Determining if this integral converge












0












$begingroup$


Does the following integral converge:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx $$



my attempt:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx = int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx ge int_1^infty frac{-x}{xsqrt{x^2+1}}dx = int_1^infty frac{-1}{sqrt{x^2+1}}dx ge -1 cdot int_1^infty frac{1}{sqrt{x^2+x^2}}dx = -frac{1}{sqrt{2}} int_1^infty frac{1}{sqrt{x^2}} = -frac{1}{sqrt{2}} int_1^infty frac{1}{x} $$



and since I know $ frac{1}{x} $ doesn't converge, the integral also doesn't??






calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    No. You've shown the value of the integral is $ge-infty$ which doesn't help.
    $endgroup$
    – saulspatz
    Jan 9 at 14:40


















0












$begingroup$


Does the following integral converge:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx $$



my attempt:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx = int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx ge int_1^infty frac{-x}{xsqrt{x^2+1}}dx = int_1^infty frac{-1}{sqrt{x^2+1}}dx ge -1 cdot int_1^infty frac{1}{sqrt{x^2+x^2}}dx = -frac{1}{sqrt{2}} int_1^infty frac{1}{sqrt{x^2}} = -frac{1}{sqrt{2}} int_1^infty frac{1}{x} $$



and since I know $ frac{1}{x} $ doesn't converge, the integral also doesn't??






calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    No. You've shown the value of the integral is $ge-infty$ which doesn't help.
    $endgroup$
    – saulspatz
    Jan 9 at 14:40
















0












0








0


1



$begingroup$


Does the following integral converge:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx $$



my attempt:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx = int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx ge int_1^infty frac{-x}{xsqrt{x^2+1}}dx = int_1^infty frac{-1}{sqrt{x^2+1}}dx ge -1 cdot int_1^infty frac{1}{sqrt{x^2+x^2}}dx = -frac{1}{sqrt{2}} int_1^infty frac{1}{sqrt{x^2}} = -frac{1}{sqrt{2}} int_1^infty frac{1}{x} $$



and since I know $ frac{1}{x} $ doesn't converge, the integral also doesn't??






calculus







share|cite|improve this question









$endgroup$




Does the following integral converge:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx $$



my attempt:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx = int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx ge int_1^infty frac{-x}{xsqrt{x^2+1}}dx = int_1^infty frac{-1}{sqrt{x^2+1}}dx ge -1 cdot int_1^infty frac{1}{sqrt{x^2+x^2}}dx = -frac{1}{sqrt{2}} int_1^infty frac{1}{sqrt{x^2}} = -frac{1}{sqrt{2}} int_1^infty frac{1}{x} $$



and since I know $ frac{1}{x} $ doesn't converge, the integral also doesn't??






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 14:36









bm1125bm1125

68516




68516








  • 1




    $begingroup$
    No. You've shown the value of the integral is $ge-infty$ which doesn't help.
    $endgroup$
    – saulspatz
    Jan 9 at 14:40
















  • 1




    $begingroup$
    No. You've shown the value of the integral is $ge-infty$ which doesn't help.
    $endgroup$
    – saulspatz
    Jan 9 at 14:40










1




1




$begingroup$
No. You've shown the value of the integral is $ge-infty$ which doesn't help.
$endgroup$
– saulspatz
Jan 9 at 14:40






$begingroup$
No. You've shown the value of the integral is $ge-infty$ which doesn't help.
$endgroup$
– saulspatz
Jan 9 at 14:40












2 Answers
2






active

oldest

votes


















1












$begingroup$

begin{eqnarray*}
int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx
& = & int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx \
& = & int_1^infty frac{1}{xsqrt{x^2+1}(sqrt{x^2+1} + x)}dx \
& stackrel{sqrt{x^2+1}geq x}{leq} & int_1^infty frac{1}{xcdot x (x+x)}dx = int_1^infty frac{1}{2x^3}dx < infty
end{eqnarray*}



So, it is convergent.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    For large $x$, we see by Taylor expansion that $$sqrt{x^2 + 1} = xsqrt{1 + frac 1 {x^2}} approx xleft(1 + frac 1 {2x^2}right).$$ Thus $$sqrt{x^2 + 1} - x approx frac1{2x},$$ so as $x to infty$, we have $$frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}} approx frac{1/(2x)}{x^2}approx frac{1}{2x^{3}}.$$ Hence the integral converges by comparison with $$int^infty_{1}frac{dx}{2x^{3}}.$$






    share|cite|improve this answer









    $endgroup$














      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%2f3067510%2fdetermining-if-this-integral-converge%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      begin{eqnarray*}
      int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx
      & = & int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx \
      & = & int_1^infty frac{1}{xsqrt{x^2+1}(sqrt{x^2+1} + x)}dx \
      & stackrel{sqrt{x^2+1}geq x}{leq} & int_1^infty frac{1}{xcdot x (x+x)}dx = int_1^infty frac{1}{2x^3}dx < infty
      end{eqnarray*}



      So, it is convergent.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        begin{eqnarray*}
        int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx
        & = & int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx \
        & = & int_1^infty frac{1}{xsqrt{x^2+1}(sqrt{x^2+1} + x)}dx \
        & stackrel{sqrt{x^2+1}geq x}{leq} & int_1^infty frac{1}{xcdot x (x+x)}dx = int_1^infty frac{1}{2x^3}dx < infty
        end{eqnarray*}



        So, it is convergent.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          begin{eqnarray*}
          int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx
          & = & int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx \
          & = & int_1^infty frac{1}{xsqrt{x^2+1}(sqrt{x^2+1} + x)}dx \
          & stackrel{sqrt{x^2+1}geq x}{leq} & int_1^infty frac{1}{xcdot x (x+x)}dx = int_1^infty frac{1}{2x^3}dx < infty
          end{eqnarray*}



          So, it is convergent.






          share|cite|improve this answer









          $endgroup$



          begin{eqnarray*}
          int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx
          & = & int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx \
          & = & int_1^infty frac{1}{xsqrt{x^2+1}(sqrt{x^2+1} + x)}dx \
          & stackrel{sqrt{x^2+1}geq x}{leq} & int_1^infty frac{1}{xcdot x (x+x)}dx = int_1^infty frac{1}{2x^3}dx < infty
          end{eqnarray*}



          So, it is convergent.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 14:58









          trancelocationtrancelocation

          13.6k1828




          13.6k1828























              2












              $begingroup$

              For large $x$, we see by Taylor expansion that $$sqrt{x^2 + 1} = xsqrt{1 + frac 1 {x^2}} approx xleft(1 + frac 1 {2x^2}right).$$ Thus $$sqrt{x^2 + 1} - x approx frac1{2x},$$ so as $x to infty$, we have $$frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}} approx frac{1/(2x)}{x^2}approx frac{1}{2x^{3}}.$$ Hence the integral converges by comparison with $$int^infty_{1}frac{dx}{2x^{3}}.$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                For large $x$, we see by Taylor expansion that $$sqrt{x^2 + 1} = xsqrt{1 + frac 1 {x^2}} approx xleft(1 + frac 1 {2x^2}right).$$ Thus $$sqrt{x^2 + 1} - x approx frac1{2x},$$ so as $x to infty$, we have $$frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}} approx frac{1/(2x)}{x^2}approx frac{1}{2x^{3}}.$$ Hence the integral converges by comparison with $$int^infty_{1}frac{dx}{2x^{3}}.$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For large $x$, we see by Taylor expansion that $$sqrt{x^2 + 1} = xsqrt{1 + frac 1 {x^2}} approx xleft(1 + frac 1 {2x^2}right).$$ Thus $$sqrt{x^2 + 1} - x approx frac1{2x},$$ so as $x to infty$, we have $$frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}} approx frac{1/(2x)}{x^2}approx frac{1}{2x^{3}}.$$ Hence the integral converges by comparison with $$int^infty_{1}frac{dx}{2x^{3}}.$$






                  share|cite|improve this answer









                  $endgroup$



                  For large $x$, we see by Taylor expansion that $$sqrt{x^2 + 1} = xsqrt{1 + frac 1 {x^2}} approx xleft(1 + frac 1 {2x^2}right).$$ Thus $$sqrt{x^2 + 1} - x approx frac1{2x},$$ so as $x to infty$, we have $$frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}} approx frac{1/(2x)}{x^2}approx frac{1}{2x^{3}}.$$ Hence the integral converges by comparison with $$int^infty_{1}frac{dx}{2x^{3}}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 9 at 14:50









                  User8128User8128

                  10.9k1622




                  10.9k1622






























                      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%2f3067510%2fdetermining-if-this-integral-converge%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

                      Bressuire

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                      Read more

                      Cabo Verde

                      Image
                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                      Read more

                      Gyllenstierna

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                      Read more