Integral of $ln(cosh x)$?












5












$begingroup$


$$int ln(cosh x) dx$$



Tried for some time, couldn't get anywhere.



This IS an elementary function, isn't it?



Thanks in advance.






calculus integration logarithms hyperbolic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the domain of integration? Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset maths here so that your posts say what you want them to, and also look good. As this question appears to be homework, please consider reading this page for information about asking effective homework-related questions. Cheers! ref: meta.math.stackexchange.com/a/6348/290189
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 11 '17 at 19:09










  • $begingroup$
    Wolfram Alpha gives a result involving the polylogarithmic function $mathrm{Li}_2.$ I don't think that's considered elementary.
    $endgroup$
    – David K
    Mar 11 '17 at 19:12










  • $begingroup$
    this integral can not expressed by the known elementary functions
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 11 '17 at 19:13










  • $begingroup$
    "This IS an elementary function, isn't it?": no. What makes you think that ?
    $endgroup$
    – Yves Daoust
    Mar 11 '17 at 19:28


















5












$begingroup$


$$int ln(cosh x) dx$$



Tried for some time, couldn't get anywhere.



This IS an elementary function, isn't it?



Thanks in advance.






calculus integration logarithms hyperbolic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the domain of integration? Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset maths here so that your posts say what you want them to, and also look good. As this question appears to be homework, please consider reading this page for information about asking effective homework-related questions. Cheers! ref: meta.math.stackexchange.com/a/6348/290189
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 11 '17 at 19:09










  • $begingroup$
    Wolfram Alpha gives a result involving the polylogarithmic function $mathrm{Li}_2.$ I don't think that's considered elementary.
    $endgroup$
    – David K
    Mar 11 '17 at 19:12










  • $begingroup$
    this integral can not expressed by the known elementary functions
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 11 '17 at 19:13










  • $begingroup$
    "This IS an elementary function, isn't it?": no. What makes you think that ?
    $endgroup$
    – Yves Daoust
    Mar 11 '17 at 19:28
















5












5








5





$begingroup$


$$int ln(cosh x) dx$$



Tried for some time, couldn't get anywhere.



This IS an elementary function, isn't it?



Thanks in advance.






calculus integration logarithms hyperbolic-functions







share|cite|improve this question











$endgroup$




$$int ln(cosh x) dx$$



Tried for some time, couldn't get anywhere.



This IS an elementary function, isn't it?



Thanks in advance.






calculus integration logarithms hyperbolic-functions




calculus integration logarithms hyperbolic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 11 '17 at 19:45









unseen_rider

462418




462418










asked Mar 11 '17 at 19:07









we3f2fwe3f2f

414




414












  • $begingroup$
    What is the domain of integration? Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset maths here so that your posts say what you want them to, and also look good. As this question appears to be homework, please consider reading this page for information about asking effective homework-related questions. Cheers! ref: meta.math.stackexchange.com/a/6348/290189
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 11 '17 at 19:09










  • $begingroup$
    Wolfram Alpha gives a result involving the polylogarithmic function $mathrm{Li}_2.$ I don't think that's considered elementary.
    $endgroup$
    – David K
    Mar 11 '17 at 19:12










  • $begingroup$
    this integral can not expressed by the known elementary functions
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 11 '17 at 19:13










  • $begingroup$
    "This IS an elementary function, isn't it?": no. What makes you think that ?
    $endgroup$
    – Yves Daoust
    Mar 11 '17 at 19:28




















  • $begingroup$
    What is the domain of integration? Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset maths here so that your posts say what you want them to, and also look good. As this question appears to be homework, please consider reading this page for information about asking effective homework-related questions. Cheers! ref: meta.math.stackexchange.com/a/6348/290189
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 11 '17 at 19:09










  • $begingroup$
    Wolfram Alpha gives a result involving the polylogarithmic function $mathrm{Li}_2.$ I don't think that's considered elementary.
    $endgroup$
    – David K
    Mar 11 '17 at 19:12










  • $begingroup$
    this integral can not expressed by the known elementary functions
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 11 '17 at 19:13










  • $begingroup$
    "This IS an elementary function, isn't it?": no. What makes you think that ?
    $endgroup$
    – Yves Daoust
    Mar 11 '17 at 19:28


















$begingroup$
What is the domain of integration? Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset maths here so that your posts say what you want them to, and also look good. As this question appears to be homework, please consider reading this page for information about asking effective homework-related questions. Cheers! ref: meta.math.stackexchange.com/a/6348/290189
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 11 '17 at 19:09




$begingroup$
What is the domain of integration? Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset maths here so that your posts say what you want them to, and also look good. As this question appears to be homework, please consider reading this page for information about asking effective homework-related questions. Cheers! ref: meta.math.stackexchange.com/a/6348/290189
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 11 '17 at 19:09












$begingroup$
Wolfram Alpha gives a result involving the polylogarithmic function $mathrm{Li}_2.$ I don't think that's considered elementary.
$endgroup$
– David K
Mar 11 '17 at 19:12




$begingroup$
Wolfram Alpha gives a result involving the polylogarithmic function $mathrm{Li}_2.$ I don't think that's considered elementary.
$endgroup$
– David K
Mar 11 '17 at 19:12












$begingroup$
this integral can not expressed by the known elementary functions
$endgroup$
– Dr. Sonnhard Graubner
Mar 11 '17 at 19:13




$begingroup$
this integral can not expressed by the known elementary functions
$endgroup$
– Dr. Sonnhard Graubner
Mar 11 '17 at 19:13












$begingroup$
"This IS an elementary function, isn't it?": no. What makes you think that ?
$endgroup$
– Yves Daoust
Mar 11 '17 at 19:28






$begingroup$
"This IS an elementary function, isn't it?": no. What makes you think that ?
$endgroup$
– Yves Daoust
Mar 11 '17 at 19:28












2 Answers
2






active

oldest

votes


















3












$begingroup$

The integral $int log(cosh(x)),dx$ cannot be expressed in terms of elementary functions. To see this, we enforce the substitution $x=log(u)$ to obtain



$$begin{align}
int log(cosh(x)),dx&=int frac{log(1+u^2)-log(u)}{u},du-log(2)x \\
&=int frac{log(1+u^2)}{u},du -frac12 x^2-log(2)x tag 1
end{align}$$



Next, we let $u=sqrt{v}$ in the integral on the right-hand side of $(1)$ to reveal



$$begin{align}
int log(cosh(x)),dx&=int frac{log(1+v)}{2v},dv -frac12 x^2-log(2)x \\
&=bbox[5px,border:2px solid #C0A000]{-frac12text{Li}_2(-e^{2x})-frac12x^2-log(2)x+C}tag 2
end{align}$$



which expresses the primitive in terms of the dilogarithm function, a special function given by $text{Li}_2(x)=-int_0^x frac{log(1-t)}{t},dt$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    $$ln(cosh(x))=lnleft(frac{e^x+e^{-x}}2right)=x-ln(2)+ln(1+e^{-2x})= x-ln(2)+sum_{k=1}^infty(-1)^{k+1}frac{e^{-2kx}}k.$$



    You can integrate term-wise and get a rapidly converging series (for positive $x$), which is asymptotic to the parabola $dfrac{x^2}2-ln(2),x+C$.






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      The integral $int log(cosh(x)),dx$ cannot be expressed in terms of elementary functions. To see this, we enforce the substitution $x=log(u)$ to obtain



      $$begin{align}
      int log(cosh(x)),dx&=int frac{log(1+u^2)-log(u)}{u},du-log(2)x \\
      &=int frac{log(1+u^2)}{u},du -frac12 x^2-log(2)x tag 1
      end{align}$$



      Next, we let $u=sqrt{v}$ in the integral on the right-hand side of $(1)$ to reveal



      $$begin{align}
      int log(cosh(x)),dx&=int frac{log(1+v)}{2v},dv -frac12 x^2-log(2)x \\
      &=bbox[5px,border:2px solid #C0A000]{-frac12text{Li}_2(-e^{2x})-frac12x^2-log(2)x+C}tag 2
      end{align}$$



      which expresses the primitive in terms of the dilogarithm function, a special function given by $text{Li}_2(x)=-int_0^x frac{log(1-t)}{t},dt$.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        The integral $int log(cosh(x)),dx$ cannot be expressed in terms of elementary functions. To see this, we enforce the substitution $x=log(u)$ to obtain



        $$begin{align}
        int log(cosh(x)),dx&=int frac{log(1+u^2)-log(u)}{u},du-log(2)x \\
        &=int frac{log(1+u^2)}{u},du -frac12 x^2-log(2)x tag 1
        end{align}$$



        Next, we let $u=sqrt{v}$ in the integral on the right-hand side of $(1)$ to reveal



        $$begin{align}
        int log(cosh(x)),dx&=int frac{log(1+v)}{2v},dv -frac12 x^2-log(2)x \\
        &=bbox[5px,border:2px solid #C0A000]{-frac12text{Li}_2(-e^{2x})-frac12x^2-log(2)x+C}tag 2
        end{align}$$



        which expresses the primitive in terms of the dilogarithm function, a special function given by $text{Li}_2(x)=-int_0^x frac{log(1-t)}{t},dt$.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          The integral $int log(cosh(x)),dx$ cannot be expressed in terms of elementary functions. To see this, we enforce the substitution $x=log(u)$ to obtain



          $$begin{align}
          int log(cosh(x)),dx&=int frac{log(1+u^2)-log(u)}{u},du-log(2)x \\
          &=int frac{log(1+u^2)}{u},du -frac12 x^2-log(2)x tag 1
          end{align}$$



          Next, we let $u=sqrt{v}$ in the integral on the right-hand side of $(1)$ to reveal



          $$begin{align}
          int log(cosh(x)),dx&=int frac{log(1+v)}{2v},dv -frac12 x^2-log(2)x \\
          &=bbox[5px,border:2px solid #C0A000]{-frac12text{Li}_2(-e^{2x})-frac12x^2-log(2)x+C}tag 2
          end{align}$$



          which expresses the primitive in terms of the dilogarithm function, a special function given by $text{Li}_2(x)=-int_0^x frac{log(1-t)}{t},dt$.






          share|cite|improve this answer











          $endgroup$



          The integral $int log(cosh(x)),dx$ cannot be expressed in terms of elementary functions. To see this, we enforce the substitution $x=log(u)$ to obtain



          $$begin{align}
          int log(cosh(x)),dx&=int frac{log(1+u^2)-log(u)}{u},du-log(2)x \\
          &=int frac{log(1+u^2)}{u},du -frac12 x^2-log(2)x tag 1
          end{align}$$



          Next, we let $u=sqrt{v}$ in the integral on the right-hand side of $(1)$ to reveal



          $$begin{align}
          int log(cosh(x)),dx&=int frac{log(1+v)}{2v},dv -frac12 x^2-log(2)x \\
          &=bbox[5px,border:2px solid #C0A000]{-frac12text{Li}_2(-e^{2x})-frac12x^2-log(2)x+C}tag 2
          end{align}$$



          which expresses the primitive in terms of the dilogarithm function, a special function given by $text{Li}_2(x)=-int_0^x frac{log(1-t)}{t},dt$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 11 '17 at 20:43

























          answered Mar 11 '17 at 19:52









          Mark ViolaMark Viola

          134k1278177




          134k1278177























              3












              $begingroup$

              $$ln(cosh(x))=lnleft(frac{e^x+e^{-x}}2right)=x-ln(2)+ln(1+e^{-2x})= x-ln(2)+sum_{k=1}^infty(-1)^{k+1}frac{e^{-2kx}}k.$$



              You can integrate term-wise and get a rapidly converging series (for positive $x$), which is asymptotic to the parabola $dfrac{x^2}2-ln(2),x+C$.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                $$ln(cosh(x))=lnleft(frac{e^x+e^{-x}}2right)=x-ln(2)+ln(1+e^{-2x})= x-ln(2)+sum_{k=1}^infty(-1)^{k+1}frac{e^{-2kx}}k.$$



                You can integrate term-wise and get a rapidly converging series (for positive $x$), which is asymptotic to the parabola $dfrac{x^2}2-ln(2),x+C$.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  $$ln(cosh(x))=lnleft(frac{e^x+e^{-x}}2right)=x-ln(2)+ln(1+e^{-2x})= x-ln(2)+sum_{k=1}^infty(-1)^{k+1}frac{e^{-2kx}}k.$$



                  You can integrate term-wise and get a rapidly converging series (for positive $x$), which is asymptotic to the parabola $dfrac{x^2}2-ln(2),x+C$.






                  share|cite|improve this answer











                  $endgroup$



                  $$ln(cosh(x))=lnleft(frac{e^x+e^{-x}}2right)=x-ln(2)+ln(1+e^{-2x})= x-ln(2)+sum_{k=1}^infty(-1)^{k+1}frac{e^{-2kx}}k.$$



                  You can integrate term-wise and get a rapidly converging series (for positive $x$), which is asymptotic to the parabola $dfrac{x^2}2-ln(2),x+C$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 29 '18 at 9:17









                  Community♦

                  1




                  1










                  answered Mar 11 '17 at 19:38









                  Yves DaoustYves Daoust

                  133k676232




                  133k676232






























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