Solve Inverse Laplace Transform Using Input Integral Theorem

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
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
    1
    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
      1
      down vote

      favorite









      up vote
      1
      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 55 mins ago









      Anthony Krivonos

      1878




      1878




















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          3
          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
            10 mins ago

















          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




















              Your Answer




              StackExchange.ifUsing("editor", function ()
              return StackExchange.using("mathjaxEditing", function ()
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              );
              );
              , "mathjax-editing");

              StackExchange.ready(function()
              var channelOptions =
              tags: "".split(" "),
              id: "69"
              ;
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function()
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled)
              StackExchange.using("snippets", function()
              createEditor();
              );

              else
              createEditor();

              );

              function createEditor()
              StackExchange.prepareEditor(
              heartbeatType: 'answer',
              convertImagesToLinks: true,
              noModals: false,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              );



              );













               

              draft saved


              draft discarded


















              StackExchange.ready(
              function ()
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2929658%2fsolve-inverse-laplace-transform-using-input-integral-theorem%23new-answer', 'question_page');

              );

              Post as a guest

















              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              3
              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
                10 mins ago














              up vote
              3
              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
                10 mins ago












              up vote
              3
              down vote










              up vote
              3
              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 47 mins ago









              Nosrati

              22.9k61951




              22.9k61951











              • +1 color text makes things easy to understand
                – Isham
                10 mins ago
















              • +1 color text makes things easy to understand
                – Isham
                10 mins ago















              +1 color text makes things easy to understand
              – Isham
              10 mins ago




              +1 color text makes things easy to understand
              – Isham
              10 mins ago










              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 48 mins ago









                  Math Lover

                  13k31333




                  13k31333




















                      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 38 mins ago









                          Mohammad Riazi-Kermani

                          32.2k41853




                          32.2k41853



























                               

                              draft saved


                              draft discarded















































                               


                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2929658%2fsolve-inverse-laplace-transform-using-input-integral-theorem%23new-answer', 'question_page');

                              );

                              Post as a guest
































































                              Comments

                              Post a Comment

                              Popular posts from this blog

                              What does second last employer means? [closed]

                              Image
                              Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
                              Read more

                              Installing NextGIS Connect into QGIS 3?

                              Image
                              Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
                              Read more

                              One-line joke

                              A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
                              Read more