Find the value of $1-frac{1}{7}+frac{1}{9}-frac{1}{15}+frac{1}{17}-frac{1}{23}+frac{1}{25}…$












9















Find the value of this :$$1-frac{1}{7}+frac{1}{9}-frac{1}{15}+frac{1}{17}-frac{1}{23}+frac{1}{25}....$$




Try: We can write the above series as



$${S} = int^{1}_{0}bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+cdotsbigg]dx$$



$$S = int^{1}_{0}(1-x^6)bigg[1+x^{8}+x^{16}+cdots cdots bigg]dx$$



So $$S = int^{1}_{0}frac{1-x^6}{1-x^{8}}dx = int^{1}_{0}frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$



Now i am struck in that integration.



Did not understand how to solve it



could some help me to solve it. Thanks in advance






integration sequences-and-series definite-integrals







share|cite|improve this question
























  • Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
    – Michael Burr
    Dec 18 at 15:24






  • 2




    Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
    – lab bhattacharjee
    Dec 18 at 15:25






  • 1




    See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
    – lab bhattacharjee
    Dec 18 at 15:26










  • Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
    – crskhr
    Dec 18 at 15:28








  • 1




    @Stockfish: Absolutely convergent series.
    – Clayton
    Dec 18 at 15:32
















9















Find the value of this :$$1-frac{1}{7}+frac{1}{9}-frac{1}{15}+frac{1}{17}-frac{1}{23}+frac{1}{25}....$$




Try: We can write the above series as



$${S} = int^{1}_{0}bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+cdotsbigg]dx$$



$$S = int^{1}_{0}(1-x^6)bigg[1+x^{8}+x^{16}+cdots cdots bigg]dx$$



So $$S = int^{1}_{0}frac{1-x^6}{1-x^{8}}dx = int^{1}_{0}frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$



Now i am struck in that integration.



Did not understand how to solve it



could some help me to solve it. Thanks in advance






integration sequences-and-series definite-integrals







share|cite|improve this question
























  • Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
    – Michael Burr
    Dec 18 at 15:24






  • 2




    Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
    – lab bhattacharjee
    Dec 18 at 15:25






  • 1




    See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
    – lab bhattacharjee
    Dec 18 at 15:26










  • Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
    – crskhr
    Dec 18 at 15:28








  • 1




    @Stockfish: Absolutely convergent series.
    – Clayton
    Dec 18 at 15:32














9












9








9


2






Find the value of this :$$1-frac{1}{7}+frac{1}{9}-frac{1}{15}+frac{1}{17}-frac{1}{23}+frac{1}{25}....$$




Try: We can write the above series as



$${S} = int^{1}_{0}bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+cdotsbigg]dx$$



$$S = int^{1}_{0}(1-x^6)bigg[1+x^{8}+x^{16}+cdots cdots bigg]dx$$



So $$S = int^{1}_{0}frac{1-x^6}{1-x^{8}}dx = int^{1}_{0}frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$



Now i am struck in that integration.



Did not understand how to solve it



could some help me to solve it. Thanks in advance






integration sequences-and-series definite-integrals







share|cite|improve this question
















Find the value of this :$$1-frac{1}{7}+frac{1}{9}-frac{1}{15}+frac{1}{17}-frac{1}{23}+frac{1}{25}....$$




Try: We can write the above series as



$${S} = int^{1}_{0}bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+cdotsbigg]dx$$



$$S = int^{1}_{0}(1-x^6)bigg[1+x^{8}+x^{16}+cdots cdots bigg]dx$$



So $$S = int^{1}_{0}frac{1-x^6}{1-x^{8}}dx = int^{1}_{0}frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$



Now i am struck in that integration.



Did not understand how to solve it



could some help me to solve it. Thanks in advance






integration sequences-and-series definite-integrals




integration sequences-and-series definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 at 6:04









Asaf Karagila♦

301k32424755




301k32424755










asked Dec 18 at 15:20









D Tiwari

5,3052630




5,3052630












  • Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
    – Michael Burr
    Dec 18 at 15:24






  • 2




    Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
    – lab bhattacharjee
    Dec 18 at 15:25






  • 1




    See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
    – lab bhattacharjee
    Dec 18 at 15:26










  • Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
    – crskhr
    Dec 18 at 15:28








  • 1




    @Stockfish: Absolutely convergent series.
    – Clayton
    Dec 18 at 15:32


















  • Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
    – Michael Burr
    Dec 18 at 15:24






  • 2




    Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
    – lab bhattacharjee
    Dec 18 at 15:25






  • 1




    See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
    – lab bhattacharjee
    Dec 18 at 15:26










  • Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
    – crskhr
    Dec 18 at 15:28








  • 1




    @Stockfish: Absolutely convergent series.
    – Clayton
    Dec 18 at 15:32
















Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
– Michael Burr
Dec 18 at 15:24




Hint: Partial fraction decomposition ($x^4+1$ can be factored over the reals).
– Michael Burr
Dec 18 at 15:24




2




2




Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
– lab bhattacharjee
Dec 18 at 15:25




Replace $x^2=y$ and use mathworld.wolfram.com/PartialFractionDecomposition.html and replace back $x$
– lab bhattacharjee
Dec 18 at 15:25




1




1




See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
– lab bhattacharjee
Dec 18 at 15:26




See also math.stackexchange.com/questions/2363639/int-fracx2x4x21-dx/…
– lab bhattacharjee
Dec 18 at 15:26












Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
– crskhr
Dec 18 at 15:28






Hey, is this a problem from Joseph edwards integral calculus for beginners? I remember solving this. The answer should equal something like $frac{pi}{8}(1+sqrt{2)}$ or something.
– crskhr
Dec 18 at 15:28






1




1




@Stockfish: Absolutely convergent series.
– Clayton
Dec 18 at 15:32




@Stockfish: Absolutely convergent series.
– Clayton
Dec 18 at 15:32










4 Answers
4






active

oldest

votes


















4














Another approach, which could be useful to check yours through integrals,
is same as per the similar post, using the Digamma function $psi (z)$.
$$
eqalign{
& 1 - {1 over {8 - 1}} + {1 over {8 + 1}} - {1 over {2 cdot 8 - 1}} + {1 over {2 cdot 8 + 1}} + cdots = cr
& = 1 - sumlimits_{1, le ,n} {{1 over {8n - 1}}} + sumlimits_{1, le ,n} {{1 over {8n + 1}}} = cr
& = 1 - {1 over 8}left( {sumlimits_{1, le ,n} {{1 over {n - 1/8}}} - sumlimits_{1, le ,n} {{1 over {n + 1/8}}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;infty } {{1 over n}} - sumnolimits_{;n = 9/8}^{;infty } {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;9/8} {{1 over n}} } right) = cr
& = 1 - {1 over 8}left( {psi (9/8) - psi (7/8)} right) = cr
& = 1 - {1 over 8}left( {psi left( {1 + 1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = 1 - {1 over 8}left( {{1 over {1/8}} + psi left( {1/8} right) - psi left( {1 - 1/8} right)} right) = cr
& = {1 over 8}left( {psi left( {1 - 1/8} right) - psi left( {1/8} right)} right) = cr
& = {1 over 8}left( {pi cot left( {{pi over 8}} right)} right) = cr
& = {{pi left( {1 + sqrt 2 } right)} over 8} cr}
$$
where:



$$
Delta _{,z} psi left( z right) = psi left( {z + 1} right) - psi left( z right) = {1 over z}
$$
is the functional equation for the Digamma;

which implies that Digamma is the Antidelta of $1/z$
$$
eqalign{
& psi left( z right) = Delta _{,z} ^{ - 1} left( {{1 over z}} right) = sumnolimits_z {{1 over z}} quad Rightarrow cr
& Rightarrow quad sumnolimits_{;z = a}^{;b} {{1 over z}} = psi left( b right) - psi left( a right) cr}
$$
and we used the Reflection formula for Digamma
$$
psi left( {1 - z} right) = psi left( z right) + pi cot left( {pi z} right)
$$




Now, the above, suggests a way to solve the integral.




Let's replace $x^8$ with $y$
$$
int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} quad mathop
= limits^{x^{,8} = y} quad {1 over 8}int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy}
$$
then consider that we have
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = cr
& = left( {left( {1 - y} right)^{ - 1} y^{, - 7/8} - left( {1 - y} right)^{ - 1} y^{, - 1/8} } right) cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1}
- left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} } right) cr}
$$
so we are ready to use the integral and Gamma representation for the Beta function
$$
eqalign{
& 8 ;int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} = int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy} = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1} dy}
- int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} dy} } right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{rm B}left( {1/8,varepsilon } right) - {rm B}left( {7/8,varepsilon } right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {Gamma left( {1/8 + varepsilon } right)}}
- {{Gamma left( {7/8} right)Gamma left( varepsilon right)} over {Gamma left( {7/8 + varepsilon } right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
over {psi left( {1/8} right)varepsilon ;Gamma left( {1/8} right)
+ Gamma left( {1/8} right)}} - {{Gamma left( {7/8} right)Gamma left( varepsilon right)}
over {psi left( {7/8} right)varepsilon ;Gamma left( {7/8} right) + Gamma left( {7/8} right)}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( varepsilon right)} over {psi left( {1/8} right)varepsilon ; + 1}}
- {{Gamma left( varepsilon right)} over {psi left( {7/8} right)varepsilon ; + 1}}} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} Gamma left( varepsilon right)left( {left( {1 - psi left( {1/8} right)varepsilon ;} right)
- left( {1 - psi left( {7/8} right)varepsilon ;} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} varepsilon ,Gamma left( varepsilon right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = mathop {lim }limits_{varepsilon , to ,0} ,Gamma left( {1 + varepsilon } right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
& = psi left( {7/8} right) - psi left( {1/8} right) cr}
$$
which is the same result as above.




Let me add a more straight derivation of the above.




We rewrite the integrand as
$$
eqalign{
& {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = {{y^{, - 7/8} - y^{, - 1/8} } over {left( {1 - y} right)}} = cr
& = {{1 - y^{, - 1/8} - left( {1 - y^{, - 7/8} } right)} over {left( {1 - y} right)}} cr}
$$
and compare with the integral representation of Digamma
$$
psi (s + 1) = - gamma + int_0^1 {{{1 - x^{,s} } over {1 - x}}dx}
$$






share|cite|improve this answer































    10














    HINT: Notice that
    $$frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=frac{1}{4}frac{1}{x^2+xsqrt{2}+1}+frac{1}{4}frac{1}{x^2-xsqrt{2}+1}+frac{1}{2}frac{1}{x^2+1}$$
    Now we just have to evaluate the integrals
    $$I_1=frac{1}{4}int_0^1 frac{dx}{x^2+xsqrt{2}+1}$$
    $$I_2=frac{1}{4}int_0^1 frac{dx}{x^2-xsqrt{2}+1}$$
    $$I_3=frac{1}{2}int_0^1 frac{dx}{x^2+1}$$
    and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?






    share|cite|improve this answer





























      2














      The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+sqrt 2 x+1)(x^2-sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $frac{1}{4}left(2tan^{-1}x+sqrt{2}left(tan^{-1}left(sqrt{2}x+1right)-tan^{-1}left(1-sqrt{2}xright)right)right)$






      share|cite|improve this answer





























        1














        Hint: your fraction is $(x^2+1)^{-1}+frac{x^2}{x^4+1}$, and $frac{x^2}{x^4+1}=frac{ax+b}{x^2-sqrt{2}x+1}+frac{-ax-b}{x^2+sqrt{2}x+1}$ for some real numbers $a,b$.






        share|cite|improve this answer





















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          4 Answers
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          4 Answers
          4






          active

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          active

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          active

          oldest

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          4














          Another approach, which could be useful to check yours through integrals,
          is same as per the similar post, using the Digamma function $psi (z)$.
          $$
          eqalign{
          & 1 - {1 over {8 - 1}} + {1 over {8 + 1}} - {1 over {2 cdot 8 - 1}} + {1 over {2 cdot 8 + 1}} + cdots = cr
          & = 1 - sumlimits_{1, le ,n} {{1 over {8n - 1}}} + sumlimits_{1, le ,n} {{1 over {8n + 1}}} = cr
          & = 1 - {1 over 8}left( {sumlimits_{1, le ,n} {{1 over {n - 1/8}}} - sumlimits_{1, le ,n} {{1 over {n + 1/8}}} } right) = cr
          & = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;infty } {{1 over n}} - sumnolimits_{;n = 9/8}^{;infty } {{1 over n}} } right) = cr
          & = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;9/8} {{1 over n}} } right) = cr
          & = 1 - {1 over 8}left( {psi (9/8) - psi (7/8)} right) = cr
          & = 1 - {1 over 8}left( {psi left( {1 + 1/8} right) - psi left( {1 - 1/8} right)} right) = cr
          & = 1 - {1 over 8}left( {{1 over {1/8}} + psi left( {1/8} right) - psi left( {1 - 1/8} right)} right) = cr
          & = {1 over 8}left( {psi left( {1 - 1/8} right) - psi left( {1/8} right)} right) = cr
          & = {1 over 8}left( {pi cot left( {{pi over 8}} right)} right) = cr
          & = {{pi left( {1 + sqrt 2 } right)} over 8} cr}
          $$
          where:



          $$
          Delta _{,z} psi left( z right) = psi left( {z + 1} right) - psi left( z right) = {1 over z}
          $$
          is the functional equation for the Digamma;

          which implies that Digamma is the Antidelta of $1/z$
          $$
          eqalign{
          & psi left( z right) = Delta _{,z} ^{ - 1} left( {{1 over z}} right) = sumnolimits_z {{1 over z}} quad Rightarrow cr
          & Rightarrow quad sumnolimits_{;z = a}^{;b} {{1 over z}} = psi left( b right) - psi left( a right) cr}
          $$
          and we used the Reflection formula for Digamma
          $$
          psi left( {1 - z} right) = psi left( z right) + pi cot left( {pi z} right)
          $$




          Now, the above, suggests a way to solve the integral.




          Let's replace $x^8$ with $y$
          $$
          int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} quad mathop
          = limits^{x^{,8} = y} quad {1 over 8}int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy}
          $$
          then consider that we have
          $$
          eqalign{
          & {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = cr
          & = left( {left( {1 - y} right)^{ - 1} y^{, - 7/8} - left( {1 - y} right)^{ - 1} y^{, - 1/8} } right) cr
          & = mathop {lim }limits_{varepsilon , to ,0} left( {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1}
          - left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} } right) cr}
          $$
          so we are ready to use the integral and Gamma representation for the Beta function
          $$
          eqalign{
          & 8 ;int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} = int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy} = cr
          & = mathop {lim }limits_{varepsilon , to ,0} left( {int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1} dy}
          - int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} dy} } right) = cr
          & = mathop {lim }limits_{varepsilon , to ,0} left( {{rm B}left( {1/8,varepsilon } right) - {rm B}left( {7/8,varepsilon } right)} right) = cr
          & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
          over {Gamma left( {1/8 + varepsilon } right)}}
          - {{Gamma left( {7/8} right)Gamma left( varepsilon right)} over {Gamma left( {7/8 + varepsilon } right)}}} right) = cr
          & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
          over {psi left( {1/8} right)varepsilon ;Gamma left( {1/8} right)
          + Gamma left( {1/8} right)}} - {{Gamma left( {7/8} right)Gamma left( varepsilon right)}
          over {psi left( {7/8} right)varepsilon ;Gamma left( {7/8} right) + Gamma left( {7/8} right)}}} right) = cr
          & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( varepsilon right)} over {psi left( {1/8} right)varepsilon ; + 1}}
          - {{Gamma left( varepsilon right)} over {psi left( {7/8} right)varepsilon ; + 1}}} right) = cr
          & = mathop {lim }limits_{varepsilon , to ,0} Gamma left( varepsilon right)left( {left( {1 - psi left( {1/8} right)varepsilon ;} right)
          - left( {1 - psi left( {7/8} right)varepsilon ;} right)} right) = cr
          & = mathop {lim }limits_{varepsilon , to ,0} varepsilon ,Gamma left( varepsilon right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
          & = mathop {lim }limits_{varepsilon , to ,0} ,Gamma left( {1 + varepsilon } right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
          & = psi left( {7/8} right) - psi left( {1/8} right) cr}
          $$
          which is the same result as above.




          Let me add a more straight derivation of the above.




          We rewrite the integrand as
          $$
          eqalign{
          & {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = {{y^{, - 7/8} - y^{, - 1/8} } over {left( {1 - y} right)}} = cr
          & = {{1 - y^{, - 1/8} - left( {1 - y^{, - 7/8} } right)} over {left( {1 - y} right)}} cr}
          $$
          and compare with the integral representation of Digamma
          $$
          psi (s + 1) = - gamma + int_0^1 {{{1 - x^{,s} } over {1 - x}}dx}
          $$






          share|cite|improve this answer




























            4














            Another approach, which could be useful to check yours through integrals,
            is same as per the similar post, using the Digamma function $psi (z)$.
            $$
            eqalign{
            & 1 - {1 over {8 - 1}} + {1 over {8 + 1}} - {1 over {2 cdot 8 - 1}} + {1 over {2 cdot 8 + 1}} + cdots = cr
            & = 1 - sumlimits_{1, le ,n} {{1 over {8n - 1}}} + sumlimits_{1, le ,n} {{1 over {8n + 1}}} = cr
            & = 1 - {1 over 8}left( {sumlimits_{1, le ,n} {{1 over {n - 1/8}}} - sumlimits_{1, le ,n} {{1 over {n + 1/8}}} } right) = cr
            & = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;infty } {{1 over n}} - sumnolimits_{;n = 9/8}^{;infty } {{1 over n}} } right) = cr
            & = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;9/8} {{1 over n}} } right) = cr
            & = 1 - {1 over 8}left( {psi (9/8) - psi (7/8)} right) = cr
            & = 1 - {1 over 8}left( {psi left( {1 + 1/8} right) - psi left( {1 - 1/8} right)} right) = cr
            & = 1 - {1 over 8}left( {{1 over {1/8}} + psi left( {1/8} right) - psi left( {1 - 1/8} right)} right) = cr
            & = {1 over 8}left( {psi left( {1 - 1/8} right) - psi left( {1/8} right)} right) = cr
            & = {1 over 8}left( {pi cot left( {{pi over 8}} right)} right) = cr
            & = {{pi left( {1 + sqrt 2 } right)} over 8} cr}
            $$
            where:



            $$
            Delta _{,z} psi left( z right) = psi left( {z + 1} right) - psi left( z right) = {1 over z}
            $$
            is the functional equation for the Digamma;

            which implies that Digamma is the Antidelta of $1/z$
            $$
            eqalign{
            & psi left( z right) = Delta _{,z} ^{ - 1} left( {{1 over z}} right) = sumnolimits_z {{1 over z}} quad Rightarrow cr
            & Rightarrow quad sumnolimits_{;z = a}^{;b} {{1 over z}} = psi left( b right) - psi left( a right) cr}
            $$
            and we used the Reflection formula for Digamma
            $$
            psi left( {1 - z} right) = psi left( z right) + pi cot left( {pi z} right)
            $$




            Now, the above, suggests a way to solve the integral.




            Let's replace $x^8$ with $y$
            $$
            int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} quad mathop
            = limits^{x^{,8} = y} quad {1 over 8}int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy}
            $$
            then consider that we have
            $$
            eqalign{
            & {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = cr
            & = left( {left( {1 - y} right)^{ - 1} y^{, - 7/8} - left( {1 - y} right)^{ - 1} y^{, - 1/8} } right) cr
            & = mathop {lim }limits_{varepsilon , to ,0} left( {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1}
            - left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} } right) cr}
            $$
            so we are ready to use the integral and Gamma representation for the Beta function
            $$
            eqalign{
            & 8 ;int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} = int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy} = cr
            & = mathop {lim }limits_{varepsilon , to ,0} left( {int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1} dy}
            - int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} dy} } right) = cr
            & = mathop {lim }limits_{varepsilon , to ,0} left( {{rm B}left( {1/8,varepsilon } right) - {rm B}left( {7/8,varepsilon } right)} right) = cr
            & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
            over {Gamma left( {1/8 + varepsilon } right)}}
            - {{Gamma left( {7/8} right)Gamma left( varepsilon right)} over {Gamma left( {7/8 + varepsilon } right)}}} right) = cr
            & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
            over {psi left( {1/8} right)varepsilon ;Gamma left( {1/8} right)
            + Gamma left( {1/8} right)}} - {{Gamma left( {7/8} right)Gamma left( varepsilon right)}
            over {psi left( {7/8} right)varepsilon ;Gamma left( {7/8} right) + Gamma left( {7/8} right)}}} right) = cr
            & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( varepsilon right)} over {psi left( {1/8} right)varepsilon ; + 1}}
            - {{Gamma left( varepsilon right)} over {psi left( {7/8} right)varepsilon ; + 1}}} right) = cr
            & = mathop {lim }limits_{varepsilon , to ,0} Gamma left( varepsilon right)left( {left( {1 - psi left( {1/8} right)varepsilon ;} right)
            - left( {1 - psi left( {7/8} right)varepsilon ;} right)} right) = cr
            & = mathop {lim }limits_{varepsilon , to ,0} varepsilon ,Gamma left( varepsilon right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
            & = mathop {lim }limits_{varepsilon , to ,0} ,Gamma left( {1 + varepsilon } right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
            & = psi left( {7/8} right) - psi left( {1/8} right) cr}
            $$
            which is the same result as above.




            Let me add a more straight derivation of the above.




            We rewrite the integrand as
            $$
            eqalign{
            & {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = {{y^{, - 7/8} - y^{, - 1/8} } over {left( {1 - y} right)}} = cr
            & = {{1 - y^{, - 1/8} - left( {1 - y^{, - 7/8} } right)} over {left( {1 - y} right)}} cr}
            $$
            and compare with the integral representation of Digamma
            $$
            psi (s + 1) = - gamma + int_0^1 {{{1 - x^{,s} } over {1 - x}}dx}
            $$






            share|cite|improve this answer


























              4












              4








              4






              Another approach, which could be useful to check yours through integrals,
              is same as per the similar post, using the Digamma function $psi (z)$.
              $$
              eqalign{
              & 1 - {1 over {8 - 1}} + {1 over {8 + 1}} - {1 over {2 cdot 8 - 1}} + {1 over {2 cdot 8 + 1}} + cdots = cr
              & = 1 - sumlimits_{1, le ,n} {{1 over {8n - 1}}} + sumlimits_{1, le ,n} {{1 over {8n + 1}}} = cr
              & = 1 - {1 over 8}left( {sumlimits_{1, le ,n} {{1 over {n - 1/8}}} - sumlimits_{1, le ,n} {{1 over {n + 1/8}}} } right) = cr
              & = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;infty } {{1 over n}} - sumnolimits_{;n = 9/8}^{;infty } {{1 over n}} } right) = cr
              & = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;9/8} {{1 over n}} } right) = cr
              & = 1 - {1 over 8}left( {psi (9/8) - psi (7/8)} right) = cr
              & = 1 - {1 over 8}left( {psi left( {1 + 1/8} right) - psi left( {1 - 1/8} right)} right) = cr
              & = 1 - {1 over 8}left( {{1 over {1/8}} + psi left( {1/8} right) - psi left( {1 - 1/8} right)} right) = cr
              & = {1 over 8}left( {psi left( {1 - 1/8} right) - psi left( {1/8} right)} right) = cr
              & = {1 over 8}left( {pi cot left( {{pi over 8}} right)} right) = cr
              & = {{pi left( {1 + sqrt 2 } right)} over 8} cr}
              $$
              where:



              $$
              Delta _{,z} psi left( z right) = psi left( {z + 1} right) - psi left( z right) = {1 over z}
              $$
              is the functional equation for the Digamma;

              which implies that Digamma is the Antidelta of $1/z$
              $$
              eqalign{
              & psi left( z right) = Delta _{,z} ^{ - 1} left( {{1 over z}} right) = sumnolimits_z {{1 over z}} quad Rightarrow cr
              & Rightarrow quad sumnolimits_{;z = a}^{;b} {{1 over z}} = psi left( b right) - psi left( a right) cr}
              $$
              and we used the Reflection formula for Digamma
              $$
              psi left( {1 - z} right) = psi left( z right) + pi cot left( {pi z} right)
              $$




              Now, the above, suggests a way to solve the integral.




              Let's replace $x^8$ with $y$
              $$
              int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} quad mathop
              = limits^{x^{,8} = y} quad {1 over 8}int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy}
              $$
              then consider that we have
              $$
              eqalign{
              & {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = cr
              & = left( {left( {1 - y} right)^{ - 1} y^{, - 7/8} - left( {1 - y} right)^{ - 1} y^{, - 1/8} } right) cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1}
              - left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} } right) cr}
              $$
              so we are ready to use the integral and Gamma representation for the Beta function
              $$
              eqalign{
              & 8 ;int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} = int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy} = cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1} dy}
              - int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} dy} } right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {{rm B}left( {1/8,varepsilon } right) - {rm B}left( {7/8,varepsilon } right)} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
              over {Gamma left( {1/8 + varepsilon } right)}}
              - {{Gamma left( {7/8} right)Gamma left( varepsilon right)} over {Gamma left( {7/8 + varepsilon } right)}}} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
              over {psi left( {1/8} right)varepsilon ;Gamma left( {1/8} right)
              + Gamma left( {1/8} right)}} - {{Gamma left( {7/8} right)Gamma left( varepsilon right)}
              over {psi left( {7/8} right)varepsilon ;Gamma left( {7/8} right) + Gamma left( {7/8} right)}}} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( varepsilon right)} over {psi left( {1/8} right)varepsilon ; + 1}}
              - {{Gamma left( varepsilon right)} over {psi left( {7/8} right)varepsilon ; + 1}}} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} Gamma left( varepsilon right)left( {left( {1 - psi left( {1/8} right)varepsilon ;} right)
              - left( {1 - psi left( {7/8} right)varepsilon ;} right)} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} varepsilon ,Gamma left( varepsilon right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} ,Gamma left( {1 + varepsilon } right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
              & = psi left( {7/8} right) - psi left( {1/8} right) cr}
              $$
              which is the same result as above.




              Let me add a more straight derivation of the above.




              We rewrite the integrand as
              $$
              eqalign{
              & {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = {{y^{, - 7/8} - y^{, - 1/8} } over {left( {1 - y} right)}} = cr
              & = {{1 - y^{, - 1/8} - left( {1 - y^{, - 7/8} } right)} over {left( {1 - y} right)}} cr}
              $$
              and compare with the integral representation of Digamma
              $$
              psi (s + 1) = - gamma + int_0^1 {{{1 - x^{,s} } over {1 - x}}dx}
              $$






              share|cite|improve this answer














              Another approach, which could be useful to check yours through integrals,
              is same as per the similar post, using the Digamma function $psi (z)$.
              $$
              eqalign{
              & 1 - {1 over {8 - 1}} + {1 over {8 + 1}} - {1 over {2 cdot 8 - 1}} + {1 over {2 cdot 8 + 1}} + cdots = cr
              & = 1 - sumlimits_{1, le ,n} {{1 over {8n - 1}}} + sumlimits_{1, le ,n} {{1 over {8n + 1}}} = cr
              & = 1 - {1 over 8}left( {sumlimits_{1, le ,n} {{1 over {n - 1/8}}} - sumlimits_{1, le ,n} {{1 over {n + 1/8}}} } right) = cr
              & = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;infty } {{1 over n}} - sumnolimits_{;n = 9/8}^{;infty } {{1 over n}} } right) = cr
              & = 1 - {1 over 8}left( {sumnolimits_{;n = 7/8}^{;9/8} {{1 over n}} } right) = cr
              & = 1 - {1 over 8}left( {psi (9/8) - psi (7/8)} right) = cr
              & = 1 - {1 over 8}left( {psi left( {1 + 1/8} right) - psi left( {1 - 1/8} right)} right) = cr
              & = 1 - {1 over 8}left( {{1 over {1/8}} + psi left( {1/8} right) - psi left( {1 - 1/8} right)} right) = cr
              & = {1 over 8}left( {psi left( {1 - 1/8} right) - psi left( {1/8} right)} right) = cr
              & = {1 over 8}left( {pi cot left( {{pi over 8}} right)} right) = cr
              & = {{pi left( {1 + sqrt 2 } right)} over 8} cr}
              $$
              where:



              $$
              Delta _{,z} psi left( z right) = psi left( {z + 1} right) - psi left( z right) = {1 over z}
              $$
              is the functional equation for the Digamma;

              which implies that Digamma is the Antidelta of $1/z$
              $$
              eqalign{
              & psi left( z right) = Delta _{,z} ^{ - 1} left( {{1 over z}} right) = sumnolimits_z {{1 over z}} quad Rightarrow cr
              & Rightarrow quad sumnolimits_{;z = a}^{;b} {{1 over z}} = psi left( b right) - psi left( a right) cr}
              $$
              and we used the Reflection formula for Digamma
              $$
              psi left( {1 - z} right) = psi left( z right) + pi cot left( {pi z} right)
              $$




              Now, the above, suggests a way to solve the integral.




              Let's replace $x^8$ with $y$
              $$
              int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} quad mathop
              = limits^{x^{,8} = y} quad {1 over 8}int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy}
              $$
              then consider that we have
              $$
              eqalign{
              & {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = cr
              & = left( {left( {1 - y} right)^{ - 1} y^{, - 7/8} - left( {1 - y} right)^{ - 1} y^{, - 1/8} } right) cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1}
              - left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} } right) cr}
              $$
              so we are ready to use the integral and Gamma representation for the Beta function
              $$
              eqalign{
              & 8 ;int_{x = 0}^{,1} {{{1 - x^{,6} } over {1 - x^{,8} }}dx} = int_{y = 0}^{,1} {{{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }}dy} = cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,1/8 - 1} dy}
              - int_{y = 0}^{,1} {left( {1 - y} right)^{varepsilon - 1} y^{,,7/8 - 1} dy} } right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {{rm B}left( {1/8,varepsilon } right) - {rm B}left( {7/8,varepsilon } right)} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
              over {Gamma left( {1/8 + varepsilon } right)}}
              - {{Gamma left( {7/8} right)Gamma left( varepsilon right)} over {Gamma left( {7/8 + varepsilon } right)}}} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( {1/8} right)Gamma left( varepsilon right)}
              over {psi left( {1/8} right)varepsilon ;Gamma left( {1/8} right)
              + Gamma left( {1/8} right)}} - {{Gamma left( {7/8} right)Gamma left( varepsilon right)}
              over {psi left( {7/8} right)varepsilon ;Gamma left( {7/8} right) + Gamma left( {7/8} right)}}} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} left( {{{Gamma left( varepsilon right)} over {psi left( {1/8} right)varepsilon ; + 1}}
              - {{Gamma left( varepsilon right)} over {psi left( {7/8} right)varepsilon ; + 1}}} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} Gamma left( varepsilon right)left( {left( {1 - psi left( {1/8} right)varepsilon ;} right)
              - left( {1 - psi left( {7/8} right)varepsilon ;} right)} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} varepsilon ,Gamma left( varepsilon right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
              & = mathop {lim }limits_{varepsilon , to ,0} ,Gamma left( {1 + varepsilon } right)left( {psi left( {7/8} right) - psi left( {1/8} right)} right) = cr
              & = psi left( {7/8} right) - psi left( {1/8} right) cr}
              $$
              which is the same result as above.




              Let me add a more straight derivation of the above.




              We rewrite the integrand as
              $$
              eqalign{
              & {{1 - y^{,3/4} } over {left( {1 - y} right)y^{,7/8} }} = {{y^{, - 7/8} - y^{, - 1/8} } over {left( {1 - y} right)}} = cr
              & = {{1 - y^{, - 1/8} - left( {1 - y^{, - 7/8} } right)} over {left( {1 - y} right)}} cr}
              $$
              and compare with the integral representation of Digamma
              $$
              psi (s + 1) = - gamma + int_0^1 {{{1 - x^{,s} } over {1 - x}}dx}
              $$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 18 at 22:24

























              answered Dec 18 at 16:19









              G Cab

              17.9k31237




              17.9k31237























                  10














                  HINT: Notice that
                  $$frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=frac{1}{4}frac{1}{x^2+xsqrt{2}+1}+frac{1}{4}frac{1}{x^2-xsqrt{2}+1}+frac{1}{2}frac{1}{x^2+1}$$
                  Now we just have to evaluate the integrals
                  $$I_1=frac{1}{4}int_0^1 frac{dx}{x^2+xsqrt{2}+1}$$
                  $$I_2=frac{1}{4}int_0^1 frac{dx}{x^2-xsqrt{2}+1}$$
                  $$I_3=frac{1}{2}int_0^1 frac{dx}{x^2+1}$$
                  and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?






                  share|cite|improve this answer


























                    10














                    HINT: Notice that
                    $$frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=frac{1}{4}frac{1}{x^2+xsqrt{2}+1}+frac{1}{4}frac{1}{x^2-xsqrt{2}+1}+frac{1}{2}frac{1}{x^2+1}$$
                    Now we just have to evaluate the integrals
                    $$I_1=frac{1}{4}int_0^1 frac{dx}{x^2+xsqrt{2}+1}$$
                    $$I_2=frac{1}{4}int_0^1 frac{dx}{x^2-xsqrt{2}+1}$$
                    $$I_3=frac{1}{2}int_0^1 frac{dx}{x^2+1}$$
                    and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?






                    share|cite|improve this answer
























                      10












                      10








                      10






                      HINT: Notice that
                      $$frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=frac{1}{4}frac{1}{x^2+xsqrt{2}+1}+frac{1}{4}frac{1}{x^2-xsqrt{2}+1}+frac{1}{2}frac{1}{x^2+1}$$
                      Now we just have to evaluate the integrals
                      $$I_1=frac{1}{4}int_0^1 frac{dx}{x^2+xsqrt{2}+1}$$
                      $$I_2=frac{1}{4}int_0^1 frac{dx}{x^2-xsqrt{2}+1}$$
                      $$I_3=frac{1}{2}int_0^1 frac{dx}{x^2+1}$$
                      and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?






                      share|cite|improve this answer












                      HINT: Notice that
                      $$frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=frac{1}{4}frac{1}{x^2+xsqrt{2}+1}+frac{1}{4}frac{1}{x^2-xsqrt{2}+1}+frac{1}{2}frac{1}{x^2+1}$$
                      Now we just have to evaluate the integrals
                      $$I_1=frac{1}{4}int_0^1 frac{dx}{x^2+xsqrt{2}+1}$$
                      $$I_2=frac{1}{4}int_0^1 frac{dx}{x^2-xsqrt{2}+1}$$
                      $$I_3=frac{1}{2}int_0^1 frac{dx}{x^2+1}$$
                      and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 18 at 15:26









                      Frpzzd

                      21.9k839107




                      21.9k839107























                          2














                          The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+sqrt 2 x+1)(x^2-sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $frac{1}{4}left(2tan^{-1}x+sqrt{2}left(tan^{-1}left(sqrt{2}x+1right)-tan^{-1}left(1-sqrt{2}xright)right)right)$






                          share|cite|improve this answer


























                            2














                            The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+sqrt 2 x+1)(x^2-sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $frac{1}{4}left(2tan^{-1}x+sqrt{2}left(tan^{-1}left(sqrt{2}x+1right)-tan^{-1}left(1-sqrt{2}xright)right)right)$






                            share|cite|improve this answer
























                              2












                              2








                              2






                              The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+sqrt 2 x+1)(x^2-sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $frac{1}{4}left(2tan^{-1}x+sqrt{2}left(tan^{-1}left(sqrt{2}x+1right)-tan^{-1}left(1-sqrt{2}xright)right)right)$






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                              The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+sqrt 2 x+1)(x^2-sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $frac{1}{4}left(2tan^{-1}x+sqrt{2}left(tan^{-1}left(sqrt{2}x+1right)-tan^{-1}left(1-sqrt{2}xright)right)right)$







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                              answered Dec 18 at 15:30









                              william122

                              52412




                              52412























                                  1














                                  Hint: your fraction is $(x^2+1)^{-1}+frac{x^2}{x^4+1}$, and $frac{x^2}{x^4+1}=frac{ax+b}{x^2-sqrt{2}x+1}+frac{-ax-b}{x^2+sqrt{2}x+1}$ for some real numbers $a,b$.






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                                    1














                                    Hint: your fraction is $(x^2+1)^{-1}+frac{x^2}{x^4+1}$, and $frac{x^2}{x^4+1}=frac{ax+b}{x^2-sqrt{2}x+1}+frac{-ax-b}{x^2+sqrt{2}x+1}$ for some real numbers $a,b$.






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                                      1












                                      1








                                      1






                                      Hint: your fraction is $(x^2+1)^{-1}+frac{x^2}{x^4+1}$, and $frac{x^2}{x^4+1}=frac{ax+b}{x^2-sqrt{2}x+1}+frac{-ax-b}{x^2+sqrt{2}x+1}$ for some real numbers $a,b$.






                                      share|cite|improve this answer












                                      Hint: your fraction is $(x^2+1)^{-1}+frac{x^2}{x^4+1}$, and $frac{x^2}{x^4+1}=frac{ax+b}{x^2-sqrt{2}x+1}+frac{-ax-b}{x^2+sqrt{2}x+1}$ for some real numbers $a,b$.







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                                      answered Dec 18 at 15:28









                                      Mindlack

                                      1,27717




                                      1,27717






























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