Prove $int_0^inftyoperatorname{sech} x,dx=pi/2$, and deduce $int_0^1operatorname{sech}^{-1}x,dx$












0















Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$




I can prove the first statement (see below), but I was unable to deduce the value of the second integral.










share|cite|improve this question
























  • Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
    – Matti P.
    Jan 2 at 14:23










  • For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
    – Matti P.
    Jan 2 at 14:30










  • I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
    – Abec
    Jan 2 at 22:38
















0















Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$




I can prove the first statement (see below), but I was unable to deduce the value of the second integral.










share|cite|improve this question
























  • Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
    – Matti P.
    Jan 2 at 14:23










  • For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
    – Matti P.
    Jan 2 at 14:30










  • I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
    – Abec
    Jan 2 at 22:38














0












0








0








Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$




I can prove the first statement (see below), but I was unable to deduce the value of the second integral.










share|cite|improve this question
















Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$




I can prove the first statement (see below), but I was unable to deduce the value of the second integral.







improper-integrals trigonometric-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 1:41







Abec

















asked Jan 2 at 14:10









AbecAbec

145




145












  • Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
    – Matti P.
    Jan 2 at 14:23










  • For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
    – Matti P.
    Jan 2 at 14:30










  • I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
    – Abec
    Jan 2 at 22:38


















  • Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
    – Matti P.
    Jan 2 at 14:23










  • For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
    – Matti P.
    Jan 2 at 14:30










  • I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
    – Abec
    Jan 2 at 22:38
















Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
– Matti P.
Jan 2 at 14:23




Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
– Matti P.
Jan 2 at 14:23












For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
– Matti P.
Jan 2 at 14:30




For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
– Matti P.
Jan 2 at 14:30












I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
– Abec
Jan 2 at 22:38




I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
– Abec
Jan 2 at 22:38










2 Answers
2






active

oldest

votes


















0














Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}

then by integration by parts you should be done.






share|cite|improve this answer































    0














    When it comes to the second part of your question:



    begin{equation}
    I = int f^{-1}(t):dt
    end{equation}



    Let $x = f^{-1}(t)$ or $t = f(x)$, then



    begin{equation}
    I = int x cdot f'(x):dx
    end{equation}



    Integrate by parts:



    begin{align}
    v'(x) &= f'(x) & u(x) &= x \
    v(x) &= f(x) & u'(x) &= 1
    end{align}



    Thus,



    begin{align}
    I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
    &= x cdot f(x) - F(x)
    end{align}



    Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have



    begin{align}
    I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
    end{align}



    So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.



    begin{equation}
    F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
    end{equation}



    Hence,



    begin{equation}
    Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
    end{equation}



    And thus:



    begin{equation}
    int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
    end{equation}



    Where $C$ is the constant of integration.






    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',
      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%2f3059517%2fprove-int-0-infty-operatornamesech-x-dx-pi-2-and-deduce-int-01-opera%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









      0














      Note that by setting $u = operatorname{sech}^{-1} x$ we have that
      begin{align}
      int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
      end{align}

      then by integration by parts you should be done.






      share|cite|improve this answer




























        0














        Note that by setting $u = operatorname{sech}^{-1} x$ we have that
        begin{align}
        int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
        end{align}

        then by integration by parts you should be done.






        share|cite|improve this answer


























          0












          0








          0






          Note that by setting $u = operatorname{sech}^{-1} x$ we have that
          begin{align}
          int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
          end{align}

          then by integration by parts you should be done.






          share|cite|improve this answer














          Note that by setting $u = operatorname{sech}^{-1} x$ we have that
          begin{align}
          int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
          end{align}

          then by integration by parts you should be done.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 23:13

























          answered Jan 2 at 22:42









          Jacky ChongJacky Chong

          17.8k21128




          17.8k21128























              0














              When it comes to the second part of your question:



              begin{equation}
              I = int f^{-1}(t):dt
              end{equation}



              Let $x = f^{-1}(t)$ or $t = f(x)$, then



              begin{equation}
              I = int x cdot f'(x):dx
              end{equation}



              Integrate by parts:



              begin{align}
              v'(x) &= f'(x) & u(x) &= x \
              v(x) &= f(x) & u'(x) &= 1
              end{align}



              Thus,



              begin{align}
              I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
              &= x cdot f(x) - F(x)
              end{align}



              Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have



              begin{align}
              I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
              end{align}



              So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.



              begin{equation}
              F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
              end{equation}



              Hence,



              begin{equation}
              Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
              end{equation}



              And thus:



              begin{equation}
              int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
              end{equation}



              Where $C$ is the constant of integration.






              share|cite|improve this answer


























                0














                When it comes to the second part of your question:



                begin{equation}
                I = int f^{-1}(t):dt
                end{equation}



                Let $x = f^{-1}(t)$ or $t = f(x)$, then



                begin{equation}
                I = int x cdot f'(x):dx
                end{equation}



                Integrate by parts:



                begin{align}
                v'(x) &= f'(x) & u(x) &= x \
                v(x) &= f(x) & u'(x) &= 1
                end{align}



                Thus,



                begin{align}
                I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
                &= x cdot f(x) - F(x)
                end{align}



                Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have



                begin{align}
                I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
                end{align}



                So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.



                begin{equation}
                F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
                end{equation}



                Hence,



                begin{equation}
                Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
                end{equation}



                And thus:



                begin{equation}
                int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
                end{equation}



                Where $C$ is the constant of integration.






                share|cite|improve this answer
























                  0












                  0








                  0






                  When it comes to the second part of your question:



                  begin{equation}
                  I = int f^{-1}(t):dt
                  end{equation}



                  Let $x = f^{-1}(t)$ or $t = f(x)$, then



                  begin{equation}
                  I = int x cdot f'(x):dx
                  end{equation}



                  Integrate by parts:



                  begin{align}
                  v'(x) &= f'(x) & u(x) &= x \
                  v(x) &= f(x) & u'(x) &= 1
                  end{align}



                  Thus,



                  begin{align}
                  I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
                  &= x cdot f(x) - F(x)
                  end{align}



                  Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have



                  begin{align}
                  I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
                  end{align}



                  So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.



                  begin{equation}
                  F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
                  end{equation}



                  Hence,



                  begin{equation}
                  Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
                  end{equation}



                  And thus:



                  begin{equation}
                  int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
                  end{equation}



                  Where $C$ is the constant of integration.






                  share|cite|improve this answer












                  When it comes to the second part of your question:



                  begin{equation}
                  I = int f^{-1}(t):dt
                  end{equation}



                  Let $x = f^{-1}(t)$ or $t = f(x)$, then



                  begin{equation}
                  I = int x cdot f'(x):dx
                  end{equation}



                  Integrate by parts:



                  begin{align}
                  v'(x) &= f'(x) & u(x) &= x \
                  v(x) &= f(x) & u'(x) &= 1
                  end{align}



                  Thus,



                  begin{align}
                  I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
                  &= x cdot f(x) - F(x)
                  end{align}



                  Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have



                  begin{align}
                  I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
                  end{align}



                  So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.



                  begin{equation}
                  F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
                  end{equation}



                  Hence,



                  begin{equation}
                  Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
                  end{equation}



                  And thus:



                  begin{equation}
                  int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
                  end{equation}



                  Where $C$ is the constant of integration.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 8:37









                  DavidGDavidG

                  1,830620




                  1,830620






























                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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.


                      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%2f3059517%2fprove-int-0-infty-operatornamesech-x-dx-pi-2-and-deduce-int-01-opera%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

                      1300-talet

                      1300-talet

                      Display a custom attribute below product name in the front-end Magento 1.9.3.8