Integral $int_0^1 frac{dx}{prod_{n=1}^infty (1+x^n)}$












26












$begingroup$


The following integral appeared this summer on AoPS. However it received no answer until today.
$$I=lim_{nto infty } int_0^1frac{dx}{(1+x)(1+x^2)dots(1+x^n)}=int_0^1 frac{dx}{prod_{n=1}^infty (1+x^n)}$$
I have learnt recently from here that: $$frac{1}{prod_{n=1}^infty (1+x^n)}=prod_{n=1}^inftyleft(1-x^{2n-1}right)Rightarrow I=int_0^1prod_{n=1}^inftyleft(1-x^{2n-1}right)dx$$
I suspect this has a closed form since a similar integral to the last equality appeared here on MSE before; however this one is a bit different since the product goes only on odd powers and I don't see how to make a connection between the two of them, so I will appreciate some help with that.






integration definite-integrals closed-form infinite-product







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    The series expansion of the integrand is here. An alternate form of the problem is $$I = int_0^1 (q;q^2)_infty ,dq,$$ where $(a;q)_infty$ is the $q$-Pochhammer symbol.
    $endgroup$
    – user153012
    Dec 16 '18 at 12:11






  • 4




    $begingroup$
    With a lot of help from Mathematica I can only obtain a (conjectural) infinite series form for the integral $$I=8 pisqrt{3} sum _{n=0}^{infty } frac{p(n) sinh left(frac{1}{6} pi sqrt{48 n+23}right)}{sqrt{48 n+23} left(2 cosh left(frac{1}{3} pi sqrt{48 n+23}right)-1right)}$$ where $p(n)$ is the number of unrestricted partitions of the integer n. This leads me to suspect that $I$ does not have a closed form.
    $endgroup$
    – James Arathoon
    Dec 19 '18 at 0:12








  • 1




    $begingroup$
    By numeric computation, this integral is about $0.42888151226615025372$. If you put it in WolframAlpha or RIES, you will see that this does not look like anything with a closed form.
    $endgroup$
    – ablmf
    Jan 14 at 19:52










  • $begingroup$
    $I=intlimits_0^1 dfrac2{(-1;x)_infty},mathrm dxapprox 0.428882.$
    $endgroup$
    – Yuri Negometyanov
    Jan 15 at 17:20
















26












$begingroup$


The following integral appeared this summer on AoPS. However it received no answer until today.
$$I=lim_{nto infty } int_0^1frac{dx}{(1+x)(1+x^2)dots(1+x^n)}=int_0^1 frac{dx}{prod_{n=1}^infty (1+x^n)}$$
I have learnt recently from here that: $$frac{1}{prod_{n=1}^infty (1+x^n)}=prod_{n=1}^inftyleft(1-x^{2n-1}right)Rightarrow I=int_0^1prod_{n=1}^inftyleft(1-x^{2n-1}right)dx$$
I suspect this has a closed form since a similar integral to the last equality appeared here on MSE before; however this one is a bit different since the product goes only on odd powers and I don't see how to make a connection between the two of them, so I will appreciate some help with that.






integration definite-integrals closed-form infinite-product







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    The series expansion of the integrand is here. An alternate form of the problem is $$I = int_0^1 (q;q^2)_infty ,dq,$$ where $(a;q)_infty$ is the $q$-Pochhammer symbol.
    $endgroup$
    – user153012
    Dec 16 '18 at 12:11






  • 4




    $begingroup$
    With a lot of help from Mathematica I can only obtain a (conjectural) infinite series form for the integral $$I=8 pisqrt{3} sum _{n=0}^{infty } frac{p(n) sinh left(frac{1}{6} pi sqrt{48 n+23}right)}{sqrt{48 n+23} left(2 cosh left(frac{1}{3} pi sqrt{48 n+23}right)-1right)}$$ where $p(n)$ is the number of unrestricted partitions of the integer n. This leads me to suspect that $I$ does not have a closed form.
    $endgroup$
    – James Arathoon
    Dec 19 '18 at 0:12








  • 1




    $begingroup$
    By numeric computation, this integral is about $0.42888151226615025372$. If you put it in WolframAlpha or RIES, you will see that this does not look like anything with a closed form.
    $endgroup$
    – ablmf
    Jan 14 at 19:52










  • $begingroup$
    $I=intlimits_0^1 dfrac2{(-1;x)_infty},mathrm dxapprox 0.428882.$
    $endgroup$
    – Yuri Negometyanov
    Jan 15 at 17:20














26












26








26


17



$begingroup$


The following integral appeared this summer on AoPS. However it received no answer until today.
$$I=lim_{nto infty } int_0^1frac{dx}{(1+x)(1+x^2)dots(1+x^n)}=int_0^1 frac{dx}{prod_{n=1}^infty (1+x^n)}$$
I have learnt recently from here that: $$frac{1}{prod_{n=1}^infty (1+x^n)}=prod_{n=1}^inftyleft(1-x^{2n-1}right)Rightarrow I=int_0^1prod_{n=1}^inftyleft(1-x^{2n-1}right)dx$$
I suspect this has a closed form since a similar integral to the last equality appeared here on MSE before; however this one is a bit different since the product goes only on odd powers and I don't see how to make a connection between the two of them, so I will appreciate some help with that.






integration definite-integrals closed-form infinite-product







share|cite|improve this question











$endgroup$




The following integral appeared this summer on AoPS. However it received no answer until today.
$$I=lim_{nto infty } int_0^1frac{dx}{(1+x)(1+x^2)dots(1+x^n)}=int_0^1 frac{dx}{prod_{n=1}^infty (1+x^n)}$$
I have learnt recently from here that: $$frac{1}{prod_{n=1}^infty (1+x^n)}=prod_{n=1}^inftyleft(1-x^{2n-1}right)Rightarrow I=int_0^1prod_{n=1}^inftyleft(1-x^{2n-1}right)dx$$
I suspect this has a closed form since a similar integral to the last equality appeared here on MSE before; however this one is a bit different since the product goes only on odd powers and I don't see how to make a connection between the two of them, so I will appreciate some help with that.






integration definite-integrals closed-form infinite-product




integration definite-integrals closed-form infinite-product






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 18:36









Michael Wang

186215




186215










asked Dec 15 '18 at 10:28









ZackyZacky

6,9251961




6,9251961








  • 5




    $begingroup$
    The series expansion of the integrand is here. An alternate form of the problem is $$I = int_0^1 (q;q^2)_infty ,dq,$$ where $(a;q)_infty$ is the $q$-Pochhammer symbol.
    $endgroup$
    – user153012
    Dec 16 '18 at 12:11






  • 4




    $begingroup$
    With a lot of help from Mathematica I can only obtain a (conjectural) infinite series form for the integral $$I=8 pisqrt{3} sum _{n=0}^{infty } frac{p(n) sinh left(frac{1}{6} pi sqrt{48 n+23}right)}{sqrt{48 n+23} left(2 cosh left(frac{1}{3} pi sqrt{48 n+23}right)-1right)}$$ where $p(n)$ is the number of unrestricted partitions of the integer n. This leads me to suspect that $I$ does not have a closed form.
    $endgroup$
    – James Arathoon
    Dec 19 '18 at 0:12








  • 1




    $begingroup$
    By numeric computation, this integral is about $0.42888151226615025372$. If you put it in WolframAlpha or RIES, you will see that this does not look like anything with a closed form.
    $endgroup$
    – ablmf
    Jan 14 at 19:52










  • $begingroup$
    $I=intlimits_0^1 dfrac2{(-1;x)_infty},mathrm dxapprox 0.428882.$
    $endgroup$
    – Yuri Negometyanov
    Jan 15 at 17:20














  • 5




    $begingroup$
    The series expansion of the integrand is here. An alternate form of the problem is $$I = int_0^1 (q;q^2)_infty ,dq,$$ where $(a;q)_infty$ is the $q$-Pochhammer symbol.
    $endgroup$
    – user153012
    Dec 16 '18 at 12:11






  • 4




    $begingroup$
    With a lot of help from Mathematica I can only obtain a (conjectural) infinite series form for the integral $$I=8 pisqrt{3} sum _{n=0}^{infty } frac{p(n) sinh left(frac{1}{6} pi sqrt{48 n+23}right)}{sqrt{48 n+23} left(2 cosh left(frac{1}{3} pi sqrt{48 n+23}right)-1right)}$$ where $p(n)$ is the number of unrestricted partitions of the integer n. This leads me to suspect that $I$ does not have a closed form.
    $endgroup$
    – James Arathoon
    Dec 19 '18 at 0:12








  • 1




    $begingroup$
    By numeric computation, this integral is about $0.42888151226615025372$. If you put it in WolframAlpha or RIES, you will see that this does not look like anything with a closed form.
    $endgroup$
    – ablmf
    Jan 14 at 19:52










  • $begingroup$
    $I=intlimits_0^1 dfrac2{(-1;x)_infty},mathrm dxapprox 0.428882.$
    $endgroup$
    – Yuri Negometyanov
    Jan 15 at 17:20








5




5




$begingroup$
The series expansion of the integrand is here. An alternate form of the problem is $$I = int_0^1 (q;q^2)_infty ,dq,$$ where $(a;q)_infty$ is the $q$-Pochhammer symbol.
$endgroup$
– user153012
Dec 16 '18 at 12:11




$begingroup$
The series expansion of the integrand is here. An alternate form of the problem is $$I = int_0^1 (q;q^2)_infty ,dq,$$ where $(a;q)_infty$ is the $q$-Pochhammer symbol.
$endgroup$
– user153012
Dec 16 '18 at 12:11




4




4




$begingroup$
With a lot of help from Mathematica I can only obtain a (conjectural) infinite series form for the integral $$I=8 pisqrt{3} sum _{n=0}^{infty } frac{p(n) sinh left(frac{1}{6} pi sqrt{48 n+23}right)}{sqrt{48 n+23} left(2 cosh left(frac{1}{3} pi sqrt{48 n+23}right)-1right)}$$ where $p(n)$ is the number of unrestricted partitions of the integer n. This leads me to suspect that $I$ does not have a closed form.
$endgroup$
– James Arathoon
Dec 19 '18 at 0:12






$begingroup$
With a lot of help from Mathematica I can only obtain a (conjectural) infinite series form for the integral $$I=8 pisqrt{3} sum _{n=0}^{infty } frac{p(n) sinh left(frac{1}{6} pi sqrt{48 n+23}right)}{sqrt{48 n+23} left(2 cosh left(frac{1}{3} pi sqrt{48 n+23}right)-1right)}$$ where $p(n)$ is the number of unrestricted partitions of the integer n. This leads me to suspect that $I$ does not have a closed form.
$endgroup$
– James Arathoon
Dec 19 '18 at 0:12






1




1




$begingroup$
By numeric computation, this integral is about $0.42888151226615025372$. If you put it in WolframAlpha or RIES, you will see that this does not look like anything with a closed form.
$endgroup$
– ablmf
Jan 14 at 19:52




$begingroup$
By numeric computation, this integral is about $0.42888151226615025372$. If you put it in WolframAlpha or RIES, you will see that this does not look like anything with a closed form.
$endgroup$
– ablmf
Jan 14 at 19:52












$begingroup$
$I=intlimits_0^1 dfrac2{(-1;x)_infty},mathrm dxapprox 0.428882.$
$endgroup$
– Yuri Negometyanov
Jan 15 at 17:20




$begingroup$
$I=intlimits_0^1 dfrac2{(-1;x)_infty},mathrm dxapprox 0.428882.$
$endgroup$
– Yuri Negometyanov
Jan 15 at 17:20










1 Answer
1






active

oldest

votes


















1












$begingroup$

$color{brown}{textbf{Analysis of the production.}}$



Let us consider the production
$$p(x)=prodlimits_{n=0}^{infty}dfrac1{1+x^n},quad xin(0,1).$$



First,
$$prodlimits_{k=0}^{infty}(1-x^{2k+1})cdotprodlimits_{k=0}^{infty}log(1-x^{2k}) = prodlimits_{k=0}^{infty}log(1-x^k),$$
so
$$p(x)=prodlimits_{n=0}^{infty}dfrac1{1+x^n} = prodlimits_{k=0}^{infty}(1-x^{2k+1}),quad xin(0,1).tag1$$



At the second, looks right the prove
$$sumlimits_{k=0}^{infty}ln(1-x^{2k+1})
= -sumlimits_{m=1}^{infty}sumlimits_{k=0}^{infty}dfrac{x^{(2k+1)m}}{m}
= -sumlimits_{m=1}^{infty}dfrac{x^m}{m(1-x^{2m})}\
= -sumlimits_{m=1}^{infty}dfrac1{2m}left(dfrac1{1-x^m} + dfrac1{1+x^m}right)
= -dfrac12sumlimits_{m=1}^{infty}sumlimits_{k=0}^{infty} left(dfrac{x^{km}}{m}+dfrac{(-x)^{km}}{m}right),$$
$$ln p(x)= dfrac12sumlimits_{k=0}^{infty}left(ln(1-x^k)+ln(1+x^k)right),$$
$$ln p(x)= dfrac13sumlimits_{k=0}^{infty}ln(1-x^k),$$
$$p(x)=sqrt[3]{prodlimits_{k=0}^{infty}(1-x^k)} = sqrt[3]{(x;x)_infty}, tag{*}$$
where $(x,x)_infty$ is q-Pochhammer symbol.



However, identity $(*)$ $color{red}{textrm{is wrong}}$ (see Wolfram Alpha counterexample).



$color{brown}{textbf{Results.}}$



Right identity is
$$p(x)=dfrac2{(-1;x)_infty}tag2$$
(see also Wolfram Alpha example).



There are not detalized information about $q$-Pochhammer symbols, so the value of integral is calculated numerically, wherein



$$boxed{I=intlimits_0^1 p(x),mathrm dx approx 0.428882.}$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @Zacky A possible error in the proof is a permutation of the members of a two-dimensional alternating signs sequence.
    $endgroup$
    – Yuri Negometyanov
    Jan 16 at 10:44













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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

$color{brown}{textbf{Analysis of the production.}}$



Let us consider the production
$$p(x)=prodlimits_{n=0}^{infty}dfrac1{1+x^n},quad xin(0,1).$$



First,
$$prodlimits_{k=0}^{infty}(1-x^{2k+1})cdotprodlimits_{k=0}^{infty}log(1-x^{2k}) = prodlimits_{k=0}^{infty}log(1-x^k),$$
so
$$p(x)=prodlimits_{n=0}^{infty}dfrac1{1+x^n} = prodlimits_{k=0}^{infty}(1-x^{2k+1}),quad xin(0,1).tag1$$



At the second, looks right the prove
$$sumlimits_{k=0}^{infty}ln(1-x^{2k+1})
= -sumlimits_{m=1}^{infty}sumlimits_{k=0}^{infty}dfrac{x^{(2k+1)m}}{m}
= -sumlimits_{m=1}^{infty}dfrac{x^m}{m(1-x^{2m})}\
= -sumlimits_{m=1}^{infty}dfrac1{2m}left(dfrac1{1-x^m} + dfrac1{1+x^m}right)
= -dfrac12sumlimits_{m=1}^{infty}sumlimits_{k=0}^{infty} left(dfrac{x^{km}}{m}+dfrac{(-x)^{km}}{m}right),$$
$$ln p(x)= dfrac12sumlimits_{k=0}^{infty}left(ln(1-x^k)+ln(1+x^k)right),$$
$$ln p(x)= dfrac13sumlimits_{k=0}^{infty}ln(1-x^k),$$
$$p(x)=sqrt[3]{prodlimits_{k=0}^{infty}(1-x^k)} = sqrt[3]{(x;x)_infty}, tag{*}$$
where $(x,x)_infty$ is q-Pochhammer symbol.



However, identity $(*)$ $color{red}{textrm{is wrong}}$ (see Wolfram Alpha counterexample).



$color{brown}{textbf{Results.}}$



Right identity is
$$p(x)=dfrac2{(-1;x)_infty}tag2$$
(see also Wolfram Alpha example).



There are not detalized information about $q$-Pochhammer symbols, so the value of integral is calculated numerically, wherein



$$boxed{I=intlimits_0^1 p(x),mathrm dx approx 0.428882.}$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @Zacky A possible error in the proof is a permutation of the members of a two-dimensional alternating signs sequence.
    $endgroup$
    – Yuri Negometyanov
    Jan 16 at 10:44


















1












$begingroup$

$color{brown}{textbf{Analysis of the production.}}$



Let us consider the production
$$p(x)=prodlimits_{n=0}^{infty}dfrac1{1+x^n},quad xin(0,1).$$



First,
$$prodlimits_{k=0}^{infty}(1-x^{2k+1})cdotprodlimits_{k=0}^{infty}log(1-x^{2k}) = prodlimits_{k=0}^{infty}log(1-x^k),$$
so
$$p(x)=prodlimits_{n=0}^{infty}dfrac1{1+x^n} = prodlimits_{k=0}^{infty}(1-x^{2k+1}),quad xin(0,1).tag1$$



At the second, looks right the prove
$$sumlimits_{k=0}^{infty}ln(1-x^{2k+1})
= -sumlimits_{m=1}^{infty}sumlimits_{k=0}^{infty}dfrac{x^{(2k+1)m}}{m}
= -sumlimits_{m=1}^{infty}dfrac{x^m}{m(1-x^{2m})}\
= -sumlimits_{m=1}^{infty}dfrac1{2m}left(dfrac1{1-x^m} + dfrac1{1+x^m}right)
= -dfrac12sumlimits_{m=1}^{infty}sumlimits_{k=0}^{infty} left(dfrac{x^{km}}{m}+dfrac{(-x)^{km}}{m}right),$$
$$ln p(x)= dfrac12sumlimits_{k=0}^{infty}left(ln(1-x^k)+ln(1+x^k)right),$$
$$ln p(x)= dfrac13sumlimits_{k=0}^{infty}ln(1-x^k),$$
$$p(x)=sqrt[3]{prodlimits_{k=0}^{infty}(1-x^k)} = sqrt[3]{(x;x)_infty}, tag{*}$$
where $(x,x)_infty$ is q-Pochhammer symbol.



However, identity $(*)$ $color{red}{textrm{is wrong}}$ (see Wolfram Alpha counterexample).



$color{brown}{textbf{Results.}}$



Right identity is
$$p(x)=dfrac2{(-1;x)_infty}tag2$$
(see also Wolfram Alpha example).



There are not detalized information about $q$-Pochhammer symbols, so the value of integral is calculated numerically, wherein



$$boxed{I=intlimits_0^1 p(x),mathrm dx approx 0.428882.}$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @Zacky A possible error in the proof is a permutation of the members of a two-dimensional alternating signs sequence.
    $endgroup$
    – Yuri Negometyanov
    Jan 16 at 10:44
















1












1








1





$begingroup$

$color{brown}{textbf{Analysis of the production.}}$



Let us consider the production
$$p(x)=prodlimits_{n=0}^{infty}dfrac1{1+x^n},quad xin(0,1).$$



First,
$$prodlimits_{k=0}^{infty}(1-x^{2k+1})cdotprodlimits_{k=0}^{infty}log(1-x^{2k}) = prodlimits_{k=0}^{infty}log(1-x^k),$$
so
$$p(x)=prodlimits_{n=0}^{infty}dfrac1{1+x^n} = prodlimits_{k=0}^{infty}(1-x^{2k+1}),quad xin(0,1).tag1$$



At the second, looks right the prove
$$sumlimits_{k=0}^{infty}ln(1-x^{2k+1})
= -sumlimits_{m=1}^{infty}sumlimits_{k=0}^{infty}dfrac{x^{(2k+1)m}}{m}
= -sumlimits_{m=1}^{infty}dfrac{x^m}{m(1-x^{2m})}\
= -sumlimits_{m=1}^{infty}dfrac1{2m}left(dfrac1{1-x^m} + dfrac1{1+x^m}right)
= -dfrac12sumlimits_{m=1}^{infty}sumlimits_{k=0}^{infty} left(dfrac{x^{km}}{m}+dfrac{(-x)^{km}}{m}right),$$
$$ln p(x)= dfrac12sumlimits_{k=0}^{infty}left(ln(1-x^k)+ln(1+x^k)right),$$
$$ln p(x)= dfrac13sumlimits_{k=0}^{infty}ln(1-x^k),$$
$$p(x)=sqrt[3]{prodlimits_{k=0}^{infty}(1-x^k)} = sqrt[3]{(x;x)_infty}, tag{*}$$
where $(x,x)_infty$ is q-Pochhammer symbol.



However, identity $(*)$ $color{red}{textrm{is wrong}}$ (see Wolfram Alpha counterexample).



$color{brown}{textbf{Results.}}$



Right identity is
$$p(x)=dfrac2{(-1;x)_infty}tag2$$
(see also Wolfram Alpha example).



There are not detalized information about $q$-Pochhammer symbols, so the value of integral is calculated numerically, wherein



$$boxed{I=intlimits_0^1 p(x),mathrm dx approx 0.428882.}$$






share|cite|improve this answer











$endgroup$



$color{brown}{textbf{Analysis of the production.}}$



Let us consider the production
$$p(x)=prodlimits_{n=0}^{infty}dfrac1{1+x^n},quad xin(0,1).$$



First,
$$prodlimits_{k=0}^{infty}(1-x^{2k+1})cdotprodlimits_{k=0}^{infty}log(1-x^{2k}) = prodlimits_{k=0}^{infty}log(1-x^k),$$
so
$$p(x)=prodlimits_{n=0}^{infty}dfrac1{1+x^n} = prodlimits_{k=0}^{infty}(1-x^{2k+1}),quad xin(0,1).tag1$$



At the second, looks right the prove
$$sumlimits_{k=0}^{infty}ln(1-x^{2k+1})
= -sumlimits_{m=1}^{infty}sumlimits_{k=0}^{infty}dfrac{x^{(2k+1)m}}{m}
= -sumlimits_{m=1}^{infty}dfrac{x^m}{m(1-x^{2m})}\
= -sumlimits_{m=1}^{infty}dfrac1{2m}left(dfrac1{1-x^m} + dfrac1{1+x^m}right)
= -dfrac12sumlimits_{m=1}^{infty}sumlimits_{k=0}^{infty} left(dfrac{x^{km}}{m}+dfrac{(-x)^{km}}{m}right),$$
$$ln p(x)= dfrac12sumlimits_{k=0}^{infty}left(ln(1-x^k)+ln(1+x^k)right),$$
$$ln p(x)= dfrac13sumlimits_{k=0}^{infty}ln(1-x^k),$$
$$p(x)=sqrt[3]{prodlimits_{k=0}^{infty}(1-x^k)} = sqrt[3]{(x;x)_infty}, tag{*}$$
where $(x,x)_infty$ is q-Pochhammer symbol.



However, identity $(*)$ $color{red}{textrm{is wrong}}$ (see Wolfram Alpha counterexample).



$color{brown}{textbf{Results.}}$



Right identity is
$$p(x)=dfrac2{(-1;x)_infty}tag2$$
(see also Wolfram Alpha example).



There are not detalized information about $q$-Pochhammer symbols, so the value of integral is calculated numerically, wherein



$$boxed{I=intlimits_0^1 p(x),mathrm dx approx 0.428882.}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 15 at 18:32

























answered Jan 13 at 13:15









Yuri NegometyanovYuri Negometyanov

11.8k1729




11.8k1729








  • 1




    $begingroup$
    @Zacky A possible error in the proof is a permutation of the members of a two-dimensional alternating signs sequence.
    $endgroup$
    – Yuri Negometyanov
    Jan 16 at 10:44
















  • 1




    $begingroup$
    @Zacky A possible error in the proof is a permutation of the members of a two-dimensional alternating signs sequence.
    $endgroup$
    – Yuri Negometyanov
    Jan 16 at 10:44










1




1




$begingroup$
@Zacky A possible error in the proof is a permutation of the members of a two-dimensional alternating signs sequence.
$endgroup$
– Yuri Negometyanov
Jan 16 at 10:44






$begingroup$
@Zacky A possible error in the proof is a permutation of the members of a two-dimensional alternating signs sequence.
$endgroup$
– Yuri Negometyanov
Jan 16 at 10:44




















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