Transformation Jacobian for integration by substitution in single and multiple variables?
$begingroup$
I wonder how to reconcile the transformation rules for integration by substitution in a single variable vs several variables.
In the single-variable case
the Jacobian factor $varphi'(x)$ is just a derivative and may take on negative as well as positive values depending on the particular function and integration range.
However, in the multi-variable case
the Jacobian determinant factor Det$(Dvarphi)(u)$ is taken in absolute value, which allows this transformation Jacobian to take on positive values only.
If we take the multi-variable expression and consider the single-variable special case, then the partial derivative determinant again reduces to simply $varphi'(x)$. However, the additional absolute value of the determinant seems to introduce a discrepancy between the purely single-variable treatment and this special case.
Moreover, it is easy to find examples showing that the absolute value indeed should not be there in the single-variable case. How to reconcile the two? Is the absolute value truly needed and correct in the multi-variable case?
EDIT:
Since some doubt was expressed in the comments on whether $varphi'(x)$ may or may not be negative in the single-variable case, consider the following example:
$$int_0^1 du,u^2=frac{1}{3}$$
Now take $uequivvarphi(x)=-x$. If we consider the absolute value of the Jacobian to be true, we get:
$$int_{0}^{-1}left|frac{dvarphi(x)}{dx}right|dx,(-x)^2=int_0^{-1}dx,x^2=-frac{1}{3}$$
which gives a wrong result; while if we do not take the absolute value, we recover:
$$int_{0}^{-1} frac{dvarphi(x)}{dx}dx,(-x)^2=-int_0^{-1}dx,x^2=frac{1}{3}$$
which is correct.
calculus integration multivariable-calculus definite-integrals substitution
asked Jan 10 at 15:42
KagaratschKagaratsch
1,044617
$endgroup$
add a comment |
$begingroup$
I wonder how to reconcile the transformation rules for integration by substitution in a single variable vs several variables.
In the single-variable case
the Jacobian factor $varphi'(x)$ is just a derivative and may take on negative as well as positive values depending on the particular function and integration range.
However, in the multi-variable case
the Jacobian determinant factor Det$(Dvarphi)(u)$ is taken in absolute value, which allows this transformation Jacobian to take on positive values only.
If we take the multi-variable expression and consider the single-variable special case, then the partial derivative determinant again reduces to simply $varphi'(x)$. However, the additional absolute value of the determinant seems to introduce a discrepancy between the purely single-variable treatment and this special case.
Moreover, it is easy to find examples showing that the absolute value indeed should not be there in the single-variable case. How to reconcile the two? Is the absolute value truly needed and correct in the multi-variable case?
EDIT:
Since some doubt was expressed in the comments on whether $varphi'(x)$ may or may not be negative in the single-variable case, consider the following example:
$$int_0^1 du,u^2=frac{1}{3}$$
Now take $uequivvarphi(x)=-x$. If we consider the absolute value of the Jacobian to be true, we get:
$$int_{0}^{-1}left|frac{dvarphi(x)}{dx}right|dx,(-x)^2=int_0^{-1}dx,x^2=-frac{1}{3}$$
which gives a wrong result; while if we do not take the absolute value, we recover:
$$int_{0}^{-1} frac{dvarphi(x)}{dx}dx,(-x)^2=-int_0^{-1}dx,x^2=frac{1}{3}$$
which is correct.
calculus integration multivariable-calculus definite-integrals substitution
asked Jan 10 at 15:42
KagaratschKagaratsch
1,044617
$endgroup$
$begingroup$
I think that for one variable case, $phi '(x)$ can't be negative. You can check the statement of the theorem again, it needs that $phi$ is strictly increasing on [a,b].
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:10
$begingroup$
@扁頭科學麵KevinLee please see the edit in the question. Do you agree with this example?
$endgroup$
– Kagaratsch
Jan 10 at 16:21
$begingroup$
@扁頭科學麵KevinLee I see, so the absolute value is written only for substitutions that satisfy positivity constraints? Then for more general substitutions dropping the absolute value gives the correct generalization? Is that the case in the multi-variable case as well?
$endgroup$
– Kagaratsch
Jan 10 at 16:35
$begingroup$
For the example you gave: Let $g(x)=x^2$, $int_0^1 x^2 dx=frac{1}{3}$, if you want to use the Theorem, you can replace $x$ by a parametrized function, for instance, $x=phi (y)=y^2$, $y$ from 0 to 1, you can check that $y^2$ in the range [0,1] is stricktly increasing.Replace $x$ by $phi (y)$ in the following formula $int_0^1 x^2 dx= int_0^1 (y^2)^2 frac{dy^2}{dy} dy=int_0^1 2y^5 dy =frac{1}{3}$. The key point is that when you choose the parametrized curve, it needs the the derivative is positive. As to the multi variable case, we have to take determinant and then the absolte value.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:51
$begingroup$
Ya, you are right, it need to satisfy positivety constraints. If you drop the absolute sign, then we can't insure that the theorem still be right, maybe the example you give can explain why if we don't add the abs sign then the answer will be wrong.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:57
add a comment |
$begingroup$
I wonder how to reconcile the transformation rules for integration by substitution in a single variable vs several variables.
In the single-variable case
the Jacobian factor $varphi'(x)$ is just a derivative and may take on negative as well as positive values depending on the particular function and integration range.
However, in the multi-variable case
the Jacobian determinant factor Det$(Dvarphi)(u)$ is taken in absolute value, which allows this transformation Jacobian to take on positive values only.
If we take the multi-variable expression and consider the single-variable special case, then the partial derivative determinant again reduces to simply $varphi'(x)$. However, the additional absolute value of the determinant seems to introduce a discrepancy between the purely single-variable treatment and this special case.
Moreover, it is easy to find examples showing that the absolute value indeed should not be there in the single-variable case. How to reconcile the two? Is the absolute value truly needed and correct in the multi-variable case?
EDIT:
Since some doubt was expressed in the comments on whether $varphi'(x)$ may or may not be negative in the single-variable case, consider the following example:
$$int_0^1 du,u^2=frac{1}{3}$$
Now take $uequivvarphi(x)=-x$. If we consider the absolute value of the Jacobian to be true, we get:
$$int_{0}^{-1}left|frac{dvarphi(x)}{dx}right|dx,(-x)^2=int_0^{-1}dx,x^2=-frac{1}{3}$$
which gives a wrong result; while if we do not take the absolute value, we recover:
$$int_{0}^{-1} frac{dvarphi(x)}{dx}dx,(-x)^2=-int_0^{-1}dx,x^2=frac{1}{3}$$
which is correct.
calculus integration multivariable-calculus definite-integrals substitution
asked Jan 10 at 15:42
KagaratschKagaratsch
1,044617
$endgroup$
I wonder how to reconcile the transformation rules for integration by substitution in a single variable vs several variables.
In the single-variable case
the Jacobian factor $varphi'(x)$ is just a derivative and may take on negative as well as positive values depending on the particular function and integration range.
However, in the multi-variable case
the Jacobian determinant factor Det$(Dvarphi)(u)$ is taken in absolute value, which allows this transformation Jacobian to take on positive values only.
If we take the multi-variable expression and consider the single-variable special case, then the partial derivative determinant again reduces to simply $varphi'(x)$. However, the additional absolute value of the determinant seems to introduce a discrepancy between the purely single-variable treatment and this special case.
Moreover, it is easy to find examples showing that the absolute value indeed should not be there in the single-variable case. How to reconcile the two? Is the absolute value truly needed and correct in the multi-variable case?
EDIT:
Since some doubt was expressed in the comments on whether $varphi'(x)$ may or may not be negative in the single-variable case, consider the following example:
$$int_0^1 du,u^2=frac{1}{3}$$
Now take $uequivvarphi(x)=-x$. If we consider the absolute value of the Jacobian to be true, we get:
$$int_{0}^{-1}left|frac{dvarphi(x)}{dx}right|dx,(-x)^2=int_0^{-1}dx,x^2=-frac{1}{3}$$
which gives a wrong result; while if we do not take the absolute value, we recover:
$$int_{0}^{-1} frac{dvarphi(x)}{dx}dx,(-x)^2=-int_0^{-1}dx,x^2=frac{1}{3}$$
which is correct.
calculus integration multivariable-calculus definite-integrals substitution
calculus integration multivariable-calculus definite-integrals substitution
asked Jan 10 at 15:42
KagaratschKagaratsch
1,044617
asked Jan 10 at 15:42
KagaratschKagaratsch
1,044617
edited Jan 10 at 16:32
Kagaratsch
asked Jan 10 at 15:42
KagaratschKagaratsch
1,044617
asked Jan 10 at 15:42
KagaratschKagaratsch
1,044617
asked Jan 10 at 15:42
KagaratschKagaratsch
1,044617
1,044617
$begingroup$
I think that for one variable case, $phi '(x)$ can't be negative. You can check the statement of the theorem again, it needs that $phi$ is strictly increasing on [a,b].
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:10
$begingroup$
@扁頭科學麵KevinLee please see the edit in the question. Do you agree with this example?
$endgroup$
– Kagaratsch
Jan 10 at 16:21
$begingroup$
@扁頭科學麵KevinLee I see, so the absolute value is written only for substitutions that satisfy positivity constraints? Then for more general substitutions dropping the absolute value gives the correct generalization? Is that the case in the multi-variable case as well?
$endgroup$
– Kagaratsch
Jan 10 at 16:35
$begingroup$
For the example you gave: Let $g(x)=x^2$, $int_0^1 x^2 dx=frac{1}{3}$, if you want to use the Theorem, you can replace $x$ by a parametrized function, for instance, $x=phi (y)=y^2$, $y$ from 0 to 1, you can check that $y^2$ in the range [0,1] is stricktly increasing.Replace $x$ by $phi (y)$ in the following formula $int_0^1 x^2 dx= int_0^1 (y^2)^2 frac{dy^2}{dy} dy=int_0^1 2y^5 dy =frac{1}{3}$. The key point is that when you choose the parametrized curve, it needs the the derivative is positive. As to the multi variable case, we have to take determinant and then the absolte value.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:51
$begingroup$
Ya, you are right, it need to satisfy positivety constraints. If you drop the absolute sign, then we can't insure that the theorem still be right, maybe the example you give can explain why if we don't add the abs sign then the answer will be wrong.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:57
add a comment |
$begingroup$
I think that for one variable case, $phi '(x)$ can't be negative. You can check the statement of the theorem again, it needs that $phi$ is strictly increasing on [a,b].
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:10
$begingroup$
@扁頭科學麵KevinLee please see the edit in the question. Do you agree with this example?
$endgroup$
– Kagaratsch
Jan 10 at 16:21
$begingroup$
@扁頭科學麵KevinLee I see, so the absolute value is written only for substitutions that satisfy positivity constraints? Then for more general substitutions dropping the absolute value gives the correct generalization? Is that the case in the multi-variable case as well?
$endgroup$
– Kagaratsch
Jan 10 at 16:35
$begingroup$
For the example you gave: Let $g(x)=x^2$, $int_0^1 x^2 dx=frac{1}{3}$, if you want to use the Theorem, you can replace $x$ by a parametrized function, for instance, $x=phi (y)=y^2$, $y$ from 0 to 1, you can check that $y^2$ in the range [0,1] is stricktly increasing.Replace $x$ by $phi (y)$ in the following formula $int_0^1 x^2 dx= int_0^1 (y^2)^2 frac{dy^2}{dy} dy=int_0^1 2y^5 dy =frac{1}{3}$. The key point is that when you choose the parametrized curve, it needs the the derivative is positive. As to the multi variable case, we have to take determinant and then the absolte value.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:51
$begingroup$
Ya, you are right, it need to satisfy positivety constraints. If you drop the absolute sign, then we can't insure that the theorem still be right, maybe the example you give can explain why if we don't add the abs sign then the answer will be wrong.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:57
$begingroup$
I think that for one variable case, $phi '(x)$ can't be negative. You can check the statement of the theorem again, it needs that $phi$ is strictly increasing on [a,b].
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:10
$begingroup$
I think that for one variable case, $phi '(x)$ can't be negative. You can check the statement of the theorem again, it needs that $phi$ is strictly increasing on [a,b].
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:10
$begingroup$
@扁頭科學麵KevinLee please see the edit in the question. Do you agree with this example?
$endgroup$
– Kagaratsch
Jan 10 at 16:21
$begingroup$
@扁頭科學麵KevinLee please see the edit in the question. Do you agree with this example?
$endgroup$
– Kagaratsch
Jan 10 at 16:21
$begingroup$
@扁頭科學麵KevinLee I see, so the absolute value is written only for substitutions that satisfy positivity constraints? Then for more general substitutions dropping the absolute value gives the correct generalization? Is that the case in the multi-variable case as well?
$endgroup$
– Kagaratsch
Jan 10 at 16:35
$begingroup$
@扁頭科學麵KevinLee I see, so the absolute value is written only for substitutions that satisfy positivity constraints? Then for more general substitutions dropping the absolute value gives the correct generalization? Is that the case in the multi-variable case as well?
$endgroup$
– Kagaratsch
Jan 10 at 16:35
$begingroup$
For the example you gave: Let $g(x)=x^2$, $int_0^1 x^2 dx=frac{1}{3}$, if you want to use the Theorem, you can replace $x$ by a parametrized function, for instance, $x=phi (y)=y^2$, $y$ from 0 to 1, you can check that $y^2$ in the range [0,1] is stricktly increasing.Replace $x$ by $phi (y)$ in the following formula $int_0^1 x^2 dx= int_0^1 (y^2)^2 frac{dy^2}{dy} dy=int_0^1 2y^5 dy =frac{1}{3}$. The key point is that when you choose the parametrized curve, it needs the the derivative is positive. As to the multi variable case, we have to take determinant and then the absolte value.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:51
$begingroup$
For the example you gave: Let $g(x)=x^2$, $int_0^1 x^2 dx=frac{1}{3}$, if you want to use the Theorem, you can replace $x$ by a parametrized function, for instance, $x=phi (y)=y^2$, $y$ from 0 to 1, you can check that $y^2$ in the range [0,1] is stricktly increasing.Replace $x$ by $phi (y)$ in the following formula $int_0^1 x^2 dx= int_0^1 (y^2)^2 frac{dy^2}{dy} dy=int_0^1 2y^5 dy =frac{1}{3}$. The key point is that when you choose the parametrized curve, it needs the the derivative is positive. As to the multi variable case, we have to take determinant and then the absolte value.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:51
$begingroup$
Ya, you are right, it need to satisfy positivety constraints. If you drop the absolute sign, then we can't insure that the theorem still be right, maybe the example you give can explain why if we don't add the abs sign then the answer will be wrong.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:57
$begingroup$
Ya, you are right, it need to satisfy positivety constraints. If you drop the absolute sign, then we can't insure that the theorem still be right, maybe the example you give can explain why if we don't add the abs sign then the answer will be wrong.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:57
add a comment |
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$begingroup$
I think that for one variable case, $phi '(x)$ can't be negative. You can check the statement of the theorem again, it needs that $phi$ is strictly increasing on [a,b].
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:10
$begingroup$
@扁頭科學麵KevinLee please see the edit in the question. Do you agree with this example?
$endgroup$
– Kagaratsch
Jan 10 at 16:21
$begingroup$
@扁頭科學麵KevinLee I see, so the absolute value is written only for substitutions that satisfy positivity constraints? Then for more general substitutions dropping the absolute value gives the correct generalization? Is that the case in the multi-variable case as well?
$endgroup$
– Kagaratsch
Jan 10 at 16:35
$begingroup$
For the example you gave: Let $g(x)=x^2$, $int_0^1 x^2 dx=frac{1}{3}$, if you want to use the Theorem, you can replace $x$ by a parametrized function, for instance, $x=phi (y)=y^2$, $y$ from 0 to 1, you can check that $y^2$ in the range [0,1] is stricktly increasing.Replace $x$ by $phi (y)$ in the following formula $int_0^1 x^2 dx= int_0^1 (y^2)^2 frac{dy^2}{dy} dy=int_0^1 2y^5 dy =frac{1}{3}$. The key point is that when you choose the parametrized curve, it needs the the derivative is positive. As to the multi variable case, we have to take determinant and then the absolte value.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:51
$begingroup$
Ya, you are right, it need to satisfy positivety constraints. If you drop the absolute sign, then we can't insure that the theorem still be right, maybe the example you give can explain why if we don't add the abs sign then the answer will be wrong.
$endgroup$
– 扁頭科學麵Kevin Lee
Jan 10 at 16:57