Mixed Integral Problem

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite












Q. If $$sum_n=1^infty frac1n^2 = fracπ^26 $$ then, $$ int_0^1 fracln x1-x^2 dx = -fracπ^2lambda $$ , then the value of $lambda$ equals,



My attempt- I tried using integrating by parts to the integral. As a result, I'm left with $$-frac12 int_0^1 fraclnleft(frac1+x1-xright)x dx $$ Now I'm stuck in here! I don't know how to move on from here! Or maybe integrating by parts was a bad option? If it is, please guide me to a solution or Please help me on how to continue from here!
Any help would be appreciated.






integration definite-integrals summation







share|cite|improve this question



























    up vote
    2
    down vote

    favorite












    Q. If $$sum_n=1^infty frac1n^2 = fracπ^26 $$ then, $$ int_0^1 fracln x1-x^2 dx = -fracπ^2lambda $$ , then the value of $lambda$ equals,



    My attempt- I tried using integrating by parts to the integral. As a result, I'm left with $$-frac12 int_0^1 fraclnleft(frac1+x1-xright)x dx $$ Now I'm stuck in here! I don't know how to move on from here! Or maybe integrating by parts was a bad option? If it is, please guide me to a solution or Please help me on how to continue from here!
    Any help would be appreciated.






    integration definite-integrals summation







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Q. If $$sum_n=1^infty frac1n^2 = fracπ^26 $$ then, $$ int_0^1 fracln x1-x^2 dx = -fracπ^2lambda $$ , then the value of $lambda$ equals,



      My attempt- I tried using integrating by parts to the integral. As a result, I'm left with $$-frac12 int_0^1 fraclnleft(frac1+x1-xright)x dx $$ Now I'm stuck in here! I don't know how to move on from here! Or maybe integrating by parts was a bad option? If it is, please guide me to a solution or Please help me on how to continue from here!
      Any help would be appreciated.






      integration definite-integrals summation







      share|cite|improve this question















      Q. If $$sum_n=1^infty frac1n^2 = fracπ^26 $$ then, $$ int_0^1 fracln x1-x^2 dx = -fracπ^2lambda $$ , then the value of $lambda$ equals,



      My attempt- I tried using integrating by parts to the integral. As a result, I'm left with $$-frac12 int_0^1 fraclnleft(frac1+x1-xright)x dx $$ Now I'm stuck in here! I don't know how to move on from here! Or maybe integrating by parts was a bad option? If it is, please guide me to a solution or Please help me on how to continue from here!
      Any help would be appreciated.






      integration definite-integrals summation




      integration definite-integrals summation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Yuval Gat

      1018




      1018










      asked 1 hour ago









      Creep Anonymous

      1549




      1549




















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          3
          down vote













          Naaah. Since $frac11-x^2=1+x^2+x^4+ldots$ over $(0,1)$ and $int_0^1x^2nlog(x),dx = -frac1(2n+1)^2$,



          $$ int_0^1fraclog x1-x^2,dx = -sum_ngeq 0frac1(2n+1)^2 = -left[zeta(2)-tfrac14zeta(2)right] $$
          and $lambda=colorred8$.






          share|cite|improve this answer




















          • Wow! Thanks , I just have one query, How did you conclude the last step?? How did you convert the summation into the function of the given data?
            – Creep Anonymous
            1 hour ago







          • 2




            Over $(0,1)$ we have $fraclog x1-x^2=sum_ngeq 0x^2nlog x$ and the exchange of $sum_ngeq 0$ and $int_0^1(ldots),dx$ is allowed by the dominated/monotone convergence theorem.
            – Jack D'Aurizio
            1 hour ago











          • Thanks for your help
            – Creep Anonymous
            1 hour ago

















          up vote
          1
          down vote













          Use the expansion
          $$lnfrac1+x1-x=2left(x+fracx^33+fracx^55+cdotsright)$$
          then
          beginalign
          -frac12 int_0^1 fraclnfrac1+x1-xx dx
          &= -frac12 int_0^1 dfrac1x2left(x+fracx^33+fracx^55+cdotsright) dx \
          &= -left(1+frac13^2+frac15^2+cdotsright) \
          &= colorblue-dfracpi^28
          endalign






          share|cite|improve this answer




















          • Thanks! Now I can assume integrating by parts wasn't a waste at all!
            – Creep Anonymous
            21 mins ago











          • You are welcome, yes, your method is useful :)
            – Nosrati
            17 mins ago

















          up vote
          1
          down vote













          For another approach, take $f(x)=x^2$ on $[0,2pi]$ and extend periodically. Then the Fourier series for $f$ converges to $frac12(textjump)$ where "jump" is the value of the difference of the left-and right-handed limit as $xto 2pi.$ For this $f$, jump=$4pi^2$. Therefore, we have, after computing the Fourier series for $f$ and substituting $x=2pi,$



          $frac4pi^23+sum^infty_n=1frac4cos(2pi n)n^2=2pi^2Rightarrowsum^infty_n=1frac1n^2=fracpi^26$






          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',
            convertImagesToLinks: true,
            noModals: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            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%2f2974841%2fmixed-integral-problem%23new-answer', 'question_page');

            );

            Post as a guest

















            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote













            Naaah. Since $frac11-x^2=1+x^2+x^4+ldots$ over $(0,1)$ and $int_0^1x^2nlog(x),dx = -frac1(2n+1)^2$,



            $$ int_0^1fraclog x1-x^2,dx = -sum_ngeq 0frac1(2n+1)^2 = -left[zeta(2)-tfrac14zeta(2)right] $$
            and $lambda=colorred8$.






            share|cite|improve this answer




















            • Wow! Thanks , I just have one query, How did you conclude the last step?? How did you convert the summation into the function of the given data?
              – Creep Anonymous
              1 hour ago







            • 2




              Over $(0,1)$ we have $fraclog x1-x^2=sum_ngeq 0x^2nlog x$ and the exchange of $sum_ngeq 0$ and $int_0^1(ldots),dx$ is allowed by the dominated/monotone convergence theorem.
              – Jack D'Aurizio
              1 hour ago











            • Thanks for your help
              – Creep Anonymous
              1 hour ago














            up vote
            3
            down vote













            Naaah. Since $frac11-x^2=1+x^2+x^4+ldots$ over $(0,1)$ and $int_0^1x^2nlog(x),dx = -frac1(2n+1)^2$,



            $$ int_0^1fraclog x1-x^2,dx = -sum_ngeq 0frac1(2n+1)^2 = -left[zeta(2)-tfrac14zeta(2)right] $$
            and $lambda=colorred8$.






            share|cite|improve this answer




















            • Wow! Thanks , I just have one query, How did you conclude the last step?? How did you convert the summation into the function of the given data?
              – Creep Anonymous
              1 hour ago







            • 2




              Over $(0,1)$ we have $fraclog x1-x^2=sum_ngeq 0x^2nlog x$ and the exchange of $sum_ngeq 0$ and $int_0^1(ldots),dx$ is allowed by the dominated/monotone convergence theorem.
              – Jack D'Aurizio
              1 hour ago











            • Thanks for your help
              – Creep Anonymous
              1 hour ago












            up vote
            3
            down vote










            up vote
            3
            down vote









            Naaah. Since $frac11-x^2=1+x^2+x^4+ldots$ over $(0,1)$ and $int_0^1x^2nlog(x),dx = -frac1(2n+1)^2$,



            $$ int_0^1fraclog x1-x^2,dx = -sum_ngeq 0frac1(2n+1)^2 = -left[zeta(2)-tfrac14zeta(2)right] $$
            and $lambda=colorred8$.






            share|cite|improve this answer












            Naaah. Since $frac11-x^2=1+x^2+x^4+ldots$ over $(0,1)$ and $int_0^1x^2nlog(x),dx = -frac1(2n+1)^2$,



            $$ int_0^1fraclog x1-x^2,dx = -sum_ngeq 0frac1(2n+1)^2 = -left[zeta(2)-tfrac14zeta(2)right] $$
            and $lambda=colorred8$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Jack D'Aurizio

            279k33272648




            279k33272648











            • Wow! Thanks , I just have one query, How did you conclude the last step?? How did you convert the summation into the function of the given data?
              – Creep Anonymous
              1 hour ago







            • 2




              Over $(0,1)$ we have $fraclog x1-x^2=sum_ngeq 0x^2nlog x$ and the exchange of $sum_ngeq 0$ and $int_0^1(ldots),dx$ is allowed by the dominated/monotone convergence theorem.
              – Jack D'Aurizio
              1 hour ago











            • Thanks for your help
              – Creep Anonymous
              1 hour ago
















            • Wow! Thanks , I just have one query, How did you conclude the last step?? How did you convert the summation into the function of the given data?
              – Creep Anonymous
              1 hour ago







            • 2




              Over $(0,1)$ we have $fraclog x1-x^2=sum_ngeq 0x^2nlog x$ and the exchange of $sum_ngeq 0$ and $int_0^1(ldots),dx$ is allowed by the dominated/monotone convergence theorem.
              – Jack D'Aurizio
              1 hour ago











            • Thanks for your help
              – Creep Anonymous
              1 hour ago















            Wow! Thanks , I just have one query, How did you conclude the last step?? How did you convert the summation into the function of the given data?
            – Creep Anonymous
            1 hour ago





            Wow! Thanks , I just have one query, How did you conclude the last step?? How did you convert the summation into the function of the given data?
            – Creep Anonymous
            1 hour ago





            2




            2




            Over $(0,1)$ we have $fraclog x1-x^2=sum_ngeq 0x^2nlog x$ and the exchange of $sum_ngeq 0$ and $int_0^1(ldots),dx$ is allowed by the dominated/monotone convergence theorem.
            – Jack D'Aurizio
            1 hour ago





            Over $(0,1)$ we have $fraclog x1-x^2=sum_ngeq 0x^2nlog x$ and the exchange of $sum_ngeq 0$ and $int_0^1(ldots),dx$ is allowed by the dominated/monotone convergence theorem.
            – Jack D'Aurizio
            1 hour ago













            Thanks for your help
            – Creep Anonymous
            1 hour ago




            Thanks for your help
            – Creep Anonymous
            1 hour ago










            up vote
            1
            down vote













            Use the expansion
            $$lnfrac1+x1-x=2left(x+fracx^33+fracx^55+cdotsright)$$
            then
            beginalign
            -frac12 int_0^1 fraclnfrac1+x1-xx dx
            &= -frac12 int_0^1 dfrac1x2left(x+fracx^33+fracx^55+cdotsright) dx \
            &= -left(1+frac13^2+frac15^2+cdotsright) \
            &= colorblue-dfracpi^28
            endalign






            share|cite|improve this answer




















            • Thanks! Now I can assume integrating by parts wasn't a waste at all!
              – Creep Anonymous
              21 mins ago











            • You are welcome, yes, your method is useful :)
              – Nosrati
              17 mins ago














            up vote
            1
            down vote













            Use the expansion
            $$lnfrac1+x1-x=2left(x+fracx^33+fracx^55+cdotsright)$$
            then
            beginalign
            -frac12 int_0^1 fraclnfrac1+x1-xx dx
            &= -frac12 int_0^1 dfrac1x2left(x+fracx^33+fracx^55+cdotsright) dx \
            &= -left(1+frac13^2+frac15^2+cdotsright) \
            &= colorblue-dfracpi^28
            endalign






            share|cite|improve this answer




















            • Thanks! Now I can assume integrating by parts wasn't a waste at all!
              – Creep Anonymous
              21 mins ago











            • You are welcome, yes, your method is useful :)
              – Nosrati
              17 mins ago












            up vote
            1
            down vote










            up vote
            1
            down vote









            Use the expansion
            $$lnfrac1+x1-x=2left(x+fracx^33+fracx^55+cdotsright)$$
            then
            beginalign
            -frac12 int_0^1 fraclnfrac1+x1-xx dx
            &= -frac12 int_0^1 dfrac1x2left(x+fracx^33+fracx^55+cdotsright) dx \
            &= -left(1+frac13^2+frac15^2+cdotsright) \
            &= colorblue-dfracpi^28
            endalign






            share|cite|improve this answer












            Use the expansion
            $$lnfrac1+x1-x=2left(x+fracx^33+fracx^55+cdotsright)$$
            then
            beginalign
            -frac12 int_0^1 fraclnfrac1+x1-xx dx
            &= -frac12 int_0^1 dfrac1x2left(x+fracx^33+fracx^55+cdotsright) dx \
            &= -left(1+frac13^2+frac15^2+cdotsright) \
            &= colorblue-dfracpi^28
            endalign







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Nosrati

            24.7k62052




            24.7k62052











            • Thanks! Now I can assume integrating by parts wasn't a waste at all!
              – Creep Anonymous
              21 mins ago











            • You are welcome, yes, your method is useful :)
              – Nosrati
              17 mins ago
















            • Thanks! Now I can assume integrating by parts wasn't a waste at all!
              – Creep Anonymous
              21 mins ago











            • You are welcome, yes, your method is useful :)
              – Nosrati
              17 mins ago















            Thanks! Now I can assume integrating by parts wasn't a waste at all!
            – Creep Anonymous
            21 mins ago





            Thanks! Now I can assume integrating by parts wasn't a waste at all!
            – Creep Anonymous
            21 mins ago













            You are welcome, yes, your method is useful :)
            – Nosrati
            17 mins ago




            You are welcome, yes, your method is useful :)
            – Nosrati
            17 mins ago










            up vote
            1
            down vote













            For another approach, take $f(x)=x^2$ on $[0,2pi]$ and extend periodically. Then the Fourier series for $f$ converges to $frac12(textjump)$ where "jump" is the value of the difference of the left-and right-handed limit as $xto 2pi.$ For this $f$, jump=$4pi^2$. Therefore, we have, after computing the Fourier series for $f$ and substituting $x=2pi,$



            $frac4pi^23+sum^infty_n=1frac4cos(2pi n)n^2=2pi^2Rightarrowsum^infty_n=1frac1n^2=fracpi^26$






            share|cite|improve this answer
























              up vote
              1
              down vote













              For another approach, take $f(x)=x^2$ on $[0,2pi]$ and extend periodically. Then the Fourier series for $f$ converges to $frac12(textjump)$ where "jump" is the value of the difference of the left-and right-handed limit as $xto 2pi.$ For this $f$, jump=$4pi^2$. Therefore, we have, after computing the Fourier series for $f$ and substituting $x=2pi,$



              $frac4pi^23+sum^infty_n=1frac4cos(2pi n)n^2=2pi^2Rightarrowsum^infty_n=1frac1n^2=fracpi^26$






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                For another approach, take $f(x)=x^2$ on $[0,2pi]$ and extend periodically. Then the Fourier series for $f$ converges to $frac12(textjump)$ where "jump" is the value of the difference of the left-and right-handed limit as $xto 2pi.$ For this $f$, jump=$4pi^2$. Therefore, we have, after computing the Fourier series for $f$ and substituting $x=2pi,$



                $frac4pi^23+sum^infty_n=1frac4cos(2pi n)n^2=2pi^2Rightarrowsum^infty_n=1frac1n^2=fracpi^26$






                share|cite|improve this answer












                For another approach, take $f(x)=x^2$ on $[0,2pi]$ and extend periodically. Then the Fourier series for $f$ converges to $frac12(textjump)$ where "jump" is the value of the difference of the left-and right-handed limit as $xto 2pi.$ For this $f$, jump=$4pi^2$. Therefore, we have, after computing the Fourier series for $f$ and substituting $x=2pi,$



                $frac4pi^23+sum^infty_n=1frac4cos(2pi n)n^2=2pi^2Rightarrowsum^infty_n=1frac1n^2=fracpi^26$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Matematleta

                8,9312918




                8,9312918



























                     

                    draft saved


                    draft discarded















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2974841%2fmixed-integral-problem%23new-answer', 'question_page');

                    );

                    Post as a guest
































































                    Comments

                    Post a Comment

                    Popular posts from this blog

                    Long meetings (6-7 hours a day): Being “babysat” by supervisor

                    Image
                    Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 31 down vote favorite 4 I hold a PhD in mechanical engineering with a 7 year background in self-guided, self-paced research including 2 years as a graduate level course instructor and a subject matter expert for various engineering projects in a university setting in the United States. For the last 10-11 months, I am working in a research industry in France where my task is to debug C++ spaghetti code written over a period of several years by my current supervisor. It is a team of 2: just I and my supervisor. Diurnally, I am posed with a "bug of the day" to get out of this C++ code. My supervisor generally gives me about 15-30 minutes to solve it and then accosts me about whether or not the task is completed and then rinses and repeats this until the end of the day. Off-late, my supervisor has been holding 6-7 hour long "meetin
                    Read more

                    Is the Concept of Multiple Fantasy Races Scientifically Flawed? [closed]

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Many fantasy worlds have multiple humanoid species (elves, orcs, humans, etc) living in the same world, in addition to that, they often have the ability to interbreed. And I can't help but question their believability... fantasy-races reproduction interspecies-relations share | improve this question edited Aug 9 at 14:46 Michael Kjörling ♦ 21.1k 10 85 161 asked Aug 9 at 13:13 Chlodio 118 6 closed as unclear what you're asking by MichaelK, Gryphon, John, Vincent, Frostfyre Aug 9 at 15:25 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. 3 Stack Exchange exist to
                    Read more

                    Confectionery

                    Image
                    Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of
                    Read more