Separate double integrals with min/max
$begingroup$
Question: Is it possible to separate integrals of multiple functions which contain min/max functions?
In my case in particular I have an integral in the following form:
$$
begin{equation}
I = iint_0^infty f(x)g(y)frac{min(x,y)^L}{max(x,y)^{L+1}}dxdy
end{equation}
$$
Where L is some integer. I have a numerical solver which seems to give reasonable results, but I would like to have a closed-form solution, if possible. What I envision (if it is possible) is to have some sort of solution of the form:
$$
begin{equation}
I = I_{x>y} + I_{x<y} + I_{x=y}
end{equation}
$$
(I argue that since $x=y$ only in some small infinitesimal case, it can be neglected)
What I have tried:
Attempt 1: Solve indefinite
I tried to solve the indefinite cases where $x>y$, $x<y$ and $x=y$, then add them thus:
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty f(x)x^{-L-1}dxint_0^infty g(y)y^Ldy +...
end{equation}
$$
However this solution seems to deviate from the numerical one. For L=0, it seems to be almost exactly a factor of 10, at L=1 it seems to be a factor of 2, etc. (This solution is especially tempting as I have closed form solutions for all of the integrals).
Attempt 2: Change bounds of integration
In this attempt, I utilized the min/max to change the bounds of integration of each of the cases. So for example when $x>y$, I only integrate $y$ from 0 to $x$, something like:
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty f(x)x^{-L-1} left(int_0^x g(y)y^Ldyright)dx + ...
end{equation}
$$
This doesn't seem to correspond to the numerical solution, but also doesn't seem to be off by some clean factor like attempt 1.
Attempt 3: Add new bound of integration
In this attempt I again use the min/max to split the equation, but this time I add a new variable and integrate over all possible values of this new variable. So for $x>y$ I integrate $y$ from 0 to $z$, then I integrate $x$ from $z$ to $infty$, then I integrate $z$ from 0 to $infty$.
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty left(int_z^infty f(x)x^{-L-1}dxint_0^z g(y)y^Ldyright)dz + ...
end{equation}
$$
Again, this doesn't seem to correspond to the numerical solution, but also doesn't seem to be off by some nice clear factors like attempt 1.
calculus integration maxima-minima
edited Jan 14 at 15:58
Paul Frost
12.7k31035
asked Jan 11 at 19:48
DaceDace
112
$endgroup$
add a comment |
$begingroup$
Question: Is it possible to separate integrals of multiple functions which contain min/max functions?
In my case in particular I have an integral in the following form:
$$
begin{equation}
I = iint_0^infty f(x)g(y)frac{min(x,y)^L}{max(x,y)^{L+1}}dxdy
end{equation}
$$
Where L is some integer. I have a numerical solver which seems to give reasonable results, but I would like to have a closed-form solution, if possible. What I envision (if it is possible) is to have some sort of solution of the form:
$$
begin{equation}
I = I_{x>y} + I_{x<y} + I_{x=y}
end{equation}
$$
(I argue that since $x=y$ only in some small infinitesimal case, it can be neglected)
What I have tried:
Attempt 1: Solve indefinite
I tried to solve the indefinite cases where $x>y$, $x<y$ and $x=y$, then add them thus:
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty f(x)x^{-L-1}dxint_0^infty g(y)y^Ldy +...
end{equation}
$$
However this solution seems to deviate from the numerical one. For L=0, it seems to be almost exactly a factor of 10, at L=1 it seems to be a factor of 2, etc. (This solution is especially tempting as I have closed form solutions for all of the integrals).
Attempt 2: Change bounds of integration
In this attempt, I utilized the min/max to change the bounds of integration of each of the cases. So for example when $x>y$, I only integrate $y$ from 0 to $x$, something like:
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty f(x)x^{-L-1} left(int_0^x g(y)y^Ldyright)dx + ...
end{equation}
$$
This doesn't seem to correspond to the numerical solution, but also doesn't seem to be off by some clean factor like attempt 1.
Attempt 3: Add new bound of integration
In this attempt I again use the min/max to split the equation, but this time I add a new variable and integrate over all possible values of this new variable. So for $x>y$ I integrate $y$ from 0 to $z$, then I integrate $x$ from $z$ to $infty$, then I integrate $z$ from 0 to $infty$.
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty left(int_z^infty f(x)x^{-L-1}dxint_0^z g(y)y^Ldyright)dz + ...
end{equation}
$$
Again, this doesn't seem to correspond to the numerical solution, but also doesn't seem to be off by some nice clear factors like attempt 1.
calculus integration maxima-minima
edited Jan 14 at 15:58
Paul Frost
12.7k31035
asked Jan 11 at 19:48
DaceDace
112
$endgroup$
$begingroup$
Attempt 2 should yield the right result.
$endgroup$
– RcnSc
Jan 14 at 16:07
$begingroup$
@RcnSc I'll double-check my attempt that, but in the meantime, do you have some basis for your assertion?
$endgroup$
– Dace
Jan 22 at 19:55
$begingroup$
You haven't posted the entire computation but every step you have posted is legal. Perhaps you can post all of it and we can look into it together?
$endgroup$
– RcnSc
Jan 23 at 10:01
$begingroup$
I redid Attempt 2, but starting from a new Python script and carefully redid everything and sure enough it worked (I was missing a term in the original). What is it you mean that you need to know the form of f(x) and g(y) ?
$endgroup$
– Dace
Jan 24 at 19:11
$begingroup$
I mostly wanted to see the other term of the integral to make sure there wasn't an oversight with the bounds. If you managed to make it match with your script then I assume it's all good.
$endgroup$
– RcnSc
Jan 25 at 8:45
add a comment |
$begingroup$
Question: Is it possible to separate integrals of multiple functions which contain min/max functions?
In my case in particular I have an integral in the following form:
$$
begin{equation}
I = iint_0^infty f(x)g(y)frac{min(x,y)^L}{max(x,y)^{L+1}}dxdy
end{equation}
$$
Where L is some integer. I have a numerical solver which seems to give reasonable results, but I would like to have a closed-form solution, if possible. What I envision (if it is possible) is to have some sort of solution of the form:
$$
begin{equation}
I = I_{x>y} + I_{x<y} + I_{x=y}
end{equation}
$$
(I argue that since $x=y$ only in some small infinitesimal case, it can be neglected)
What I have tried:
Attempt 1: Solve indefinite
I tried to solve the indefinite cases where $x>y$, $x<y$ and $x=y$, then add them thus:
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty f(x)x^{-L-1}dxint_0^infty g(y)y^Ldy +...
end{equation}
$$
However this solution seems to deviate from the numerical one. For L=0, it seems to be almost exactly a factor of 10, at L=1 it seems to be a factor of 2, etc. (This solution is especially tempting as I have closed form solutions for all of the integrals).
Attempt 2: Change bounds of integration
In this attempt, I utilized the min/max to change the bounds of integration of each of the cases. So for example when $x>y$, I only integrate $y$ from 0 to $x$, something like:
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty f(x)x^{-L-1} left(int_0^x g(y)y^Ldyright)dx + ...
end{equation}
$$
This doesn't seem to correspond to the numerical solution, but also doesn't seem to be off by some clean factor like attempt 1.
Attempt 3: Add new bound of integration
In this attempt I again use the min/max to split the equation, but this time I add a new variable and integrate over all possible values of this new variable. So for $x>y$ I integrate $y$ from 0 to $z$, then I integrate $x$ from $z$ to $infty$, then I integrate $z$ from 0 to $infty$.
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty left(int_z^infty f(x)x^{-L-1}dxint_0^z g(y)y^Ldyright)dz + ...
end{equation}
$$
Again, this doesn't seem to correspond to the numerical solution, but also doesn't seem to be off by some nice clear factors like attempt 1.
calculus integration maxima-minima
edited Jan 14 at 15:58
Paul Frost
12.7k31035
asked Jan 11 at 19:48
DaceDace
112
$endgroup$
Question: Is it possible to separate integrals of multiple functions which contain min/max functions?
In my case in particular I have an integral in the following form:
$$
begin{equation}
I = iint_0^infty f(x)g(y)frac{min(x,y)^L}{max(x,y)^{L+1}}dxdy
end{equation}
$$
Where L is some integer. I have a numerical solver which seems to give reasonable results, but I would like to have a closed-form solution, if possible. What I envision (if it is possible) is to have some sort of solution of the form:
$$
begin{equation}
I = I_{x>y} + I_{x<y} + I_{x=y}
end{equation}
$$
(I argue that since $x=y$ only in some small infinitesimal case, it can be neglected)
What I have tried:
Attempt 1: Solve indefinite
I tried to solve the indefinite cases where $x>y$, $x<y$ and $x=y$, then add them thus:
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty f(x)x^{-L-1}dxint_0^infty g(y)y^Ldy +...
end{equation}
$$
However this solution seems to deviate from the numerical one. For L=0, it seems to be almost exactly a factor of 10, at L=1 it seems to be a factor of 2, etc. (This solution is especially tempting as I have closed form solutions for all of the integrals).
Attempt 2: Change bounds of integration
In this attempt, I utilized the min/max to change the bounds of integration of each of the cases. So for example when $x>y$, I only integrate $y$ from 0 to $x$, something like:
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty f(x)x^{-L-1} left(int_0^x g(y)y^Ldyright)dx + ...
end{equation}
$$
This doesn't seem to correspond to the numerical solution, but also doesn't seem to be off by some clean factor like attempt 1.
Attempt 3: Add new bound of integration
In this attempt I again use the min/max to split the equation, but this time I add a new variable and integrate over all possible values of this new variable. So for $x>y$ I integrate $y$ from 0 to $z$, then I integrate $x$ from $z$ to $infty$, then I integrate $z$ from 0 to $infty$.
$$
begin{equation}
I = I_{x>y} + I_{x<y}=int_0^infty left(int_z^infty f(x)x^{-L-1}dxint_0^z g(y)y^Ldyright)dz + ...
end{equation}
$$
Again, this doesn't seem to correspond to the numerical solution, but also doesn't seem to be off by some nice clear factors like attempt 1.
calculus integration maxima-minima
calculus integration maxima-minima
edited Jan 14 at 15:58
Paul Frost
12.7k31035
asked Jan 11 at 19:48
DaceDace
112
edited Jan 14 at 15:58
Paul Frost
12.7k31035
asked Jan 11 at 19:48
DaceDace
112
edited Jan 14 at 15:58
Paul Frost
12.7k31035
edited Jan 14 at 15:58
Paul Frost
12.7k31035
edited Jan 14 at 15:58
Paul Frost
12.7k31035
12.7k31035
asked Jan 11 at 19:48
DaceDace
112
asked Jan 11 at 19:48
DaceDace
112
asked Jan 11 at 19:48
DaceDace
112
112
$begingroup$
Attempt 2 should yield the right result.
$endgroup$
– RcnSc
Jan 14 at 16:07
$begingroup$
@RcnSc I'll double-check my attempt that, but in the meantime, do you have some basis for your assertion?
$endgroup$
– Dace
Jan 22 at 19:55
$begingroup$
You haven't posted the entire computation but every step you have posted is legal. Perhaps you can post all of it and we can look into it together?
$endgroup$
– RcnSc
Jan 23 at 10:01
$begingroup$
I redid Attempt 2, but starting from a new Python script and carefully redid everything and sure enough it worked (I was missing a term in the original). What is it you mean that you need to know the form of f(x) and g(y) ?
$endgroup$
– Dace
Jan 24 at 19:11
$begingroup$
I mostly wanted to see the other term of the integral to make sure there wasn't an oversight with the bounds. If you managed to make it match with your script then I assume it's all good.
$endgroup$
– RcnSc
Jan 25 at 8:45
add a comment |
$begingroup$
Attempt 2 should yield the right result.
$endgroup$
– RcnSc
Jan 14 at 16:07
$begingroup$
@RcnSc I'll double-check my attempt that, but in the meantime, do you have some basis for your assertion?
$endgroup$
– Dace
Jan 22 at 19:55
$begingroup$
You haven't posted the entire computation but every step you have posted is legal. Perhaps you can post all of it and we can look into it together?
$endgroup$
– RcnSc
Jan 23 at 10:01
$begingroup$
I redid Attempt 2, but starting from a new Python script and carefully redid everything and sure enough it worked (I was missing a term in the original). What is it you mean that you need to know the form of f(x) and g(y) ?
$endgroup$
– Dace
Jan 24 at 19:11
$begingroup$
I mostly wanted to see the other term of the integral to make sure there wasn't an oversight with the bounds. If you managed to make it match with your script then I assume it's all good.
$endgroup$
– RcnSc
Jan 25 at 8:45
$begingroup$
Attempt 2 should yield the right result.
$endgroup$
– RcnSc
Jan 14 at 16:07
$begingroup$
Attempt 2 should yield the right result.
$endgroup$
– RcnSc
Jan 14 at 16:07
$begingroup$
@RcnSc I'll double-check my attempt that, but in the meantime, do you have some basis for your assertion?
$endgroup$
– Dace
Jan 22 at 19:55
$begingroup$
@RcnSc I'll double-check my attempt that, but in the meantime, do you have some basis for your assertion?
$endgroup$
– Dace
Jan 22 at 19:55
$begingroup$
You haven't posted the entire computation but every step you have posted is legal. Perhaps you can post all of it and we can look into it together?
$endgroup$
– RcnSc
Jan 23 at 10:01
$begingroup$
You haven't posted the entire computation but every step you have posted is legal. Perhaps you can post all of it and we can look into it together?
$endgroup$
– RcnSc
Jan 23 at 10:01
$begingroup$
I redid Attempt 2, but starting from a new Python script and carefully redid everything and sure enough it worked (I was missing a term in the original). What is it you mean that you need to know the form of f(x) and g(y) ?
$endgroup$
– Dace
Jan 24 at 19:11
$begingroup$
I redid Attempt 2, but starting from a new Python script and carefully redid everything and sure enough it worked (I was missing a term in the original). What is it you mean that you need to know the form of f(x) and g(y) ?
$endgroup$
– Dace
Jan 24 at 19:11
$begingroup$
I mostly wanted to see the other term of the integral to make sure there wasn't an oversight with the bounds. If you managed to make it match with your script then I assume it's all good.
$endgroup$
– RcnSc
Jan 25 at 8:45
$begingroup$
I mostly wanted to see the other term of the integral to make sure there wasn't an oversight with the bounds. If you managed to make it match with your script then I assume it's all good.
$endgroup$
– RcnSc
Jan 25 at 8:45
add a comment |
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$begingroup$
Attempt 2 should yield the right result.
$endgroup$
– RcnSc
Jan 14 at 16:07
$begingroup$
@RcnSc I'll double-check my attempt that, but in the meantime, do you have some basis for your assertion?
$endgroup$
– Dace
Jan 22 at 19:55
$begingroup$
You haven't posted the entire computation but every step you have posted is legal. Perhaps you can post all of it and we can look into it together?
$endgroup$
– RcnSc
Jan 23 at 10:01
$begingroup$
I redid Attempt 2, but starting from a new Python script and carefully redid everything and sure enough it worked (I was missing a term in the original). What is it you mean that you need to know the form of f(x) and g(y) ?
$endgroup$
– Dace
Jan 24 at 19:11
$begingroup$
I mostly wanted to see the other term of the integral to make sure there wasn't an oversight with the bounds. If you managed to make it match with your script then I assume it's all good.
$endgroup$
– RcnSc
Jan 25 at 8:45