did you google "input integral principle" ?
                – Nosrati
                Sep 25 at 4:00










              • Actually I haven't ever heard this name, but look at merounak.files.wordpress.com/2016/02/… page 190.
                – Nosrati
                Sep 25 at 4:08















              up vote
              4
              down vote













              Compare these
              $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) $$
              $$ mathscrL left[ int_0^t colorredsin u du right] (s) = frac1s colorredfrac1(s^2+1) $$






              share|cite|improve this answer




















              • +1 color text makes things easy to understand
                – Isham
                Sep 25 at 3:27










              • You really saved my as(ymptote) here––super descriptive, and I was able to just take the inverse Laplace transform of both sides to quickly get the solution described by the other answerers. Also, thanks for just providing a hint, as I was not looking for the full solution. Lastly, do you mind answering the last part of the question, at least to the best of your ability and with accuracy, before I mark this as the accepted answer? Thanks!
                – Anthony Krivonos
                Sep 25 at 3:55










              • You are welcome and did you google "input integral principle" ?
                – Nosrati
                Sep 25 at 4:00










              • Actually I haven't ever heard this name, but look at merounak.files.wordpress.com/2016/02/… page 190.
                – Nosrati
                Sep 25 at 4:08













              up vote
              4
              down vote










              up vote
              4
              down vote









              Compare these
              $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) $$
              $$ mathscrL left[ int_0^t colorredsin u du right] (s) = frac1s colorredfrac1(s^2+1) $$






              share|cite|improve this answer












              Compare these
              $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) $$
              $$ mathscrL left[ int_0^t colorredsin u du right] (s) = frac1s colorredfrac1(s^2+1) $$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Sep 25 at 2:50









              Nosrati

              23.5k61952




              23.5k61952











              • +1 color text makes things easy to understand
                – Isham
                Sep 25 at 3:27










              • You really saved my as(ymptote) here––super descriptive, and I was able to just take the inverse Laplace transform of both sides to quickly get the solution described by the other answerers. Also, thanks for just providing a hint, as I was not looking for the full solution. Lastly, do you mind answering the last part of the question, at least to the best of your ability and with accuracy, before I mark this as the accepted answer? Thanks!
                – Anthony Krivonos
                Sep 25 at 3:55










              • You are welcome and did you google "input integral principle" ?
                – Nosrati
                Sep 25 at 4:00










              • Actually I haven't ever heard this name, but look at merounak.files.wordpress.com/2016/02/… page 190.
                – Nosrati
                Sep 25 at 4:08

















              • +1 color text makes things easy to understand
                – Isham
                Sep 25 at 3:27










              • You really saved my as(ymptote) here––super descriptive, and I was able to just take the inverse Laplace transform of both sides to quickly get the solution described by the other answerers. Also, thanks for just providing a hint, as I was not looking for the full solution. Lastly, do you mind answering the last part of the question, at least to the best of your ability and with accuracy, before I mark this as the accepted answer? Thanks!
                – Anthony Krivonos
                Sep 25 at 3:55










              • You are welcome and did you google "input integral principle" ?
                – Nosrati
                Sep 25 at 4:00










              • Actually I haven't ever heard this name, but look at merounak.files.wordpress.com/2016/02/… page 190.
                – Nosrati
                Sep 25 at 4:08
















              +1 color text makes things easy to understand
              – Isham
              Sep 25 at 3:27




              +1 color text makes things easy to understand
              – Isham
              Sep 25 at 3:27












              You really saved my as(ymptote) here––super descriptive, and I was able to just take the inverse Laplace transform of both sides to quickly get the solution described by the other answerers. Also, thanks for just providing a hint, as I was not looking for the full solution. Lastly, do you mind answering the last part of the question, at least to the best of your ability and with accuracy, before I mark this as the accepted answer? Thanks!
              – Anthony Krivonos
              Sep 25 at 3:55




              You really saved my as(ymptote) here––super descriptive, and I was able to just take the inverse Laplace transform of both sides to quickly get the solution described by the other answerers. Also, thanks for just providing a hint, as I was not looking for the full solution. Lastly, do you mind answering the last part of the question, at least to the best of your ability and with accuracy, before I mark this as the accepted answer? Thanks!
              – Anthony Krivonos
              Sep 25 at 3:55












              You are welcome and did you google "input integral principle" ?
              – Nosrati
              Sep 25 at 4:00




              You are welcome and did you google "input integral principle" ?
              – Nosrati
              Sep 25 at 4:00












              Actually I haven't ever heard this name, but look at merounak.files.wordpress.com/2016/02/… page 190.
              – Nosrati
              Sep 25 at 4:08





              Actually I haven't ever heard this name, but look at merounak.files.wordpress.com/2016/02/… page 190.
              – Nosrati
              Sep 25 at 4:08











              up vote
              1
              down vote













              Hint: The implied way is
              $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) = frac1sfrac1s^2+1 implies mathscrL left[ f(t) right] (s) = frac1s^2+1.$$






              share|cite|improve this answer
























                up vote
                1
                down vote













                Hint: The implied way is
                $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) = frac1sfrac1s^2+1 implies mathscrL left[ f(t) right] (s) = frac1s^2+1.$$






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Hint: The implied way is
                  $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) = frac1sfrac1s^2+1 implies mathscrL left[ f(t) right] (s) = frac1s^2+1.$$






                  share|cite|improve this answer












                  Hint: The implied way is
                  $$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) = frac1sfrac1s^2+1 implies mathscrL left[ f(t) right] (s) = frac1s^2+1.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 25 at 2:49









                  Math Lover

                  13.1k31333




                  13.1k31333




















                      up vote
                      1
                      down vote













                      Note that your $$f(t)= sin t $$ so $$ mathscrL^-1 left[ frac1s(s^2+1) right](t)=int _0^t sin u du = 1- cos t$$






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













                        Note that your $$f(t)= sin t $$ so $$ mathscrL^-1 left[ frac1s(s^2+1) right](t)=int _0^t sin u du = 1- cos t$$






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Note that your $$f(t)= sin t $$ so $$ mathscrL^-1 left[ frac1s(s^2+1) right](t)=int _0^t sin u du = 1- cos t$$






                          share|cite|improve this answer












                          Note that your $$f(t)= sin t $$ so $$ mathscrL^-1 left[ frac1s(s^2+1) right](t)=int _0^t sin u du = 1- cos t$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Sep 25 at 2:59









                          Mohammad Riazi-Kermani

                          33.4k41855




                          33.4k41855



























                               

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