Solving an integral for $pi^2/6$

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So the above value is the result of the integral
$$
int_0^1 int_0^1 fracrm dx rm dy1-xy , .
$$
I'm playing around to find a substitution for which this integral is actually solvable in terms of elementary functions.



Does anyone know of one or more?



edit: I actually found one substitution in Proof3 of http://math.cmu.edu/~bwsulliv/basel-problem.pdf, but I think there are others.




calculus integration definite-integrals




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  • You can integrate $x$ first to get $-int_0^1 log(1-y)/y , dy$, and then use a power series expansion for $log(1-y)$. The resulting series sums to $pi^2/6$. Not sure if that qualifies as elementary functions.
    – Jeff
    Jul 23 at 0:19











  • No it does not, because that is what I want to calculate. I think I have actually seen some fancy transformation involving something like $x=tan(u)tan(v)$ and likewise something for $y$, but I dont remember.
    – Diger
    Jul 23 at 0:24











  • This is $zeta(2)$ and it's not an elementary functions.
    – Nosrati
    Jul 23 at 2:09










  • Look in the link and you'll know what I mean by "elementary"
    – Diger
    Jul 23 at 2:26






  • 1




    It has many pages, you may simply write $$int_0^1int_0^1frac11-xydxdy=int_0^1frac-ln(1-x)xdx=int_0^1frac1xsum_ngeqslant1fracx^nndx=sum_ngeqslant1frac1n^2=zeta(2)$$
    – Nosrati
    Jul 23 at 2:40















up vote
1
down vote

favorite
2












So the above value is the result of the integral
$$
int_0^1 int_0^1 fracrm dx rm dy1-xy , .
$$
I'm playing around to find a substitution for which this integral is actually solvable in terms of elementary functions.



Does anyone know of one or more?



edit: I actually found one substitution in Proof3 of http://math.cmu.edu/~bwsulliv/basel-problem.pdf, but I think there are others.




calculus integration definite-integrals




share|cite|improve this question





















  • You can integrate $x$ first to get $-int_0^1 log(1-y)/y , dy$, and then use a power series expansion for $log(1-y)$. The resulting series sums to $pi^2/6$. Not sure if that qualifies as elementary functions.
    – Jeff
    Jul 23 at 0:19











  • No it does not, because that is what I want to calculate. I think I have actually seen some fancy transformation involving something like $x=tan(u)tan(v)$ and likewise something for $y$, but I dont remember.
    – Diger
    Jul 23 at 0:24











  • This is $zeta(2)$ and it's not an elementary functions.
    – Nosrati
    Jul 23 at 2:09










  • Look in the link and you'll know what I mean by "elementary"
    – Diger
    Jul 23 at 2:26






  • 1




    It has many pages, you may simply write $$int_0^1int_0^1frac11-xydxdy=int_0^1frac-ln(1-x)xdx=int_0^1frac1xsum_ngeqslant1fracx^nndx=sum_ngeqslant1frac1n^2=zeta(2)$$
    – Nosrati
    Jul 23 at 2:40













up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2





So the above value is the result of the integral
$$
int_0^1 int_0^1 fracrm dx rm dy1-xy , .
$$
I'm playing around to find a substitution for which this integral is actually solvable in terms of elementary functions.



Does anyone know of one or more?



edit: I actually found one substitution in Proof3 of http://math.cmu.edu/~bwsulliv/basel-problem.pdf, but I think there are others.




calculus integration definite-integrals




share|cite|improve this question













So the above value is the result of the integral
$$
int_0^1 int_0^1 fracrm dx rm dy1-xy , .
$$
I'm playing around to find a substitution for which this integral is actually solvable in terms of elementary functions.



Does anyone know of one or more?



edit: I actually found one substitution in Proof3 of http://math.cmu.edu/~bwsulliv/basel-problem.pdf, but I think there are others.





calculus integration definite-integrals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 23 at 0:48
























asked Jul 23 at 0:12









Diger

53029




53029











  • You can integrate $x$ first to get $-int_0^1 log(1-y)/y , dy$, and then use a power series expansion for $log(1-y)$. The resulting series sums to $pi^2/6$. Not sure if that qualifies as elementary functions.
    – Jeff
    Jul 23 at 0:19











  • No it does not, because that is what I want to calculate. I think I have actually seen some fancy transformation involving something like $x=tan(u)tan(v)$ and likewise something for $y$, but I dont remember.
    – Diger
    Jul 23 at 0:24











  • This is $zeta(2)$ and it's not an elementary functions.
    – Nosrati
    Jul 23 at 2:09










  • Look in the link and you'll know what I mean by "elementary"
    – Diger
    Jul 23 at 2:26






  • 1




    It has many pages, you may simply write $$int_0^1int_0^1frac11-xydxdy=int_0^1frac-ln(1-x)xdx=int_0^1frac1xsum_ngeqslant1fracx^nndx=sum_ngeqslant1frac1n^2=zeta(2)$$
    – Nosrati
    Jul 23 at 2:40

















  • You can integrate $x$ first to get $-int_0^1 log(1-y)/y , dy$, and then use a power series expansion for $log(1-y)$. The resulting series sums to $pi^2/6$. Not sure if that qualifies as elementary functions.
    – Jeff
    Jul 23 at 0:19











  • No it does not, because that is what I want to calculate. I think I have actually seen some fancy transformation involving something like $x=tan(u)tan(v)$ and likewise something for $y$, but I dont remember.
    – Diger
    Jul 23 at 0:24











  • This is $zeta(2)$ and it's not an elementary functions.
    – Nosrati
    Jul 23 at 2:09










  • Look in the link and you'll know what I mean by "elementary"
    – Diger
    Jul 23 at 2:26






  • 1




    It has many pages, you may simply write $$int_0^1int_0^1frac11-xydxdy=int_0^1frac-ln(1-x)xdx=int_0^1frac1xsum_ngeqslant1fracx^nndx=sum_ngeqslant1frac1n^2=zeta(2)$$
    – Nosrati
    Jul 23 at 2:40
















You can integrate $x$ first to get $-int_0^1 log(1-y)/y , dy$, and then use a power series expansion for $log(1-y)$. The resulting series sums to $pi^2/6$. Not sure if that qualifies as elementary functions.
– Jeff
Jul 23 at 0:19





You can integrate $x$ first to get $-int_0^1 log(1-y)/y , dy$, and then use a power series expansion for $log(1-y)$. The resulting series sums to $pi^2/6$. Not sure if that qualifies as elementary functions.
– Jeff
Jul 23 at 0:19













No it does not, because that is what I want to calculate. I think I have actually seen some fancy transformation involving something like $x=tan(u)tan(v)$ and likewise something for $y$, but I dont remember.
– Diger
Jul 23 at 0:24





No it does not, because that is what I want to calculate. I think I have actually seen some fancy transformation involving something like $x=tan(u)tan(v)$ and likewise something for $y$, but I dont remember.
– Diger
Jul 23 at 0:24













This is $zeta(2)$ and it's not an elementary functions.
– Nosrati
Jul 23 at 2:09




This is $zeta(2)$ and it's not an elementary functions.
– Nosrati
Jul 23 at 2:09












Look in the link and you'll know what I mean by "elementary"
– Diger
Jul 23 at 2:26




Look in the link and you'll know what I mean by "elementary"
– Diger
Jul 23 at 2:26




1




1




It has many pages, you may simply write $$int_0^1int_0^1frac11-xydxdy=int_0^1frac-ln(1-x)xdx=int_0^1frac1xsum_ngeqslant1fracx^nndx=sum_ngeqslant1frac1n^2=zeta(2)$$
– Nosrati
Jul 23 at 2:40





It has many pages, you may simply write $$int_0^1int_0^1frac11-xydxdy=int_0^1frac-ln(1-x)xdx=int_0^1frac1xsum_ngeqslant1fracx^nndx=sum_ngeqslant1frac1n^2=zeta(2)$$
– Nosrati
Jul 23 at 2:40











1 Answer
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I don't know if it's solvable using but only elementary functions, but I do know it can solved as follows:



$$int _0^1int _0^1frac11-x ydxdy = int_0^1 -fraclogleft(1-yright)y+fraclogleft(1right)y dy= int_0^1 -fraclogleft(1-yright)ydy$$



This can be integrated and becomes the polylogarithm function:



$$textLi_2left(yright) \ textLi_2left(1right) = fracpi^26 \ textLi_2left(0right)=0 \ fracpi^26-0=fracpi^26$$






share|cite|improve this answer





















  • polylog is not elementary in my opinion.
    – Diger
    Jul 23 at 2:28










  • OK, but I'd suggest looking here for more info and maybe leads on an answer
    – Joseph Eck
    Jul 23 at 2:51










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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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up vote
0
down vote













I don't know if it's solvable using but only elementary functions, but I do know it can solved as follows:



$$int _0^1int _0^1frac11-x ydxdy = int_0^1 -fraclogleft(1-yright)y+fraclogleft(1right)y dy= int_0^1 -fraclogleft(1-yright)ydy$$



This can be integrated and becomes the polylogarithm function:



$$textLi_2left(yright) \ textLi_2left(1right) = fracpi^26 \ textLi_2left(0right)=0 \ fracpi^26-0=fracpi^26$$






share|cite|improve this answer





















  • polylog is not elementary in my opinion.
    – Diger
    Jul 23 at 2:28










  • OK, but I'd suggest looking here for more info and maybe leads on an answer
    – Joseph Eck
    Jul 23 at 2:51














up vote
0
down vote













I don't know if it's solvable using but only elementary functions, but I do know it can solved as follows:



$$int _0^1int _0^1frac11-x ydxdy = int_0^1 -fraclogleft(1-yright)y+fraclogleft(1right)y dy= int_0^1 -fraclogleft(1-yright)ydy$$



This can be integrated and becomes the polylogarithm function:



$$textLi_2left(yright) \ textLi_2left(1right) = fracpi^26 \ textLi_2left(0right)=0 \ fracpi^26-0=fracpi^26$$






share|cite|improve this answer





















  • polylog is not elementary in my opinion.
    – Diger
    Jul 23 at 2:28










  • OK, but I'd suggest looking here for more info and maybe leads on an answer
    – Joseph Eck
    Jul 23 at 2:51












up vote
0
down vote










up vote
0
down vote









I don't know if it's solvable using but only elementary functions, but I do know it can solved as follows:



$$int _0^1int _0^1frac11-x ydxdy = int_0^1 -fraclogleft(1-yright)y+fraclogleft(1right)y dy= int_0^1 -fraclogleft(1-yright)ydy$$



This can be integrated and becomes the polylogarithm function:



$$textLi_2left(yright) \ textLi_2left(1right) = fracpi^26 \ textLi_2left(0right)=0 \ fracpi^26-0=fracpi^26$$






share|cite|improve this answer













I don't know if it's solvable using but only elementary functions, but I do know it can solved as follows:



$$int _0^1int _0^1frac11-x ydxdy = int_0^1 -fraclogleft(1-yright)y+fraclogleft(1right)y dy= int_0^1 -fraclogleft(1-yright)ydy$$



This can be integrated and becomes the polylogarithm function:



$$textLi_2left(yright) \ textLi_2left(1right) = fracpi^26 \ textLi_2left(0right)=0 \ fracpi^26-0=fracpi^26$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 23 at 1:59









Joseph Eck

570313




570313











  • polylog is not elementary in my opinion.
    – Diger
    Jul 23 at 2:28










  • OK, but I'd suggest looking here for more info and maybe leads on an answer
    – Joseph Eck
    Jul 23 at 2:51
















  • polylog is not elementary in my opinion.
    – Diger
    Jul 23 at 2:28










  • OK, but I'd suggest looking here for more info and maybe leads on an answer
    – Joseph Eck
    Jul 23 at 2:51















polylog is not elementary in my opinion.
– Diger
Jul 23 at 2:28




polylog is not elementary in my opinion.
– Diger
Jul 23 at 2:28












OK, but I'd suggest looking here for more info and maybe leads on an answer
– Joseph Eck
Jul 23 at 2:51




OK, but I'd suggest looking here for more info and maybe leads on an answer
– Joseph Eck
Jul 23 at 2:51












 

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