Mixed Integral Problem

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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







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    up vote
    2
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    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






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      edited 1 hour ago









      Yuval Gat

      1018




      1018










      asked 1 hour ago









      Creep Anonymous

      1549




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          3 Answers
          3






          active

          oldest

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          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




















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            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




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