Evaluate $int_{0}^{frac{pi}{4}} ln(sec x)dx$












2












$begingroup$


Evaluate $$P=int_{0}^{frac{pi}{4}} ln(sec x)dx$$



My try: I tried using its complimentary integral:



Let $$Q=int_{0}^{frac{pi}{4}} ln(csc x)dx$$



Adding both we get:



$$P+Q=int_{0}^{frac{pi}{4}}ln(sec xcsc x)dx$$ $implies$



$$2P+2Q=int_{0}^{frac{pi}{4}}ln(sec^2 xcsc^2 x)dx=int_{0}^{frac{pi}{4}}lnleft(frac{4}{4sin^2 xcos^2 x}right)dx$$ $implies$



$$2P+2Q=frac{pi}{4}ln 4-int_{0}^{frac{pi}{4}}lnleft(sin^2 2xright)dx$$



$$2P+2Q=frac{pi}{2}ln 2-2 int_{0}^{frac{pi}{4}}ln(sin 2x)dx$$



Using the substitution $2x=t$ we get



$$2P+2Q=frac{pi}{2}ln 2- int_{0}^{frac{pi}{2}}ln(sin t)dt$$



Using the formula:



$$int_{0}^{frac{pi}{2}}ln(sin t)dt=frac{-pi}{2}ln 2$$ we get



$$2P+2Q=pi ln 2$$



$$P+Q=frac{pi}{2}ln 2$$



Is there any way to find $P-Q$






algebra-precalculus definite-integrals logarithms







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Evaluate $$P=int_{0}^{frac{pi}{4}} ln(sec x)dx$$



    My try: I tried using its complimentary integral:



    Let $$Q=int_{0}^{frac{pi}{4}} ln(csc x)dx$$



    Adding both we get:



    $$P+Q=int_{0}^{frac{pi}{4}}ln(sec xcsc x)dx$$ $implies$



    $$2P+2Q=int_{0}^{frac{pi}{4}}ln(sec^2 xcsc^2 x)dx=int_{0}^{frac{pi}{4}}lnleft(frac{4}{4sin^2 xcos^2 x}right)dx$$ $implies$



    $$2P+2Q=frac{pi}{4}ln 4-int_{0}^{frac{pi}{4}}lnleft(sin^2 2xright)dx$$



    $$2P+2Q=frac{pi}{2}ln 2-2 int_{0}^{frac{pi}{4}}ln(sin 2x)dx$$



    Using the substitution $2x=t$ we get



    $$2P+2Q=frac{pi}{2}ln 2- int_{0}^{frac{pi}{2}}ln(sin t)dt$$



    Using the formula:



    $$int_{0}^{frac{pi}{2}}ln(sin t)dt=frac{-pi}{2}ln 2$$ we get



    $$2P+2Q=pi ln 2$$



    $$P+Q=frac{pi}{2}ln 2$$



    Is there any way to find $P-Q$






    algebra-precalculus definite-integrals logarithms







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Evaluate $$P=int_{0}^{frac{pi}{4}} ln(sec x)dx$$



      My try: I tried using its complimentary integral:



      Let $$Q=int_{0}^{frac{pi}{4}} ln(csc x)dx$$



      Adding both we get:



      $$P+Q=int_{0}^{frac{pi}{4}}ln(sec xcsc x)dx$$ $implies$



      $$2P+2Q=int_{0}^{frac{pi}{4}}ln(sec^2 xcsc^2 x)dx=int_{0}^{frac{pi}{4}}lnleft(frac{4}{4sin^2 xcos^2 x}right)dx$$ $implies$



      $$2P+2Q=frac{pi}{4}ln 4-int_{0}^{frac{pi}{4}}lnleft(sin^2 2xright)dx$$



      $$2P+2Q=frac{pi}{2}ln 2-2 int_{0}^{frac{pi}{4}}ln(sin 2x)dx$$



      Using the substitution $2x=t$ we get



      $$2P+2Q=frac{pi}{2}ln 2- int_{0}^{frac{pi}{2}}ln(sin t)dt$$



      Using the formula:



      $$int_{0}^{frac{pi}{2}}ln(sin t)dt=frac{-pi}{2}ln 2$$ we get



      $$2P+2Q=pi ln 2$$



      $$P+Q=frac{pi}{2}ln 2$$



      Is there any way to find $P-Q$






      algebra-precalculus definite-integrals logarithms







      share|cite|improve this question









      $endgroup$




      Evaluate $$P=int_{0}^{frac{pi}{4}} ln(sec x)dx$$



      My try: I tried using its complimentary integral:



      Let $$Q=int_{0}^{frac{pi}{4}} ln(csc x)dx$$



      Adding both we get:



      $$P+Q=int_{0}^{frac{pi}{4}}ln(sec xcsc x)dx$$ $implies$



      $$2P+2Q=int_{0}^{frac{pi}{4}}ln(sec^2 xcsc^2 x)dx=int_{0}^{frac{pi}{4}}lnleft(frac{4}{4sin^2 xcos^2 x}right)dx$$ $implies$



      $$2P+2Q=frac{pi}{4}ln 4-int_{0}^{frac{pi}{4}}lnleft(sin^2 2xright)dx$$



      $$2P+2Q=frac{pi}{2}ln 2-2 int_{0}^{frac{pi}{4}}ln(sin 2x)dx$$



      Using the substitution $2x=t$ we get



      $$2P+2Q=frac{pi}{2}ln 2- int_{0}^{frac{pi}{2}}ln(sin t)dt$$



      Using the formula:



      $$int_{0}^{frac{pi}{2}}ln(sin t)dt=frac{-pi}{2}ln 2$$ we get



      $$2P+2Q=pi ln 2$$



      $$P+Q=frac{pi}{2}ln 2$$



      Is there any way to find $P-Q$






      algebra-precalculus definite-integrals logarithms




      algebra-precalculus definite-integrals logarithms






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 11 '18 at 17:53









      Umesh shankarUmesh shankar

      2,68031219




      2,68031219






















          2 Answers
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          $begingroup$

          In fact
          begin{eqnarray*}
          P-Q&=&int_0^{frac{pi}{4}}ln(tan t)dt\
          &=&int_0^1frac{ln u}{1+u^2}du\
          &=&int_0^1ln usum_{n=0}^infty(-1)^nu^{2n}du\
          &=&sum_{n=0}^infty(-1)^nint_0^1u^{2n}ln udu\
          &=&-sum_{n=0}^infty(-1)^nfrac{1}{(2n+1)^2}\
          &=&-C
          end{eqnarray*}
          where $C$ is the Catalan constant.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Through the properties of the logarithm, the integral can be rewritten as$$mathfrak{I}=-intlimits_0^{pi/4}mathrm dx,logcos x$$Now use the Fourier expansion for $logcos x$ which is




            $$logcos x=sumlimits_{ngeq1}frac {(-1)^{n-1}cos 2nx}{n}-log 2$$




            Now integrate each time termwise to get that$$begin{align*}mathfrak{I} & =frac {pi}4log 2+sumlimits_{ngeq1}frac {(-1)^n}{n}intlimits_0^{pi/4}mathrm dx,cos 2nx\ & =frac {pi}4log 2+frac 12sumlimits_{ngeq1}frac {(-1)^n}{n^2}sinleft(frac {pi n}2right)\ & =frac {pi}4log 2-frac 12Gend{align*}$$where $G$ is Catalan’s constant.






            share|cite|improve this answer









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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

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              active

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              active

              oldest

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              3












              $begingroup$

              In fact
              begin{eqnarray*}
              P-Q&=&int_0^{frac{pi}{4}}ln(tan t)dt\
              &=&int_0^1frac{ln u}{1+u^2}du\
              &=&int_0^1ln usum_{n=0}^infty(-1)^nu^{2n}du\
              &=&sum_{n=0}^infty(-1)^nint_0^1u^{2n}ln udu\
              &=&-sum_{n=0}^infty(-1)^nfrac{1}{(2n+1)^2}\
              &=&-C
              end{eqnarray*}
              where $C$ is the Catalan constant.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                In fact
                begin{eqnarray*}
                P-Q&=&int_0^{frac{pi}{4}}ln(tan t)dt\
                &=&int_0^1frac{ln u}{1+u^2}du\
                &=&int_0^1ln usum_{n=0}^infty(-1)^nu^{2n}du\
                &=&sum_{n=0}^infty(-1)^nint_0^1u^{2n}ln udu\
                &=&-sum_{n=0}^infty(-1)^nfrac{1}{(2n+1)^2}\
                &=&-C
                end{eqnarray*}
                where $C$ is the Catalan constant.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  In fact
                  begin{eqnarray*}
                  P-Q&=&int_0^{frac{pi}{4}}ln(tan t)dt\
                  &=&int_0^1frac{ln u}{1+u^2}du\
                  &=&int_0^1ln usum_{n=0}^infty(-1)^nu^{2n}du\
                  &=&sum_{n=0}^infty(-1)^nint_0^1u^{2n}ln udu\
                  &=&-sum_{n=0}^infty(-1)^nfrac{1}{(2n+1)^2}\
                  &=&-C
                  end{eqnarray*}
                  where $C$ is the Catalan constant.






                  share|cite|improve this answer









                  $endgroup$



                  In fact
                  begin{eqnarray*}
                  P-Q&=&int_0^{frac{pi}{4}}ln(tan t)dt\
                  &=&int_0^1frac{ln u}{1+u^2}du\
                  &=&int_0^1ln usum_{n=0}^infty(-1)^nu^{2n}du\
                  &=&sum_{n=0}^infty(-1)^nint_0^1u^{2n}ln udu\
                  &=&-sum_{n=0}^infty(-1)^nfrac{1}{(2n+1)^2}\
                  &=&-C
                  end{eqnarray*}
                  where $C$ is the Catalan constant.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 18:04









                  xpaulxpaul

                  22.7k24455




                  22.7k24455























                      2












                      $begingroup$

                      Through the properties of the logarithm, the integral can be rewritten as$$mathfrak{I}=-intlimits_0^{pi/4}mathrm dx,logcos x$$Now use the Fourier expansion for $logcos x$ which is




                      $$logcos x=sumlimits_{ngeq1}frac {(-1)^{n-1}cos 2nx}{n}-log 2$$




                      Now integrate each time termwise to get that$$begin{align*}mathfrak{I} & =frac {pi}4log 2+sumlimits_{ngeq1}frac {(-1)^n}{n}intlimits_0^{pi/4}mathrm dx,cos 2nx\ & =frac {pi}4log 2+frac 12sumlimits_{ngeq1}frac {(-1)^n}{n^2}sinleft(frac {pi n}2right)\ & =frac {pi}4log 2-frac 12Gend{align*}$$where $G$ is Catalan’s constant.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Through the properties of the logarithm, the integral can be rewritten as$$mathfrak{I}=-intlimits_0^{pi/4}mathrm dx,logcos x$$Now use the Fourier expansion for $logcos x$ which is




                        $$logcos x=sumlimits_{ngeq1}frac {(-1)^{n-1}cos 2nx}{n}-log 2$$




                        Now integrate each time termwise to get that$$begin{align*}mathfrak{I} & =frac {pi}4log 2+sumlimits_{ngeq1}frac {(-1)^n}{n}intlimits_0^{pi/4}mathrm dx,cos 2nx\ & =frac {pi}4log 2+frac 12sumlimits_{ngeq1}frac {(-1)^n}{n^2}sinleft(frac {pi n}2right)\ & =frac {pi}4log 2-frac 12Gend{align*}$$where $G$ is Catalan’s constant.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Through the properties of the logarithm, the integral can be rewritten as$$mathfrak{I}=-intlimits_0^{pi/4}mathrm dx,logcos x$$Now use the Fourier expansion for $logcos x$ which is




                          $$logcos x=sumlimits_{ngeq1}frac {(-1)^{n-1}cos 2nx}{n}-log 2$$




                          Now integrate each time termwise to get that$$begin{align*}mathfrak{I} & =frac {pi}4log 2+sumlimits_{ngeq1}frac {(-1)^n}{n}intlimits_0^{pi/4}mathrm dx,cos 2nx\ & =frac {pi}4log 2+frac 12sumlimits_{ngeq1}frac {(-1)^n}{n^2}sinleft(frac {pi n}2right)\ & =frac {pi}4log 2-frac 12Gend{align*}$$where $G$ is Catalan’s constant.






                          share|cite|improve this answer









                          $endgroup$



                          Through the properties of the logarithm, the integral can be rewritten as$$mathfrak{I}=-intlimits_0^{pi/4}mathrm dx,logcos x$$Now use the Fourier expansion for $logcos x$ which is




                          $$logcos x=sumlimits_{ngeq1}frac {(-1)^{n-1}cos 2nx}{n}-log 2$$




                          Now integrate each time termwise to get that$$begin{align*}mathfrak{I} & =frac {pi}4log 2+sumlimits_{ngeq1}frac {(-1)^n}{n}intlimits_0^{pi/4}mathrm dx,cos 2nx\ & =frac {pi}4log 2+frac 12sumlimits_{ngeq1}frac {(-1)^n}{n^2}sinleft(frac {pi n}2right)\ & =frac {pi}4log 2-frac 12Gend{align*}$$where $G$ is Catalan’s constant.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 11 '18 at 18:36









                          Frank W.Frank W.

                          3,5331321




                          3,5331321






























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