Laplace transform of complementary error function $operatorname{erfc}(1/sqrt{t})$- using infinite series












0












$begingroup$


The Laplace transform of the complementary error function $operatorname{erfc}left(frac{1}{sqrt{t}}right)$ is



$$Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}=Lleft{1-operatorname{erf}left(frac{1}{sqrt{t}}right) right}$$



That is



$$begin{aligned}
Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}
&=Lleft{1-frac{2}{sqrt{pi}} int_{0}^{1/sqrt{t}} e^{-x^2} dx right}\
&=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!} int_{0}^{1/sqrt{t}} x^{2n} dx right} \
&=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} frac{1}{t^{n+1/2}} right} \
&=frac{1}{p} - frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} Lleft{frac{1}{t^{n+1/2}} right} \
end{aligned}$$



After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!






laplace-transform







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    The Laplace transform of the complementary error function $operatorname{erfc}left(frac{1}{sqrt{t}}right)$ is



    $$Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}=Lleft{1-operatorname{erf}left(frac{1}{sqrt{t}}right) right}$$



    That is



    $$begin{aligned}
    Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}
    &=Lleft{1-frac{2}{sqrt{pi}} int_{0}^{1/sqrt{t}} e^{-x^2} dx right}\
    &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!} int_{0}^{1/sqrt{t}} x^{2n} dx right} \
    &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} frac{1}{t^{n+1/2}} right} \
    &=frac{1}{p} - frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} Lleft{frac{1}{t^{n+1/2}} right} \
    end{aligned}$$



    After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!






    laplace-transform







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      The Laplace transform of the complementary error function $operatorname{erfc}left(frac{1}{sqrt{t}}right)$ is



      $$Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}=Lleft{1-operatorname{erf}left(frac{1}{sqrt{t}}right) right}$$



      That is



      $$begin{aligned}
      Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}
      &=Lleft{1-frac{2}{sqrt{pi}} int_{0}^{1/sqrt{t}} e^{-x^2} dx right}\
      &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!} int_{0}^{1/sqrt{t}} x^{2n} dx right} \
      &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} frac{1}{t^{n+1/2}} right} \
      &=frac{1}{p} - frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} Lleft{frac{1}{t^{n+1/2}} right} \
      end{aligned}$$



      After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!






      laplace-transform







      share|cite|improve this question











      $endgroup$




      The Laplace transform of the complementary error function $operatorname{erfc}left(frac{1}{sqrt{t}}right)$ is



      $$Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}=Lleft{1-operatorname{erf}left(frac{1}{sqrt{t}}right) right}$$



      That is



      $$begin{aligned}
      Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}
      &=Lleft{1-frac{2}{sqrt{pi}} int_{0}^{1/sqrt{t}} e^{-x^2} dx right}\
      &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!} int_{0}^{1/sqrt{t}} x^{2n} dx right} \
      &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} frac{1}{t^{n+1/2}} right} \
      &=frac{1}{p} - frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} Lleft{frac{1}{t^{n+1/2}} right} \
      end{aligned}$$



      After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!






      laplace-transform




      laplace-transform






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 22 '16 at 1:29









      Bungo

      13.7k22148




      13.7k22148










      asked Aug 22 '16 at 1:08









      Venkatesan MurugesanVenkatesan Murugesan

      913




      913






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          It looks fine. Now you just have to exploit
          $$mathcal{L}(1)=frac{1}{s},qquad mathcal{L}left(frac{1}{t^{n+1/2}}right) = s^{n-frac{1}{2}}Gammaleft(frac{1}{2}-nright)=s^{n-frac{1}{2}}(-1)^nfrac{2^nsqrt{pi}}{(2n-1)!!}. $$
          Anyway, It would have been faster to exploit the properties of the Laplace transform. $text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that
          $$ mathcal{L}left(text{Erfc}left(frac{1}{sqrt{t}}right)right) = color{red}{frac{e^{-2sqrt{s}}}{s}}. $$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
            $endgroup$
            – Venkatesan Murugesan
            Aug 22 '16 at 15:52











          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',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          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%2f1899537%2flaplace-transform-of-complementary-error-function-operatornameerfc1-sqrtt%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          It looks fine. Now you just have to exploit
          $$mathcal{L}(1)=frac{1}{s},qquad mathcal{L}left(frac{1}{t^{n+1/2}}right) = s^{n-frac{1}{2}}Gammaleft(frac{1}{2}-nright)=s^{n-frac{1}{2}}(-1)^nfrac{2^nsqrt{pi}}{(2n-1)!!}. $$
          Anyway, It would have been faster to exploit the properties of the Laplace transform. $text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that
          $$ mathcal{L}left(text{Erfc}left(frac{1}{sqrt{t}}right)right) = color{red}{frac{e^{-2sqrt{s}}}{s}}. $$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
            $endgroup$
            – Venkatesan Murugesan
            Aug 22 '16 at 15:52
















          0












          $begingroup$

          It looks fine. Now you just have to exploit
          $$mathcal{L}(1)=frac{1}{s},qquad mathcal{L}left(frac{1}{t^{n+1/2}}right) = s^{n-frac{1}{2}}Gammaleft(frac{1}{2}-nright)=s^{n-frac{1}{2}}(-1)^nfrac{2^nsqrt{pi}}{(2n-1)!!}. $$
          Anyway, It would have been faster to exploit the properties of the Laplace transform. $text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that
          $$ mathcal{L}left(text{Erfc}left(frac{1}{sqrt{t}}right)right) = color{red}{frac{e^{-2sqrt{s}}}{s}}. $$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
            $endgroup$
            – Venkatesan Murugesan
            Aug 22 '16 at 15:52














          0












          0








          0





          $begingroup$

          It looks fine. Now you just have to exploit
          $$mathcal{L}(1)=frac{1}{s},qquad mathcal{L}left(frac{1}{t^{n+1/2}}right) = s^{n-frac{1}{2}}Gammaleft(frac{1}{2}-nright)=s^{n-frac{1}{2}}(-1)^nfrac{2^nsqrt{pi}}{(2n-1)!!}. $$
          Anyway, It would have been faster to exploit the properties of the Laplace transform. $text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that
          $$ mathcal{L}left(text{Erfc}left(frac{1}{sqrt{t}}right)right) = color{red}{frac{e^{-2sqrt{s}}}{s}}. $$






          share|cite|improve this answer









          $endgroup$



          It looks fine. Now you just have to exploit
          $$mathcal{L}(1)=frac{1}{s},qquad mathcal{L}left(frac{1}{t^{n+1/2}}right) = s^{n-frac{1}{2}}Gammaleft(frac{1}{2}-nright)=s^{n-frac{1}{2}}(-1)^nfrac{2^nsqrt{pi}}{(2n-1)!!}. $$
          Anyway, It would have been faster to exploit the properties of the Laplace transform. $text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that
          $$ mathcal{L}left(text{Erfc}left(frac{1}{sqrt{t}}right)right) = color{red}{frac{e^{-2sqrt{s}}}{s}}. $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 22 '16 at 1:26









          Jack D'AurizioJack D'Aurizio

          290k33282662




          290k33282662








          • 1




            $begingroup$
            Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
            $endgroup$
            – Venkatesan Murugesan
            Aug 22 '16 at 15:52














          • 1




            $begingroup$
            Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
            $endgroup$
            – Venkatesan Murugesan
            Aug 22 '16 at 15:52








          1




          1




          $begingroup$
          Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
          $endgroup$
          – Venkatesan Murugesan
          Aug 22 '16 at 15:52




          $begingroup$
          Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
          $endgroup$
          – Venkatesan Murugesan
          Aug 22 '16 at 15:52


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1899537%2flaplace-transform-of-complementary-error-function-operatornameerfc1-sqrtt%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Human spaceflight

          Image
          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
          Read more

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          Image
          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
          Read more

          File:DeusFollowingSea.jpg

          Image
          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
          Read more