Integral $intlimits_0^inftyfrac{e^{-Ak^{2}}}{k}sin(kr)dk$












4












$begingroup$


I have the following integral



$$
f(r)=int_{0}^{infty}frac{exp(-Ak^{2})}{k},sin(kr),mathrm{d}k
$$



with $A>0$ and $r>0$. I know from Wolfram that the result should be



$$
f(r)=frac{pi}{2}text{erf} left (frac{r}{2sqrt{A}} right)
$$



The problem is I have no clue how to obtain this result. I would be very curious how to do this by myself.
I know that this is a sine transform, and I should be able to use a similar way like shown here
Fourier transform of the error function, erf (x).
But already when introducing the sgn function I get lost. Any hints are appreciated.






integration definite-integrals improper-integrals fourier-transform error-function







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    I have the following integral



    $$
    f(r)=int_{0}^{infty}frac{exp(-Ak^{2})}{k},sin(kr),mathrm{d}k
    $$



    with $A>0$ and $r>0$. I know from Wolfram that the result should be



    $$
    f(r)=frac{pi}{2}text{erf} left (frac{r}{2sqrt{A}} right)
    $$



    The problem is I have no clue how to obtain this result. I would be very curious how to do this by myself.
    I know that this is a sine transform, and I should be able to use a similar way like shown here
    Fourier transform of the error function, erf (x).
    But already when introducing the sgn function I get lost. Any hints are appreciated.






    integration definite-integrals improper-integrals fourier-transform error-function







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      I have the following integral



      $$
      f(r)=int_{0}^{infty}frac{exp(-Ak^{2})}{k},sin(kr),mathrm{d}k
      $$



      with $A>0$ and $r>0$. I know from Wolfram that the result should be



      $$
      f(r)=frac{pi}{2}text{erf} left (frac{r}{2sqrt{A}} right)
      $$



      The problem is I have no clue how to obtain this result. I would be very curious how to do this by myself.
      I know that this is a sine transform, and I should be able to use a similar way like shown here
      Fourier transform of the error function, erf (x).
      But already when introducing the sgn function I get lost. Any hints are appreciated.






      integration definite-integrals improper-integrals fourier-transform error-function







      share|cite|improve this question











      $endgroup$




      I have the following integral



      $$
      f(r)=int_{0}^{infty}frac{exp(-Ak^{2})}{k},sin(kr),mathrm{d}k
      $$



      with $A>0$ and $r>0$. I know from Wolfram that the result should be



      $$
      f(r)=frac{pi}{2}text{erf} left (frac{r}{2sqrt{A}} right)
      $$



      The problem is I have no clue how to obtain this result. I would be very curious how to do this by myself.
      I know that this is a sine transform, and I should be able to use a similar way like shown here
      Fourier transform of the error function, erf (x).
      But already when introducing the sgn function I get lost. Any hints are appreciated.






      integration definite-integrals improper-integrals fourier-transform error-function




      integration definite-integrals improper-integrals fourier-transform error-function






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 1 at 13:45









      Martin Sleziak

      44.8k10119273




      44.8k10119273










      asked Jan 22 at 17:35









      zodiaczodiac

      667




      667






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Hint. By setting
          $$
          I(r):=int_{0}^{infty}frac{e^{-Ak^{2}}}{k}sin(kr)dk,qquad A>0,,r>0,
          $$          one is allowed to differentiate under the integral sign with respect to $r$ to get
          $$
          I'(r)=int_{0}^{infty}e^{-Ak^{2}}cos(kr)dk,qquad A>0,,r>0,
          $$           that is
          $$
          I'(r)=frac12sqrt{frac{pi}A}e^{large-frac{r^2}{4A}},qquad A>0,,r>0,
          $$           giving
          $$
          I(r)=frac{pi}{2}text{erf} left (frac{r}{2sqrt{A}} right),qquad A>0,,r>0.
          $$






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            I was just about to write the exact same thing! :)
            $endgroup$
            – clathratus
            Jan 22 at 17:51











          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%2f3083442%2fintegral-int-limits-0-infty-frace-ak2k-sinkrdk%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









          3












          $begingroup$

          Hint. By setting
          $$
          I(r):=int_{0}^{infty}frac{e^{-Ak^{2}}}{k}sin(kr)dk,qquad A>0,,r>0,
          $$          one is allowed to differentiate under the integral sign with respect to $r$ to get
          $$
          I'(r)=int_{0}^{infty}e^{-Ak^{2}}cos(kr)dk,qquad A>0,,r>0,
          $$           that is
          $$
          I'(r)=frac12sqrt{frac{pi}A}e^{large-frac{r^2}{4A}},qquad A>0,,r>0,
          $$           giving
          $$
          I(r)=frac{pi}{2}text{erf} left (frac{r}{2sqrt{A}} right),qquad A>0,,r>0.
          $$






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            I was just about to write the exact same thing! :)
            $endgroup$
            – clathratus
            Jan 22 at 17:51
















          3












          $begingroup$

          Hint. By setting
          $$
          I(r):=int_{0}^{infty}frac{e^{-Ak^{2}}}{k}sin(kr)dk,qquad A>0,,r>0,
          $$          one is allowed to differentiate under the integral sign with respect to $r$ to get
          $$
          I'(r)=int_{0}^{infty}e^{-Ak^{2}}cos(kr)dk,qquad A>0,,r>0,
          $$           that is
          $$
          I'(r)=frac12sqrt{frac{pi}A}e^{large-frac{r^2}{4A}},qquad A>0,,r>0,
          $$           giving
          $$
          I(r)=frac{pi}{2}text{erf} left (frac{r}{2sqrt{A}} right),qquad A>0,,r>0.
          $$






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            I was just about to write the exact same thing! :)
            $endgroup$
            – clathratus
            Jan 22 at 17:51














          3












          3








          3





          $begingroup$

          Hint. By setting
          $$
          I(r):=int_{0}^{infty}frac{e^{-Ak^{2}}}{k}sin(kr)dk,qquad A>0,,r>0,
          $$          one is allowed to differentiate under the integral sign with respect to $r$ to get
          $$
          I'(r)=int_{0}^{infty}e^{-Ak^{2}}cos(kr)dk,qquad A>0,,r>0,
          $$           that is
          $$
          I'(r)=frac12sqrt{frac{pi}A}e^{large-frac{r^2}{4A}},qquad A>0,,r>0,
          $$           giving
          $$
          I(r)=frac{pi}{2}text{erf} left (frac{r}{2sqrt{A}} right),qquad A>0,,r>0.
          $$






          share|cite|improve this answer









          $endgroup$



          Hint. By setting
          $$
          I(r):=int_{0}^{infty}frac{e^{-Ak^{2}}}{k}sin(kr)dk,qquad A>0,,r>0,
          $$          one is allowed to differentiate under the integral sign with respect to $r$ to get
          $$
          I'(r)=int_{0}^{infty}e^{-Ak^{2}}cos(kr)dk,qquad A>0,,r>0,
          $$           that is
          $$
          I'(r)=frac12sqrt{frac{pi}A}e^{large-frac{r^2}{4A}},qquad A>0,,r>0,
          $$           giving
          $$
          I(r)=frac{pi}{2}text{erf} left (frac{r}{2sqrt{A}} right),qquad A>0,,r>0.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 22 at 17:50









          Olivier OloaOlivier Oloa

          109k17178294




          109k17178294








          • 2




            $begingroup$
            I was just about to write the exact same thing! :)
            $endgroup$
            – clathratus
            Jan 22 at 17:51














          • 2




            $begingroup$
            I was just about to write the exact same thing! :)
            $endgroup$
            – clathratus
            Jan 22 at 17:51








          2




          2




          $begingroup$
          I was just about to write the exact same thing! :)
          $endgroup$
          – clathratus
          Jan 22 at 17:51




          $begingroup$
          I was just about to write the exact same thing! :)
          $endgroup$
          – clathratus
          Jan 22 at 17:51


















          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%2f3083442%2fintegral-int-limits-0-infty-frace-ak2k-sinkrdk%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

          MongoDB - Not Authorized To Execute Command

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
          Read more

          Npm cannot find a required file even through it is in the searched directory

          Image
          up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
          Read more

          in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

          Image
          1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
          Read more