How to compute a special double integral on a finite domain
$begingroup$
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
asked Dec 1 '18 at 18:13
kaffeeaufkaffeeauf
14810
$endgroup$
add a comment |
$begingroup$
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
asked Dec 1 '18 at 18:13
kaffeeaufkaffeeauf
14810
$endgroup$
1
$begingroup$
Open the parentheses in the numerator to start with.
$endgroup$
– fedja
Dec 1 '18 at 18:30
add a comment |
$begingroup$
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
asked Dec 1 '18 at 18:13
kaffeeaufkaffeeauf
14810
$endgroup$
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
asked Dec 1 '18 at 18:13
kaffeeaufkaffeeauf
14810
asked Dec 1 '18 at 18:13
kaffeeaufkaffeeauf
14810
asked Dec 1 '18 at 18:13
kaffeeaufkaffeeauf
14810
asked Dec 1 '18 at 18:13
kaffeeaufkaffeeauf
14810
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
add a comment |
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
add a comment |
1 Answer
1
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$begingroup$
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
answered Dec 1 '18 at 19:02
Henry LeeHenry Lee
1,857219
$endgroup$
$begingroup$
The inside integral could be expressed in terms of certain hypergeometric functions.
$endgroup$
– kaffeeauf
Dec 1 '18 at 20:43
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
answered Dec 1 '18 at 19:02
Henry LeeHenry Lee
1,857219
$endgroup$
$begingroup$
The inside integral could be expressed in terms of certain hypergeometric functions.
$endgroup$
– kaffeeauf
Dec 1 '18 at 20:43
add a comment |
$begingroup$
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
answered Dec 1 '18 at 19:02
Henry LeeHenry Lee
1,857219
$endgroup$
$begingroup$
The inside integral could be expressed in terms of certain hypergeometric functions.
$endgroup$
– kaffeeauf
Dec 1 '18 at 20:43
add a comment |
$begingroup$
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
answered Dec 1 '18 at 19:02
Henry LeeHenry Lee
1,857219
$endgroup$
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
answered Dec 1 '18 at 19:02
Henry LeeHenry Lee
1,857219
answered Dec 1 '18 at 19:02
Henry LeeHenry Lee
1,857219
answered Dec 1 '18 at 19:02
Henry LeeHenry Lee
1,857219
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Open the parentheses in the numerator to start with.
$endgroup$
– fedja
Dec 1 '18 at 18:30