How to compute a special double integral on a finite domain












0












$begingroup$


For any given integer $n>2$ and real constants $0leq aleq 1$ and $c>0$, I want to compute the integral:



$$ int _0^aint _0^afrac{(1+ c,x, y)^{n-2} (1+n,c, x, y)}{(1+x) (1+y)}, dx,dy $$



Wolfram Mathematica 11.3 does not give an answer.






integration multivariable-calculus







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  • 1




    $begingroup$
    Open the parentheses in the numerator to start with.
    $endgroup$
    – fedja
    Dec 1 '18 at 18:30
















0












$begingroup$


For any given integer $n>2$ and real constants $0leq aleq 1$ and $c>0$, I want to compute the integral:



$$ int _0^aint _0^afrac{(1+ c,x, y)^{n-2} (1+n,c, x, y)}{(1+x) (1+y)}, dx,dy $$



Wolfram Mathematica 11.3 does not give an answer.






integration multivariable-calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Open the parentheses in the numerator to start with.
    $endgroup$
    – fedja
    Dec 1 '18 at 18:30














0












0








0





$begingroup$


For any given integer $n>2$ and real constants $0leq aleq 1$ and $c>0$, I want to compute the integral:



$$ int _0^aint _0^afrac{(1+ c,x, y)^{n-2} (1+n,c, x, y)}{(1+x) (1+y)}, dx,dy $$



Wolfram Mathematica 11.3 does not give an answer.






integration multivariable-calculus







share|cite|improve this question









$endgroup$




For any given integer $n>2$ and real constants $0leq aleq 1$ and $c>0$, I want to compute the integral:



$$ int _0^aint _0^afrac{(1+ c,x, y)^{n-2} (1+n,c, x, y)}{(1+x) (1+y)}, dx,dy $$



Wolfram Mathematica 11.3 does not give an answer.






integration multivariable-calculus




integration multivariable-calculus






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Dec 1 '18 at 18:13









kaffeeaufkaffeeauf

14810




14810








  • 1




    $begingroup$
    Open the parentheses in the numerator to start with.
    $endgroup$
    – fedja
    Dec 1 '18 at 18:30














  • 1




    $begingroup$
    Open the parentheses in the numerator to start with.
    $endgroup$
    – fedja
    Dec 1 '18 at 18:30








1




1




$begingroup$
Open the parentheses in the numerator to start with.
$endgroup$
– fedja
Dec 1 '18 at 18:30




$begingroup$
Open the parentheses in the numerator to start with.
$endgroup$
– fedja
Dec 1 '18 at 18:30










1 Answer
1






active

oldest

votes


















1












$begingroup$

we have:
$$I=int_0^aint_0^afrac{(1+cxy)^{n-2}(1+ncxy)}{(1+x)(1+y)}dxdy$$
$$=int_0^afrac{1}{(1+y)}int_0^afrac{(1+cyx)^{n-2}+ncyx(1+cyx)^{n-2}}{(1+x)}dxdy$$
if we focus on the first one:
$$I_1=int_0^afrac{1}{(1+y)}int_0^afrac{(1+cyx)^{n-2}}{(1+x)}dxdy$$
this inside integral has no elementary derivative and neither do the others involved, so it appears that if there is a solution it will involve a double summation potentially






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The inside integral could be expressed in terms of certain hypergeometric functions.
    $endgroup$
    – kaffeeauf
    Dec 1 '18 at 20:43











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1 Answer
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active

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votes








1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









1












$begingroup$

we have:
$$I=int_0^aint_0^afrac{(1+cxy)^{n-2}(1+ncxy)}{(1+x)(1+y)}dxdy$$
$$=int_0^afrac{1}{(1+y)}int_0^afrac{(1+cyx)^{n-2}+ncyx(1+cyx)^{n-2}}{(1+x)}dxdy$$
if we focus on the first one:
$$I_1=int_0^afrac{1}{(1+y)}int_0^afrac{(1+cyx)^{n-2}}{(1+x)}dxdy$$
this inside integral has no elementary derivative and neither do the others involved, so it appears that if there is a solution it will involve a double summation potentially






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The inside integral could be expressed in terms of certain hypergeometric functions.
    $endgroup$
    – kaffeeauf
    Dec 1 '18 at 20:43
















1












$begingroup$

we have:
$$I=int_0^aint_0^afrac{(1+cxy)^{n-2}(1+ncxy)}{(1+x)(1+y)}dxdy$$
$$=int_0^afrac{1}{(1+y)}int_0^afrac{(1+cyx)^{n-2}+ncyx(1+cyx)^{n-2}}{(1+x)}dxdy$$
if we focus on the first one:
$$I_1=int_0^afrac{1}{(1+y)}int_0^afrac{(1+cyx)^{n-2}}{(1+x)}dxdy$$
this inside integral has no elementary derivative and neither do the others involved, so it appears that if there is a solution it will involve a double summation potentially






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The inside integral could be expressed in terms of certain hypergeometric functions.
    $endgroup$
    – kaffeeauf
    Dec 1 '18 at 20:43














1












1








1





$begingroup$

we have:
$$I=int_0^aint_0^afrac{(1+cxy)^{n-2}(1+ncxy)}{(1+x)(1+y)}dxdy$$
$$=int_0^afrac{1}{(1+y)}int_0^afrac{(1+cyx)^{n-2}+ncyx(1+cyx)^{n-2}}{(1+x)}dxdy$$
if we focus on the first one:
$$I_1=int_0^afrac{1}{(1+y)}int_0^afrac{(1+cyx)^{n-2}}{(1+x)}dxdy$$
this inside integral has no elementary derivative and neither do the others involved, so it appears that if there is a solution it will involve a double summation potentially






share|cite|improve this answer









$endgroup$



we have:
$$I=int_0^aint_0^afrac{(1+cxy)^{n-2}(1+ncxy)}{(1+x)(1+y)}dxdy$$
$$=int_0^afrac{1}{(1+y)}int_0^afrac{(1+cyx)^{n-2}+ncyx(1+cyx)^{n-2}}{(1+x)}dxdy$$
if we focus on the first one:
$$I_1=int_0^afrac{1}{(1+y)}int_0^afrac{(1+cyx)^{n-2}}{(1+x)}dxdy$$
this inside integral has no elementary derivative and neither do the others involved, so it appears that if there is a solution it will involve a double summation potentially







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '18 at 19:02









Henry LeeHenry Lee

1,857219




1,857219












  • $begingroup$
    The inside integral could be expressed in terms of certain hypergeometric functions.
    $endgroup$
    – kaffeeauf
    Dec 1 '18 at 20:43


















  • $begingroup$
    The inside integral could be expressed in terms of certain hypergeometric functions.
    $endgroup$
    – kaffeeauf
    Dec 1 '18 at 20:43
















$begingroup$
The inside integral could be expressed in terms of certain hypergeometric functions.
$endgroup$
– kaffeeauf
Dec 1 '18 at 20:43




$begingroup$
The inside integral could be expressed in terms of certain hypergeometric functions.
$endgroup$
– kaffeeauf
Dec 1 '18 at 20:43


















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