How to do integral $ int_0^{infty} left(frac{a}{(e^{ax}-1)}-frac{b}{(e^{bx}-1)}right)mathrm{d}x$?












2












$begingroup$


I met this integral:
$$
int_0^{infty} left(frac{a}{(e^{ax}-1)}-frac{b}{(e^{bx}-1)}right)mathrm{d}x, text{ where }
,,0<a<b .
$$
It is a problem that showed up on my final exam of the lesson Mathematics I. Would anyone help me calculate this integral? Thanks.






definite-integrals







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$endgroup$












  • $begingroup$
    Two ideas : maybe it's possible to use Frulanni's integrals ? And what about to represent $frac{1}{e^{ax} - 1} = sum $ and integrate under the sum sign ?
    $endgroup$
    – openspace
    Jan 10 at 12:21












  • $begingroup$
    I think it must be zero
    $endgroup$
    – Aditya Garg
    Jan 10 at 12:25










  • $begingroup$
    For the left side make the substitution $u=e^{ax}-1$ and for the right side make the substitution $v=e^{bx}-1$ and see where this leads you
    $endgroup$
    – Henry Lee
    Jan 10 at 12:42
















2












$begingroup$


I met this integral:
$$
int_0^{infty} left(frac{a}{(e^{ax}-1)}-frac{b}{(e^{bx}-1)}right)mathrm{d}x, text{ where }
,,0<a<b .
$$
It is a problem that showed up on my final exam of the lesson Mathematics I. Would anyone help me calculate this integral? Thanks.






definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Two ideas : maybe it's possible to use Frulanni's integrals ? And what about to represent $frac{1}{e^{ax} - 1} = sum $ and integrate under the sum sign ?
    $endgroup$
    – openspace
    Jan 10 at 12:21












  • $begingroup$
    I think it must be zero
    $endgroup$
    – Aditya Garg
    Jan 10 at 12:25










  • $begingroup$
    For the left side make the substitution $u=e^{ax}-1$ and for the right side make the substitution $v=e^{bx}-1$ and see where this leads you
    $endgroup$
    – Henry Lee
    Jan 10 at 12:42














2












2








2





$begingroup$


I met this integral:
$$
int_0^{infty} left(frac{a}{(e^{ax}-1)}-frac{b}{(e^{bx}-1)}right)mathrm{d}x, text{ where }
,,0<a<b .
$$
It is a problem that showed up on my final exam of the lesson Mathematics I. Would anyone help me calculate this integral? Thanks.






definite-integrals







share|cite|improve this question











$endgroup$




I met this integral:
$$
int_0^{infty} left(frac{a}{(e^{ax}-1)}-frac{b}{(e^{bx}-1)}right)mathrm{d}x, text{ where }
,,0<a<b .
$$
It is a problem that showed up on my final exam of the lesson Mathematics I. Would anyone help me calculate this integral? Thanks.






definite-integrals




definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 2:02









Thomas Shelby

2,453221




2,453221










asked Jan 10 at 12:19









Zhu HuanhuanZhu Huanhuan

31816




31816












  • $begingroup$
    Two ideas : maybe it's possible to use Frulanni's integrals ? And what about to represent $frac{1}{e^{ax} - 1} = sum $ and integrate under the sum sign ?
    $endgroup$
    – openspace
    Jan 10 at 12:21












  • $begingroup$
    I think it must be zero
    $endgroup$
    – Aditya Garg
    Jan 10 at 12:25










  • $begingroup$
    For the left side make the substitution $u=e^{ax}-1$ and for the right side make the substitution $v=e^{bx}-1$ and see where this leads you
    $endgroup$
    – Henry Lee
    Jan 10 at 12:42


















  • $begingroup$
    Two ideas : maybe it's possible to use Frulanni's integrals ? And what about to represent $frac{1}{e^{ax} - 1} = sum $ and integrate under the sum sign ?
    $endgroup$
    – openspace
    Jan 10 at 12:21












  • $begingroup$
    I think it must be zero
    $endgroup$
    – Aditya Garg
    Jan 10 at 12:25










  • $begingroup$
    For the left side make the substitution $u=e^{ax}-1$ and for the right side make the substitution $v=e^{bx}-1$ and see where this leads you
    $endgroup$
    – Henry Lee
    Jan 10 at 12:42
















$begingroup$
Two ideas : maybe it's possible to use Frulanni's integrals ? And what about to represent $frac{1}{e^{ax} - 1} = sum $ and integrate under the sum sign ?
$endgroup$
– openspace
Jan 10 at 12:21






$begingroup$
Two ideas : maybe it's possible to use Frulanni's integrals ? And what about to represent $frac{1}{e^{ax} - 1} = sum $ and integrate under the sum sign ?
$endgroup$
– openspace
Jan 10 at 12:21














$begingroup$
I think it must be zero
$endgroup$
– Aditya Garg
Jan 10 at 12:25




$begingroup$
I think it must be zero
$endgroup$
– Aditya Garg
Jan 10 at 12:25












$begingroup$
For the left side make the substitution $u=e^{ax}-1$ and for the right side make the substitution $v=e^{bx}-1$ and see where this leads you
$endgroup$
– Henry Lee
Jan 10 at 12:42




$begingroup$
For the left side make the substitution $u=e^{ax}-1$ and for the right side make the substitution $v=e^{bx}-1$ and see where this leads you
$endgroup$
– Henry Lee
Jan 10 at 12:42










3 Answers
3






active

oldest

votes


















3












$begingroup$

Here is an hint for solving $displaystyleint frac{a}{e^{ax}-1}dx $.



Let $e^{ax}-1=u $. This substitution will give us the integral $$intfrac{1}{u(u+1)}du=intleft(frac1u-frac{1}{(u+1)}right)du.$$ Can you take it from here?






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Hints:
    0) Prove the integral to be absolutely convergent.



    1) Integrate from $epsilon >0$ instead and prove that you may split the integral.



    2) Change variables so that integrands are the same.



    3) See what happens and remember $e^x=1+x+o(x)$ at $x rightarrow 0$.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      Another hint for solving $int frac{a}{e^{ax}-1} dx$
      $$int frac{a}{e^{ax}-1} dx = int frac{a(e^{ax}+1)}{e^{2ax}-1} dx = frac{1}{2}int frac{a}{e^{ax}-1} dx + int frac{a e^{ax}}{e^{2ax}-1} dx$$



      Reagrouping terms and a change of variable s.t. $u = e^{ax}$ and $du = a e^{ax}dx$, leads to:
      $$frac{1}{2}int frac{a}{e^{ax}-1} dx = int frac{1}{u^2-1} du= frac{1}{2} left( int frac{1}{u-1} du - int frac{1}{u+1} duright)$$



      $$
      boxed{
      int frac{a}{e^{ax}-1} dx = lnleft( e^{2ax} - 1right) - lnleft( e^{2ax} + 1right)
      }$$






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Here is an hint for solving $displaystyleint frac{a}{e^{ax}-1}dx $.



        Let $e^{ax}-1=u $. This substitution will give us the integral $$intfrac{1}{u(u+1)}du=intleft(frac1u-frac{1}{(u+1)}right)du.$$ Can you take it from here?






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          Here is an hint for solving $displaystyleint frac{a}{e^{ax}-1}dx $.



          Let $e^{ax}-1=u $. This substitution will give us the integral $$intfrac{1}{u(u+1)}du=intleft(frac1u-frac{1}{(u+1)}right)du.$$ Can you take it from here?






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            Here is an hint for solving $displaystyleint frac{a}{e^{ax}-1}dx $.



            Let $e^{ax}-1=u $. This substitution will give us the integral $$intfrac{1}{u(u+1)}du=intleft(frac1u-frac{1}{(u+1)}right)du.$$ Can you take it from here?






            share|cite|improve this answer











            $endgroup$



            Here is an hint for solving $displaystyleint frac{a}{e^{ax}-1}dx $.



            Let $e^{ax}-1=u $. This substitution will give us the integral $$intfrac{1}{u(u+1)}du=intleft(frac1u-frac{1}{(u+1)}right)du.$$ Can you take it from here?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 11 at 1:59

























            answered Jan 10 at 12:39









            Thomas ShelbyThomas Shelby

            2,453221




            2,453221























                1












                $begingroup$

                Hints:
                0) Prove the integral to be absolutely convergent.



                1) Integrate from $epsilon >0$ instead and prove that you may split the integral.



                2) Change variables so that integrands are the same.



                3) See what happens and remember $e^x=1+x+o(x)$ at $x rightarrow 0$.






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  Hints:
                  0) Prove the integral to be absolutely convergent.



                  1) Integrate from $epsilon >0$ instead and prove that you may split the integral.



                  2) Change variables so that integrands are the same.



                  3) See what happens and remember $e^x=1+x+o(x)$ at $x rightarrow 0$.






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Hints:
                    0) Prove the integral to be absolutely convergent.



                    1) Integrate from $epsilon >0$ instead and prove that you may split the integral.



                    2) Change variables so that integrands are the same.



                    3) See what happens and remember $e^x=1+x+o(x)$ at $x rightarrow 0$.






                    share|cite|improve this answer











                    $endgroup$



                    Hints:
                    0) Prove the integral to be absolutely convergent.



                    1) Integrate from $epsilon >0$ instead and prove that you may split the integral.



                    2) Change variables so that integrands are the same.



                    3) See what happens and remember $e^x=1+x+o(x)$ at $x rightarrow 0$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 10 at 12:36

























                    answered Jan 10 at 12:26









                    MindlackMindlack

                    3,00717




                    3,00717























                        1












                        $begingroup$

                        Another hint for solving $int frac{a}{e^{ax}-1} dx$
                        $$int frac{a}{e^{ax}-1} dx = int frac{a(e^{ax}+1)}{e^{2ax}-1} dx = frac{1}{2}int frac{a}{e^{ax}-1} dx + int frac{a e^{ax}}{e^{2ax}-1} dx$$



                        Reagrouping terms and a change of variable s.t. $u = e^{ax}$ and $du = a e^{ax}dx$, leads to:
                        $$frac{1}{2}int frac{a}{e^{ax}-1} dx = int frac{1}{u^2-1} du= frac{1}{2} left( int frac{1}{u-1} du - int frac{1}{u+1} duright)$$



                        $$
                        boxed{
                        int frac{a}{e^{ax}-1} dx = lnleft( e^{2ax} - 1right) - lnleft( e^{2ax} + 1right)
                        }$$






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Another hint for solving $int frac{a}{e^{ax}-1} dx$
                          $$int frac{a}{e^{ax}-1} dx = int frac{a(e^{ax}+1)}{e^{2ax}-1} dx = frac{1}{2}int frac{a}{e^{ax}-1} dx + int frac{a e^{ax}}{e^{2ax}-1} dx$$



                          Reagrouping terms and a change of variable s.t. $u = e^{ax}$ and $du = a e^{ax}dx$, leads to:
                          $$frac{1}{2}int frac{a}{e^{ax}-1} dx = int frac{1}{u^2-1} du= frac{1}{2} left( int frac{1}{u-1} du - int frac{1}{u+1} duright)$$



                          $$
                          boxed{
                          int frac{a}{e^{ax}-1} dx = lnleft( e^{2ax} - 1right) - lnleft( e^{2ax} + 1right)
                          }$$






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Another hint for solving $int frac{a}{e^{ax}-1} dx$
                            $$int frac{a}{e^{ax}-1} dx = int frac{a(e^{ax}+1)}{e^{2ax}-1} dx = frac{1}{2}int frac{a}{e^{ax}-1} dx + int frac{a e^{ax}}{e^{2ax}-1} dx$$



                            Reagrouping terms and a change of variable s.t. $u = e^{ax}$ and $du = a e^{ax}dx$, leads to:
                            $$frac{1}{2}int frac{a}{e^{ax}-1} dx = int frac{1}{u^2-1} du= frac{1}{2} left( int frac{1}{u-1} du - int frac{1}{u+1} duright)$$



                            $$
                            boxed{
                            int frac{a}{e^{ax}-1} dx = lnleft( e^{2ax} - 1right) - lnleft( e^{2ax} + 1right)
                            }$$






                            share|cite|improve this answer











                            $endgroup$



                            Another hint for solving $int frac{a}{e^{ax}-1} dx$
                            $$int frac{a}{e^{ax}-1} dx = int frac{a(e^{ax}+1)}{e^{2ax}-1} dx = frac{1}{2}int frac{a}{e^{ax}-1} dx + int frac{a e^{ax}}{e^{2ax}-1} dx$$



                            Reagrouping terms and a change of variable s.t. $u = e^{ax}$ and $du = a e^{ax}dx$, leads to:
                            $$frac{1}{2}int frac{a}{e^{ax}-1} dx = int frac{1}{u^2-1} du= frac{1}{2} left( int frac{1}{u-1} du - int frac{1}{u+1} duright)$$



                            $$
                            boxed{
                            int frac{a}{e^{ax}-1} dx = lnleft( e^{2ax} - 1right) - lnleft( e^{2ax} + 1right)
                            }$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 10 at 14:30

























                            answered Jan 10 at 12:46









                            Carlos CamposCarlos Campos

                            57439




                            57439






























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