Mixing
integration by substitution of multiple variables
$begingroup$
I have an integral
begin{equation}
int_{mathbb{R}^n}f(mathbf{B}mathbf{x})mathrm{d}mathbf{x}
end{equation}
where $f: mathbb{R}^m rightarrow mathbb{R}$ and $mathbf{B}inmathbb{R}^{mtimes n}$. I also know
begin{equation}
int_{mathbb{R}^m}f(mathbf{u})mathrm{d}mathbf{u}.
end{equation}
For $n=m$, we have
begin{equation}
int_{mathbb{R}^n}f(mathbf{B}mathbf{x})mathrm{d}mathbf{x} = frac{1}{detmathbf{B}} int_{mathbb{R}^n}f(mathbf{u})mathrm{d}mathbf{u}.
end{equation}
What do I do for $nneq m$?
calculus integration multivariable-calculus substitution
edited Jan 28 at 21:40
Yly
6,85921539
asked Jan 28 at 21:06
J. ReinhardJ. Reinhard
184
$endgroup$
add a comment |
$begingroup$
I have an integral
begin{equation}
int_{mathbb{R}^n}f(mathbf{B}mathbf{x})mathrm{d}mathbf{x}
end{equation}
where $f: mathbb{R}^m rightarrow mathbb{R}$ and $mathbf{B}inmathbb{R}^{mtimes n}$. I also know
begin{equation}
int_{mathbb{R}^m}f(mathbf{u})mathrm{d}mathbf{u}.
end{equation}
For $n=m$, we have
begin{equation}
int_{mathbb{R}^n}f(mathbf{B}mathbf{x})mathrm{d}mathbf{x} = frac{1}{detmathbf{B}} int_{mathbb{R}^n}f(mathbf{u})mathrm{d}mathbf{u}.
end{equation}
What do I do for $nneq m$?
calculus integration multivariable-calculus substitution
edited Jan 28 at 21:40
Yly
6,85921539
asked Jan 28 at 21:06
J. ReinhardJ. Reinhard
184
$endgroup$
$begingroup$
You should mention what $X$ and $U$ are
$endgroup$
– 0x539
Jan 28 at 21:20
$begingroup$
Ok, I changed the sets.
$endgroup$
– J. Reinhard
Jan 28 at 21:25
add a comment |
$begingroup$
I have an integral
begin{equation}
int_{mathbb{R}^n}f(mathbf{B}mathbf{x})mathrm{d}mathbf{x}
end{equation}
where $f: mathbb{R}^m rightarrow mathbb{R}$ and $mathbf{B}inmathbb{R}^{mtimes n}$. I also know
begin{equation}
int_{mathbb{R}^m}f(mathbf{u})mathrm{d}mathbf{u}.
end{equation}
For $n=m$, we have
begin{equation}
int_{mathbb{R}^n}f(mathbf{B}mathbf{x})mathrm{d}mathbf{x} = frac{1}{detmathbf{B}} int_{mathbb{R}^n}f(mathbf{u})mathrm{d}mathbf{u}.
end{equation}
What do I do for $nneq m$?
calculus integration multivariable-calculus substitution
edited Jan 28 at 21:40
Yly
6,85921539
asked Jan 28 at 21:06
J. ReinhardJ. Reinhard
184
$endgroup$
I have an integral
begin{equation}
int_{mathbb{R}^n}f(mathbf{B}mathbf{x})mathrm{d}mathbf{x}
end{equation}
where $f: mathbb{R}^m rightarrow mathbb{R}$ and $mathbf{B}inmathbb{R}^{mtimes n}$. I also know
begin{equation}
int_{mathbb{R}^m}f(mathbf{u})mathrm{d}mathbf{u}.
end{equation}
For $n=m$, we have
begin{equation}
int_{mathbb{R}^n}f(mathbf{B}mathbf{x})mathrm{d}mathbf{x} = frac{1}{detmathbf{B}} int_{mathbb{R}^n}f(mathbf{u})mathrm{d}mathbf{u}.
end{equation}
What do I do for $nneq m$?
calculus integration multivariable-calculus substitution
calculus integration multivariable-calculus substitution
edited Jan 28 at 21:40
Yly
6,85921539
asked Jan 28 at 21:06
J. ReinhardJ. Reinhard
184
edited Jan 28 at 21:40
Yly
6,85921539
asked Jan 28 at 21:06
J. ReinhardJ. Reinhard
184
edited Jan 28 at 21:40
Yly
6,85921539
edited Jan 28 at 21:40
Yly
6,85921539
edited Jan 28 at 21:40
Yly
6,85921539
6,85921539
asked Jan 28 at 21:06
J. ReinhardJ. Reinhard
184
asked Jan 28 at 21:06
J. ReinhardJ. Reinhard
184
asked Jan 28 at 21:06
J. ReinhardJ. Reinhard
184
184
$begingroup$
You should mention what $X$ and $U$ are
$endgroup$
– 0x539
Jan 28 at 21:20
$begingroup$
Ok, I changed the sets.
$endgroup$
– J. Reinhard
Jan 28 at 21:25
add a comment |
$begingroup$
You should mention what $X$ and $U$ are
$endgroup$
– 0x539
Jan 28 at 21:20
$begingroup$
Ok, I changed the sets.
$endgroup$
– J. Reinhard
Jan 28 at 21:25
$begingroup$
You should mention what $X$ and $U$ are
$endgroup$
– 0x539
Jan 28 at 21:20
$begingroup$
You should mention what $X$ and $U$ are
$endgroup$
– 0x539
Jan 28 at 21:20
$begingroup$
Ok, I changed the sets.
$endgroup$
– J. Reinhard
Jan 28 at 21:25
$begingroup$
Ok, I changed the sets.
$endgroup$
– J. Reinhard
Jan 28 at 21:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In short, you learn nothing about $I_1 := int_{mathbb{R}^n} f(Bx)dx$ by knowing $I_2 := int_{mathbb{R}^m} f(u)du$ when $nneq m$. This is for geometric reasons:
- If $m>n$, then $I_1$ integrates just "one slice" of the integral $I_2$, which is a set of measure zero in $mathbb{R}^m$ and so can be anything at all. For example, if $n=1$ and $m=2$ and $f$ is the characteristic function of a rectangle $[0,1]times [0,2]$, then $I_2 = 2$, but $I_1$ can be anything from $1$ to $sqrt{5}$ depending on $B$. If we made $f$ the characteristic function of the line $0times [0,1]$, then $I_2 = 0$ and $I_1 = 1$ if $B = left(1 :: 0 right)^T$ but $I_1=0$ if $B$ has any non-zero second entry.
- On the other hand, $m < n$, then the integral $I_1$ generally won't converge at all, because there are infinitely many points taking on every non-zero value of the integrand. For example if $m=1$ and $n=2$ with $f$ being the characteristic function of the interval $[0,1]$, then for a $B$ like $left(1:: 0right)$ we see that $f(Bx)$ is the characteristic function of $[0,1]times[-infty,infty]$, which has infinite area.
There is, however, some relationship between these integrals. In the case where $m>n$, we can consider comparing the integral $I_1$ with the surface integral in $mathbb{R}^m$ over the slice spanned by $B$. Here there is a formula analogous to the one you mention, involving the "Gramian" of $B$. Here it is:
$$ int_{mathbb{R}^n} f(Bx)dx = frac{1}{sqrt{det{B^T B}}} int_{R(B)} f(u)du$$
where $R(B)$ stands for the range of $B$, which is a subspace of $mathbb{R}^m$. I recommend chapter 8 of this book for reference.
answered Jan 28 at 22:06
YlyYly
6,85921539
$endgroup$
$begingroup$
Thanks! That actually makes a lot sense in my application!
$endgroup$
– J. Reinhard
Jan 28 at 22:48
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
In short, you learn nothing about $I_1 := int_{mathbb{R}^n} f(Bx)dx$ by knowing $I_2 := int_{mathbb{R}^m} f(u)du$ when $nneq m$. This is for geometric reasons:
- If $m>n$, then $I_1$ integrates just "one slice" of the integral $I_2$, which is a set of measure zero in $mathbb{R}^m$ and so can be anything at all. For example, if $n=1$ and $m=2$ and $f$ is the characteristic function of a rectangle $[0,1]times [0,2]$, then $I_2 = 2$, but $I_1$ can be anything from $1$ to $sqrt{5}$ depending on $B$. If we made $f$ the characteristic function of the line $0times [0,1]$, then $I_2 = 0$ and $I_1 = 1$ if $B = left(1 :: 0 right)^T$ but $I_1=0$ if $B$ has any non-zero second entry.
- On the other hand, $m < n$, then the integral $I_1$ generally won't converge at all, because there are infinitely many points taking on every non-zero value of the integrand. For example if $m=1$ and $n=2$ with $f$ being the characteristic function of the interval $[0,1]$, then for a $B$ like $left(1:: 0right)$ we see that $f(Bx)$ is the characteristic function of $[0,1]times[-infty,infty]$, which has infinite area.
There is, however, some relationship between these integrals. In the case where $m>n$, we can consider comparing the integral $I_1$ with the surface integral in $mathbb{R}^m$ over the slice spanned by $B$. Here there is a formula analogous to the one you mention, involving the "Gramian" of $B$. Here it is:
$$ int_{mathbb{R}^n} f(Bx)dx = frac{1}{sqrt{det{B^T B}}} int_{R(B)} f(u)du$$
where $R(B)$ stands for the range of $B$, which is a subspace of $mathbb{R}^m$. I recommend chapter 8 of this book for reference.
answered Jan 28 at 22:06
YlyYly
6,85921539
$endgroup$
$begingroup$
Thanks! That actually makes a lot sense in my application!
$endgroup$
– J. Reinhard
Jan 28 at 22:48
add a comment |
$begingroup$
In short, you learn nothing about $I_1 := int_{mathbb{R}^n} f(Bx)dx$ by knowing $I_2 := int_{mathbb{R}^m} f(u)du$ when $nneq m$. This is for geometric reasons:
- If $m>n$, then $I_1$ integrates just "one slice" of the integral $I_2$, which is a set of measure zero in $mathbb{R}^m$ and so can be anything at all. For example, if $n=1$ and $m=2$ and $f$ is the characteristic function of a rectangle $[0,1]times [0,2]$, then $I_2 = 2$, but $I_1$ can be anything from $1$ to $sqrt{5}$ depending on $B$. If we made $f$ the characteristic function of the line $0times [0,1]$, then $I_2 = 0$ and $I_1 = 1$ if $B = left(1 :: 0 right)^T$ but $I_1=0$ if $B$ has any non-zero second entry.
- On the other hand, $m < n$, then the integral $I_1$ generally won't converge at all, because there are infinitely many points taking on every non-zero value of the integrand. For example if $m=1$ and $n=2$ with $f$ being the characteristic function of the interval $[0,1]$, then for a $B$ like $left(1:: 0right)$ we see that $f(Bx)$ is the characteristic function of $[0,1]times[-infty,infty]$, which has infinite area.
There is, however, some relationship between these integrals. In the case where $m>n$, we can consider comparing the integral $I_1$ with the surface integral in $mathbb{R}^m$ over the slice spanned by $B$. Here there is a formula analogous to the one you mention, involving the "Gramian" of $B$. Here it is:
$$ int_{mathbb{R}^n} f(Bx)dx = frac{1}{sqrt{det{B^T B}}} int_{R(B)} f(u)du$$
where $R(B)$ stands for the range of $B$, which is a subspace of $mathbb{R}^m$. I recommend chapter 8 of this book for reference.
answered Jan 28 at 22:06
YlyYly
6,85921539
$endgroup$
$begingroup$
Thanks! That actually makes a lot sense in my application!
$endgroup$
– J. Reinhard
Jan 28 at 22:48
add a comment |
$begingroup$
In short, you learn nothing about $I_1 := int_{mathbb{R}^n} f(Bx)dx$ by knowing $I_2 := int_{mathbb{R}^m} f(u)du$ when $nneq m$. This is for geometric reasons:
- If $m>n$, then $I_1$ integrates just "one slice" of the integral $I_2$, which is a set of measure zero in $mathbb{R}^m$ and so can be anything at all. For example, if $n=1$ and $m=2$ and $f$ is the characteristic function of a rectangle $[0,1]times [0,2]$, then $I_2 = 2$, but $I_1$ can be anything from $1$ to $sqrt{5}$ depending on $B$. If we made $f$ the characteristic function of the line $0times [0,1]$, then $I_2 = 0$ and $I_1 = 1$ if $B = left(1 :: 0 right)^T$ but $I_1=0$ if $B$ has any non-zero second entry.
- On the other hand, $m < n$, then the integral $I_1$ generally won't converge at all, because there are infinitely many points taking on every non-zero value of the integrand. For example if $m=1$ and $n=2$ with $f$ being the characteristic function of the interval $[0,1]$, then for a $B$ like $left(1:: 0right)$ we see that $f(Bx)$ is the characteristic function of $[0,1]times[-infty,infty]$, which has infinite area.
There is, however, some relationship between these integrals. In the case where $m>n$, we can consider comparing the integral $I_1$ with the surface integral in $mathbb{R}^m$ over the slice spanned by $B$. Here there is a formula analogous to the one you mention, involving the "Gramian" of $B$. Here it is:
$$ int_{mathbb{R}^n} f(Bx)dx = frac{1}{sqrt{det{B^T B}}} int_{R(B)} f(u)du$$
where $R(B)$ stands for the range of $B$, which is a subspace of $mathbb{R}^m$. I recommend chapter 8 of this book for reference.
answered Jan 28 at 22:06
YlyYly
6,85921539
$endgroup$
In short, you learn nothing about $I_1 := int_{mathbb{R}^n} f(Bx)dx$ by knowing $I_2 := int_{mathbb{R}^m} f(u)du$ when $nneq m$. This is for geometric reasons:
- If $m>n$, then $I_1$ integrates just "one slice" of the integral $I_2$, which is a set of measure zero in $mathbb{R}^m$ and so can be anything at all. For example, if $n=1$ and $m=2$ and $f$ is the characteristic function of a rectangle $[0,1]times [0,2]$, then $I_2 = 2$, but $I_1$ can be anything from $1$ to $sqrt{5}$ depending on $B$. If we made $f$ the characteristic function of the line $0times [0,1]$, then $I_2 = 0$ and $I_1 = 1$ if $B = left(1 :: 0 right)^T$ but $I_1=0$ if $B$ has any non-zero second entry.
- On the other hand, $m < n$, then the integral $I_1$ generally won't converge at all, because there are infinitely many points taking on every non-zero value of the integrand. For example if $m=1$ and $n=2$ with $f$ being the characteristic function of the interval $[0,1]$, then for a $B$ like $left(1:: 0right)$ we see that $f(Bx)$ is the characteristic function of $[0,1]times[-infty,infty]$, which has infinite area.
There is, however, some relationship between these integrals. In the case where $m>n$, we can consider comparing the integral $I_1$ with the surface integral in $mathbb{R}^m$ over the slice spanned by $B$. Here there is a formula analogous to the one you mention, involving the "Gramian" of $B$. Here it is:
$$ int_{mathbb{R}^n} f(Bx)dx = frac{1}{sqrt{det{B^T B}}} int_{R(B)} f(u)du$$
where $R(B)$ stands for the range of $B$, which is a subspace of $mathbb{R}^m$. I recommend chapter 8 of this book for reference.
answered Jan 28 at 22:06
YlyYly
6,85921539
answered Jan 28 at 22:06
YlyYly
6,85921539
answered Jan 28 at 22:06
YlyYly
6,85921539
answered Jan 28 at 22:06
YlyYly
6,85921539
6,85921539
$begingroup$
Thanks! That actually makes a lot sense in my application!
$endgroup$
– J. Reinhard
Jan 28 at 22:48
add a comment |
$begingroup$
Thanks! That actually makes a lot sense in my application!
$endgroup$
– J. Reinhard
Jan 28 at 22:48
$begingroup$
Thanks! That actually makes a lot sense in my application!
$endgroup$
– J. Reinhard
Jan 28 at 22:48
$begingroup$
Thanks! That actually makes a lot sense in my application!
$endgroup$
– J. Reinhard
Jan 28 at 22:48
add a comment |
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$begingroup$
You should mention what $X$ and $U$ are
$endgroup$
– 0x539
Jan 28 at 21:20
$begingroup$
Ok, I changed the sets.
$endgroup$
– J. Reinhard
Jan 28 at 21:25