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Is $int_gamma f(z) dz = int_a^b f(z(t))z'(t) dt$ just a substitution trick?
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I have read that for a smooth curve $gamma$ in $mathbbC$ parametrized by $z:[a, b] rightarrow mathbbC$ and $f$ a continuous function on $gamma,$ we define the integral of $f$ along the curve as: $int_gamma f(z) dz = int_a^b f(z(t))z'(t) dt.$ Why exactly is this defined in this way? Also is this just a substitution trick? We have $dz = z'(t)dt$ so if we substitute $z$ for $z(t)$ this gives us the formula. Is this a valid substitution?
complex-analysis
asked Aug 1 at 3:12
伽罗瓦
781615
add a comment |Â
up vote
0
down vote
favorite
I have read that for a smooth curve $gamma$ in $mathbbC$ parametrized by $z:[a, b] rightarrow mathbbC$ and $f$ a continuous function on $gamma,$ we define the integral of $f$ along the curve as: $int_gamma f(z) dz = int_a^b f(z(t))z'(t) dt.$ Why exactly is this defined in this way? Also is this just a substitution trick? We have $dz = z'(t)dt$ so if we substitute $z$ for $z(t)$ this gives us the formula. Is this a valid substitution?
complex-analysis
asked Aug 1 at 3:12
伽罗瓦
781615
4
It is a definition.
– Mark Viola
Aug 1 at 3:21
Then wouldn't that be like saying we define 2 + 2 = 1 + 3? It feels kinda weird to me. Why do we define it this way?
– ä¼½ç½—瓦
Aug 1 at 3:22
The definition is meant to mimic the line integrals that one usually learns in multi variable calc.
– Dionel Jaime
Aug 1 at 3:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have read that for a smooth curve $gamma$ in $mathbbC$ parametrized by $z:[a, b] rightarrow mathbbC$ and $f$ a continuous function on $gamma,$ we define the integral of $f$ along the curve as: $int_gamma f(z) dz = int_a^b f(z(t))z'(t) dt.$ Why exactly is this defined in this way? Also is this just a substitution trick? We have $dz = z'(t)dt$ so if we substitute $z$ for $z(t)$ this gives us the formula. Is this a valid substitution?
complex-analysis
asked Aug 1 at 3:12
伽罗瓦
781615
I have read that for a smooth curve $gamma$ in $mathbbC$ parametrized by $z:[a, b] rightarrow mathbbC$ and $f$ a continuous function on $gamma,$ we define the integral of $f$ along the curve as: $int_gamma f(z) dz = int_a^b f(z(t))z'(t) dt.$ Why exactly is this defined in this way? Also is this just a substitution trick? We have $dz = z'(t)dt$ so if we substitute $z$ for $z(t)$ this gives us the formula. Is this a valid substitution?
complex-analysis
asked Aug 1 at 3:12
伽罗瓦
781615
asked Aug 1 at 3:12
伽罗瓦
781615
asked Aug 1 at 3:12
伽罗瓦
781615
asked Aug 1 at 3:12
伽罗瓦
781615
781615
4
It is a definition.
– Mark Viola
Aug 1 at 3:21
Then wouldn't that be like saying we define 2 + 2 = 1 + 3? It feels kinda weird to me. Why do we define it this way?
– ä¼½ç½—瓦
Aug 1 at 3:22
The definition is meant to mimic the line integrals that one usually learns in multi variable calc.
– Dionel Jaime
Aug 1 at 3:24
add a comment |Â
4
It is a definition.
– Mark Viola
Aug 1 at 3:21
Then wouldn't that be like saying we define 2 + 2 = 1 + 3? It feels kinda weird to me. Why do we define it this way?
– ä¼½ç½—瓦
Aug 1 at 3:22
The definition is meant to mimic the line integrals that one usually learns in multi variable calc.
– Dionel Jaime
Aug 1 at 3:24
4
4
It is a definition.
– Mark Viola
Aug 1 at 3:21
It is a definition.
– Mark Viola
Aug 1 at 3:21
Then wouldn't that be like saying we define 2 + 2 = 1 + 3? It feels kinda weird to me. Why do we define it this way?
– ä¼½ç½—瓦
Aug 1 at 3:22
Then wouldn't that be like saying we define 2 + 2 = 1 + 3? It feels kinda weird to me. Why do we define it this way?
– ä¼½ç½—瓦
Aug 1 at 3:22
The definition is meant to mimic the line integrals that one usually learns in multi variable calc.
– Dionel Jaime
Aug 1 at 3:24
The definition is meant to mimic the line integrals that one usually learns in multi variable calc.
– Dionel Jaime
Aug 1 at 3:24
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
No. Many textbooks define it this way, but...
There is a subtle issue here: why is this definition good? Consider two parametrisations $z_1,z_2$ of the same smooth curve $gamma$.
Should $int_gamma f(z) dz$ be $ int_a_1^b_1 f(z_1(t))z_1'(t) dt$ or $ int_a_2^b_2 f(z_2(t))z_2'(t) dt$?
To use this definition, one needs to prove that the definition is independent of the choice of the parametrisation, which is not trivial (but not too complicated either).
This is exactly why more rigorous textbooks usually introduce the integral as the limit of the Riemann sums. With this definition of $int_gamma f(z) dz$ the following Theorem is easy to prove, and implicitly shows the independence of the choice of parametrisation:
Theorem: Let $z: [a,b] to mathbb C$ be a piecewise smooth parametrisation of $gamma$ and $f :mathbb C to mathbb C$ be a continuous function. Then
$$int_gamma f(z) dz = int_a^b f(z(t))z'(t) dt$$
So, to sum it up, there is much more to this "definition" than just a substitution trick.
answered Aug 1 at 3:50
N. S.
97.6k5105197
Is there a good reason why we want $f$ to be continuous? In general, do line integrals require $f$ to be continuous?
– ä¼½ç½—瓦
Aug 1 at 7:34
@伽罗瓦 If $f$ is continuous then it is integrable over $gamma$. If $f$ is not continuous, it is not clear if the integral is defined.
– N. S.
Aug 1 at 14:55
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No. Many textbooks define it this way, but...
There is a subtle issue here: why is this definition good? Consider two parametrisations $z_1,z_2$ of the same smooth curve $gamma$.
Should $int_gamma f(z) dz$ be $ int_a_1^b_1 f(z_1(t))z_1'(t) dt$ or $ int_a_2^b_2 f(z_2(t))z_2'(t) dt$?
To use this definition, one needs to prove that the definition is independent of the choice of the parametrisation, which is not trivial (but not too complicated either).
This is exactly why more rigorous textbooks usually introduce the integral as the limit of the Riemann sums. With this definition of $int_gamma f(z) dz$ the following Theorem is easy to prove, and implicitly shows the independence of the choice of parametrisation:
Theorem: Let $z: [a,b] to mathbb C$ be a piecewise smooth parametrisation of $gamma$ and $f :mathbb C to mathbb C$ be a continuous function. Then
$$int_gamma f(z) dz = int_a^b f(z(t))z'(t) dt$$
So, to sum it up, there is much more to this "definition" than just a substitution trick.
answered Aug 1 at 3:50
N. S.
97.6k5105197
Is there a good reason why we want $f$ to be continuous? In general, do line integrals require $f$ to be continuous?
– ä¼½ç½—瓦
Aug 1 at 7:34
@伽罗瓦 If $f$ is continuous then it is integrable over $gamma$. If $f$ is not continuous, it is not clear if the integral is defined.
– N. S.
Aug 1 at 14:55
add a comment |Â
up vote
2
down vote
accepted
No. Many textbooks define it this way, but...
There is a subtle issue here: why is this definition good? Consider two parametrisations $z_1,z_2$ of the same smooth curve $gamma$.
Should $int_gamma f(z) dz$ be $ int_a_1^b_1 f(z_1(t))z_1'(t) dt$ or $ int_a_2^b_2 f(z_2(t))z_2'(t) dt$?
To use this definition, one needs to prove that the definition is independent of the choice of the parametrisation, which is not trivial (but not too complicated either).
This is exactly why more rigorous textbooks usually introduce the integral as the limit of the Riemann sums. With this definition of $int_gamma f(z) dz$ the following Theorem is easy to prove, and implicitly shows the independence of the choice of parametrisation:
Theorem: Let $z: [a,b] to mathbb C$ be a piecewise smooth parametrisation of $gamma$ and $f :mathbb C to mathbb C$ be a continuous function. Then
$$int_gamma f(z) dz = int_a^b f(z(t))z'(t) dt$$
So, to sum it up, there is much more to this "definition" than just a substitution trick.
answered Aug 1 at 3:50
N. S.
97.6k5105197
Is there a good reason why we want $f$ to be continuous? In general, do line integrals require $f$ to be continuous?
– ä¼½ç½—瓦
Aug 1 at 7:34
@伽罗瓦 If $f$ is continuous then it is integrable over $gamma$. If $f$ is not continuous, it is not clear if the integral is defined.
– N. S.
Aug 1 at 14:55
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No. Many textbooks define it this way, but...
There is a subtle issue here: why is this definition good? Consider two parametrisations $z_1,z_2$ of the same smooth curve $gamma$.
Should $int_gamma f(z) dz$ be $ int_a_1^b_1 f(z_1(t))z_1'(t) dt$ or $ int_a_2^b_2 f(z_2(t))z_2'(t) dt$?
To use this definition, one needs to prove that the definition is independent of the choice of the parametrisation, which is not trivial (but not too complicated either).
This is exactly why more rigorous textbooks usually introduce the integral as the limit of the Riemann sums. With this definition of $int_gamma f(z) dz$ the following Theorem is easy to prove, and implicitly shows the independence of the choice of parametrisation:
Theorem: Let $z: [a,b] to mathbb C$ be a piecewise smooth parametrisation of $gamma$ and $f :mathbb C to mathbb C$ be a continuous function. Then
$$int_gamma f(z) dz = int_a^b f(z(t))z'(t) dt$$
So, to sum it up, there is much more to this "definition" than just a substitution trick.
answered Aug 1 at 3:50
N. S.
97.6k5105197
No. Many textbooks define it this way, but...
There is a subtle issue here: why is this definition good? Consider two parametrisations $z_1,z_2$ of the same smooth curve $gamma$.
Should $int_gamma f(z) dz$ be $ int_a_1^b_1 f(z_1(t))z_1'(t) dt$ or $ int_a_2^b_2 f(z_2(t))z_2'(t) dt$?
To use this definition, one needs to prove that the definition is independent of the choice of the parametrisation, which is not trivial (but not too complicated either).
This is exactly why more rigorous textbooks usually introduce the integral as the limit of the Riemann sums. With this definition of $int_gamma f(z) dz$ the following Theorem is easy to prove, and implicitly shows the independence of the choice of parametrisation:
Theorem: Let $z: [a,b] to mathbb C$ be a piecewise smooth parametrisation of $gamma$ and $f :mathbb C to mathbb C$ be a continuous function. Then
$$int_gamma f(z) dz = int_a^b f(z(t))z'(t) dt$$
So, to sum it up, there is much more to this "definition" than just a substitution trick.
answered Aug 1 at 3:50
N. S.
97.6k5105197
answered Aug 1 at 3:50
N. S.
97.6k5105197
answered Aug 1 at 3:50
N. S.
97.6k5105197
answered Aug 1 at 3:50
N. S.
97.6k5105197
97.6k5105197
Is there a good reason why we want $f$ to be continuous? In general, do line integrals require $f$ to be continuous?
– ä¼½ç½—瓦
Aug 1 at 7:34
@伽罗瓦 If $f$ is continuous then it is integrable over $gamma$. If $f$ is not continuous, it is not clear if the integral is defined.
– N. S.
Aug 1 at 14:55
add a comment |Â
Is there a good reason why we want $f$ to be continuous? In general, do line integrals require $f$ to be continuous?
– ä¼½ç½—瓦
Aug 1 at 7:34
@伽罗瓦 If $f$ is continuous then it is integrable over $gamma$. If $f$ is not continuous, it is not clear if the integral is defined.
– N. S.
Aug 1 at 14:55
Is there a good reason why we want $f$ to be continuous? In general, do line integrals require $f$ to be continuous?
– ä¼½ç½—瓦
Aug 1 at 7:34
Is there a good reason why we want $f$ to be continuous? In general, do line integrals require $f$ to be continuous?
– ä¼½ç½—瓦
Aug 1 at 7:34
@伽罗瓦 If $f$ is continuous then it is integrable over $gamma$. If $f$ is not continuous, it is not clear if the integral is defined.
– N. S.
Aug 1 at 14:55
@伽罗瓦 If $f$ is continuous then it is integrable over $gamma$. If $f$ is not continuous, it is not clear if the integral is defined.
– N. S.
Aug 1 at 14:55
add a comment |Â
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4
It is a definition.
– Mark Viola
Aug 1 at 3:21
Then wouldn't that be like saying we define 2 + 2 = 1 + 3? It feels kinda weird to me. Why do we define it this way?
– ä¼½ç½—瓦
Aug 1 at 3:22
The definition is meant to mimic the line integrals that one usually learns in multi variable calc.
– Dionel Jaime
Aug 1 at 3:24