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







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












$begingroup$


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

3,7142525




3,7142525










asked Jan 10 at 12:19









Zhu HuanhuanZhu Huanhuan

34616




34616












  • $begingroup$
    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











      $endgroup$













        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%2f3068570%2fhow-to-do-integral-int-0-infty-left-fracaeax-1-fracbebx%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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

            3,7142525




            3,7142525























                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

                    4,885210




                    4,885210























                        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






























                                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%2f3068570%2fhow-to-do-integral-int-0-infty-left-fracaeax-1-fracbebx%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

                                Human spaceflight

                                Image
                                Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
                                Read more

                                Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                Image
                                0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
                                Read more

                                File:DeusFollowingSea.jpg

                                Image
                                File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
                                Read more