Mixing
Prove that $int_0^1Bigl(int_0^1f(x,y),dxBigr),dyneint_0^1Bigl(int_0^1f(x,y),dyBigr),dx$
$begingroup$
Let $f(x,y)=xy/(x^2+y^2)$ for $x,yin(0,1]$. Prove that $$int_0^1Bigl(int_0^1f(x,y),dxBigr),dyneint_0^1Bigl(int_0^1f(x,y),dyBigr),dx.$$
I am not sure how these are unequal though because I get $$int_0^1Bigl(int_0^1f(x,y),dxBigr),dy = int_0^1Bigl(frac{1}{2}y(log(y^2+1)-log(y^2)) Bigr)dy= frac{log(2)}{2}$$
and we have $$int_0^1Bigl(int_0^1f(x,y),dyBigr),dx = int_0^1Bigl(frac{1}{2}x(log(x^2+1)-log(x^2)) Bigr)dx= frac{log(2)}{2}$$
I must be missing something here about the conditions of the inner integral. I get for the LHS that $Re(y)!=0 ∨ Im(y)>1 ∨ Im(y)<-1$. Similarly for the RHS.
real-analysis integration
edited Jan 20 at 20:43
Shaun
9,366113684
asked Jan 20 at 19:44
MathIsHardMathIsHard
1,278516
$endgroup$
add a comment |
$begingroup$
Let $f(x,y)=xy/(x^2+y^2)$ for $x,yin(0,1]$. Prove that $$int_0^1Bigl(int_0^1f(x,y),dxBigr),dyneint_0^1Bigl(int_0^1f(x,y),dyBigr),dx.$$
I am not sure how these are unequal though because I get $$int_0^1Bigl(int_0^1f(x,y),dxBigr),dy = int_0^1Bigl(frac{1}{2}y(log(y^2+1)-log(y^2)) Bigr)dy= frac{log(2)}{2}$$
and we have $$int_0^1Bigl(int_0^1f(x,y),dyBigr),dx = int_0^1Bigl(frac{1}{2}x(log(x^2+1)-log(x^2)) Bigr)dx= frac{log(2)}{2}$$
I must be missing something here about the conditions of the inner integral. I get for the LHS that $Re(y)!=0 ∨ Im(y)>1 ∨ Im(y)<-1$. Similarly for the RHS.
real-analysis integration
edited Jan 20 at 20:43
Shaun
9,366113684
asked Jan 20 at 19:44
MathIsHardMathIsHard
1,278516
$endgroup$
1
$begingroup$
Shouldn't they be equal because $f(x,y)=f(y,x)$?
$endgroup$
– Botond
Jan 20 at 19:50
5
$begingroup$
You are right, the integrals mentioned in the title are equal (also because $f(x,y)=f(y,x)$). The intended counter-example was probably $frac{x^2-y^2}{(x^2+y^2)^2}$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 19:51
$begingroup$
The end of your message is illisible ; make an effort for it. For the integral, the expression and the domain are symetrical, it is impossible to get distinct values.
$endgroup$
– DLeMeur
Jan 20 at 19:55
$begingroup$
Thank you Jack D'Aurizio. It must have been a typo. I just couldn't figure out the problem and thought I was missing something haha.
$endgroup$
– MathIsHard
Jan 20 at 20:08
add a comment |
$begingroup$
Let $f(x,y)=xy/(x^2+y^2)$ for $x,yin(0,1]$. Prove that $$int_0^1Bigl(int_0^1f(x,y),dxBigr),dyneint_0^1Bigl(int_0^1f(x,y),dyBigr),dx.$$
I am not sure how these are unequal though because I get $$int_0^1Bigl(int_0^1f(x,y),dxBigr),dy = int_0^1Bigl(frac{1}{2}y(log(y^2+1)-log(y^2)) Bigr)dy= frac{log(2)}{2}$$
and we have $$int_0^1Bigl(int_0^1f(x,y),dyBigr),dx = int_0^1Bigl(frac{1}{2}x(log(x^2+1)-log(x^2)) Bigr)dx= frac{log(2)}{2}$$
I must be missing something here about the conditions of the inner integral. I get for the LHS that $Re(y)!=0 ∨ Im(y)>1 ∨ Im(y)<-1$. Similarly for the RHS.
real-analysis integration
edited Jan 20 at 20:43
Shaun
9,366113684
asked Jan 20 at 19:44
MathIsHardMathIsHard
1,278516
$endgroup$
Let $f(x,y)=xy/(x^2+y^2)$ for $x,yin(0,1]$. Prove that $$int_0^1Bigl(int_0^1f(x,y),dxBigr),dyneint_0^1Bigl(int_0^1f(x,y),dyBigr),dx.$$
I am not sure how these are unequal though because I get $$int_0^1Bigl(int_0^1f(x,y),dxBigr),dy = int_0^1Bigl(frac{1}{2}y(log(y^2+1)-log(y^2)) Bigr)dy= frac{log(2)}{2}$$
and we have $$int_0^1Bigl(int_0^1f(x,y),dyBigr),dx = int_0^1Bigl(frac{1}{2}x(log(x^2+1)-log(x^2)) Bigr)dx= frac{log(2)}{2}$$
I must be missing something here about the conditions of the inner integral. I get for the LHS that $Re(y)!=0 ∨ Im(y)>1 ∨ Im(y)<-1$. Similarly for the RHS.
real-analysis integration
real-analysis integration
edited Jan 20 at 20:43
Shaun
9,366113684
asked Jan 20 at 19:44
MathIsHardMathIsHard
1,278516
edited Jan 20 at 20:43
Shaun
9,366113684
asked Jan 20 at 19:44
MathIsHardMathIsHard
1,278516
edited Jan 20 at 20:43
Shaun
9,366113684
edited Jan 20 at 20:43
Shaun
9,366113684
edited Jan 20 at 20:43
Shaun
9,366113684
9,366113684
asked Jan 20 at 19:44
MathIsHardMathIsHard
1,278516
asked Jan 20 at 19:44
MathIsHardMathIsHard
1,278516
asked Jan 20 at 19:44
MathIsHardMathIsHard
1,278516
1,278516
1
$begingroup$
Shouldn't they be equal because $f(x,y)=f(y,x)$?
$endgroup$
– Botond
Jan 20 at 19:50
5
$begingroup$
You are right, the integrals mentioned in the title are equal (also because $f(x,y)=f(y,x)$). The intended counter-example was probably $frac{x^2-y^2}{(x^2+y^2)^2}$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 19:51
$begingroup$
The end of your message is illisible ; make an effort for it. For the integral, the expression and the domain are symetrical, it is impossible to get distinct values.
$endgroup$
– DLeMeur
Jan 20 at 19:55
$begingroup$
Thank you Jack D'Aurizio. It must have been a typo. I just couldn't figure out the problem and thought I was missing something haha.
$endgroup$
– MathIsHard
Jan 20 at 20:08
add a comment |
1
$begingroup$
Shouldn't they be equal because $f(x,y)=f(y,x)$?
$endgroup$
– Botond
Jan 20 at 19:50
5
$begingroup$
You are right, the integrals mentioned in the title are equal (also because $f(x,y)=f(y,x)$). The intended counter-example was probably $frac{x^2-y^2}{(x^2+y^2)^2}$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 19:51
$begingroup$
The end of your message is illisible ; make an effort for it. For the integral, the expression and the domain are symetrical, it is impossible to get distinct values.
$endgroup$
– DLeMeur
Jan 20 at 19:55
$begingroup$
Thank you Jack D'Aurizio. It must have been a typo. I just couldn't figure out the problem and thought I was missing something haha.
$endgroup$
– MathIsHard
Jan 20 at 20:08
1
1
$begingroup$
Shouldn't they be equal because $f(x,y)=f(y,x)$?
$endgroup$
– Botond
Jan 20 at 19:50
$begingroup$
Shouldn't they be equal because $f(x,y)=f(y,x)$?
$endgroup$
– Botond
Jan 20 at 19:50
5
5
$begingroup$
You are right, the integrals mentioned in the title are equal (also because $f(x,y)=f(y,x)$). The intended counter-example was probably $frac{x^2-y^2}{(x^2+y^2)^2}$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 19:51
$begingroup$
You are right, the integrals mentioned in the title are equal (also because $f(x,y)=f(y,x)$). The intended counter-example was probably $frac{x^2-y^2}{(x^2+y^2)^2}$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 19:51
$begingroup$
The end of your message is illisible ; make an effort for it. For the integral, the expression and the domain are symetrical, it is impossible to get distinct values.
$endgroup$
– DLeMeur
Jan 20 at 19:55
$begingroup$
The end of your message is illisible ; make an effort for it. For the integral, the expression and the domain are symetrical, it is impossible to get distinct values.
$endgroup$
– DLeMeur
Jan 20 at 19:55
$begingroup$
Thank you Jack D'Aurizio. It must have been a typo. I just couldn't figure out the problem and thought I was missing something haha.
$endgroup$
– MathIsHard
Jan 20 at 20:08
$begingroup$
Thank you Jack D'Aurizio. It must have been a typo. I just couldn't figure out the problem and thought I was missing something haha.
$endgroup$
– MathIsHard
Jan 20 at 20:08
add a comment |
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$begingroup$
Shouldn't they be equal because $f(x,y)=f(y,x)$?
$endgroup$
– Botond
Jan 20 at 19:50
5
$begingroup$
You are right, the integrals mentioned in the title are equal (also because $f(x,y)=f(y,x)$). The intended counter-example was probably $frac{x^2-y^2}{(x^2+y^2)^2}$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 19:51
$begingroup$
The end of your message is illisible ; make an effort for it. For the integral, the expression and the domain are symetrical, it is impossible to get distinct values.
$endgroup$
– DLeMeur
Jan 20 at 19:55
$begingroup$
Thank you Jack D'Aurizio. It must have been a typo. I just couldn't figure out the problem and thought I was missing something haha.
$endgroup$
– MathIsHard
Jan 20 at 20:08