Mixing
$int_0^1 rm dx , (-1)^x$ [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
-3
down vote
favorite
I guess everyone has thought about this integral and I just want to discuss what to think of the implications. So,
$$int_0^1 rm dx , (-1)^x = int_0^1 rm dx , left(e^ipi(2n+1)right)^x$$
where $nin mathbb Z$. Now I guess the next step implies deciding for a specific Riemann-Sheet, thus
$$
int_0^1 rm dx , e^ipi(2n+1)x = frac2ipi(2n+1) , .
$$
So the result of the integral is multi-valued, depending on which specific Riemann-Sheet I look at.
But still one is tempted to ask, what is the intuitively correct result? The one for the principal branch, or is each result equally correct?
It is also hard to incorporate this into the picture of an integral measuring some area underneath the curve.
So does anyone have some insights or has thought about it more thoroughly?
EDIT2: Apparently it is "unclear" how to interpret a complex integral as the area under a curve. Use some imagination!
Here is an example:
How would you calculate the area under a curve in a 2d case? I guess something like
$$
int_gamma f(x,y) , rm ds
$$
would qualify where $rm ds = |dotgamma(t)| rm dt$ is the arc-length of the curve, or? Consider $f(x,y)=x$ along the curve $gamma:[0,1] rightarrow mathbb R^2, t mapsto (t,t)$. Then
$$
int_gamma x , rm ds = sqrt2 int_0^1 t , rm dt = frac1sqrt2 , .
$$
What would one have guessed??? The slope in $x$-direction is $1$ and in $y$-direction $0$. Therefore the directional derivative along the bisector gives $(1,0)cdot (1/sqrt2,1/sqrt2)^t = frac1sqrt2$ and $f(s)$ is expected to be $fracssqrt2$.
In fact the arc-length is $s=sqrt2,t$, so likewise $f(t,t)=t=fracssqrt2=f(s)$ is a 1d function with a proper interpretation of area under it
$$
int_0^sqrt2 f(s) , rm ds = frac1sqrt2 int_0^sqrt2 s , rm ds = frac1sqrt2 , .
$$
Now let's go to complex analysis. It's basically the same thing.
Let $f(z)=z=x+iy$ along the curve $gamma:[0,1] rightarrow mathbb C, t mapsto t+it$. The integrals over $rm Re(f)$ and $rm Im(f)$ can be considered in the same way the example above illustrated. So lets start with $rm Re(f)$
$$
int_gamma rm Re(f) , rm dz = int_0^1 t , (1+i) , rm dt = frac1+i2
$$
whose absolute value is precisely $frac1sqrt2$. Due to symmetry the result for $rm Im(f)$ is the same, so adding together we have
$$
frac1+i2 + i , frac1+i2 = i
$$
with absolute value $1$.
As mentioned hitherto the area of a function is interpreted as in the example from $mathbb R^2$ above.
$rm Re(f)$ and $rm Im(f)$ are thus giving areas each which have to be added vectorially.
In fact this is just what happens for 2 functions $f(x,y)=x$ and $g(x,y)=y$ in $mathbb R^2$ whose areas were determined by $frac1sqrt2$. Adding vectorially leads to
$$
sqrtfrac12 + frac12 = 1
$$
which is exactly the result from the complex function.
complex-analysis definite-integrals exponential-function
edited Jul 25 at 0:41
quid♦
36.2k85090
asked Jul 24 at 11:42
Diger
52029
closed as primarily opinion-based by Did, amWhy, Leucippus, Parcly Taxel, Claude Leibovici Jul 25 at 5:55
Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |Â
up vote
-3
down vote
favorite
I guess everyone has thought about this integral and I just want to discuss what to think of the implications. So,
$$int_0^1 rm dx , (-1)^x = int_0^1 rm dx , left(e^ipi(2n+1)right)^x$$
where $nin mathbb Z$. Now I guess the next step implies deciding for a specific Riemann-Sheet, thus
$$
int_0^1 rm dx , e^ipi(2n+1)x = frac2ipi(2n+1) , .
$$
So the result of the integral is multi-valued, depending on which specific Riemann-Sheet I look at.
But still one is tempted to ask, what is the intuitively correct result? The one for the principal branch, or is each result equally correct?
It is also hard to incorporate this into the picture of an integral measuring some area underneath the curve.
So does anyone have some insights or has thought about it more thoroughly?
EDIT2: Apparently it is "unclear" how to interpret a complex integral as the area under a curve. Use some imagination!
Here is an example:
How would you calculate the area under a curve in a 2d case? I guess something like
$$
int_gamma f(x,y) , rm ds
$$
would qualify where $rm ds = |dotgamma(t)| rm dt$ is the arc-length of the curve, or? Consider $f(x,y)=x$ along the curve $gamma:[0,1] rightarrow mathbb R^2, t mapsto (t,t)$. Then
$$
int_gamma x , rm ds = sqrt2 int_0^1 t , rm dt = frac1sqrt2 , .
$$
What would one have guessed??? The slope in $x$-direction is $1$ and in $y$-direction $0$. Therefore the directional derivative along the bisector gives $(1,0)cdot (1/sqrt2,1/sqrt2)^t = frac1sqrt2$ and $f(s)$ is expected to be $fracssqrt2$.
In fact the arc-length is $s=sqrt2,t$, so likewise $f(t,t)=t=fracssqrt2=f(s)$ is a 1d function with a proper interpretation of area under it
$$
int_0^sqrt2 f(s) , rm ds = frac1sqrt2 int_0^sqrt2 s , rm ds = frac1sqrt2 , .
$$
Now let's go to complex analysis. It's basically the same thing.
Let $f(z)=z=x+iy$ along the curve $gamma:[0,1] rightarrow mathbb C, t mapsto t+it$. The integrals over $rm Re(f)$ and $rm Im(f)$ can be considered in the same way the example above illustrated. So lets start with $rm Re(f)$
$$
int_gamma rm Re(f) , rm dz = int_0^1 t , (1+i) , rm dt = frac1+i2
$$
whose absolute value is precisely $frac1sqrt2$. Due to symmetry the result for $rm Im(f)$ is the same, so adding together we have
$$
frac1+i2 + i , frac1+i2 = i
$$
with absolute value $1$.
As mentioned hitherto the area of a function is interpreted as in the example from $mathbb R^2$ above.
$rm Re(f)$ and $rm Im(f)$ are thus giving areas each which have to be added vectorially.
In fact this is just what happens for 2 functions $f(x,y)=x$ and $g(x,y)=y$ in $mathbb R^2$ whose areas were determined by $frac1sqrt2$. Adding vectorially leads to
$$
sqrtfrac12 + frac12 = 1
$$
which is exactly the result from the complex function.
complex-analysis definite-integrals exponential-function
edited Jul 25 at 0:41
quid♦
36.2k85090
asked Jul 24 at 11:42
Diger
52029
closed as primarily opinion-based by Did, amWhy, Leucippus, Parcly Taxel, Claude Leibovici Jul 25 at 5:55
Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.
2
Relevant: youtu.be/pYJQZDMayeI
– aidangallagher4
Jul 24 at 11:47
It's unclear how "area under a curve" applies to any complex integral (except when it's actually just a real integral after all).
– David K
Jul 24 at 20:44
Clearer now????
– Diger
Jul 24 at 23:12
add a comment |Â
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
I guess everyone has thought about this integral and I just want to discuss what to think of the implications. So,
$$int_0^1 rm dx , (-1)^x = int_0^1 rm dx , left(e^ipi(2n+1)right)^x$$
where $nin mathbb Z$. Now I guess the next step implies deciding for a specific Riemann-Sheet, thus
$$
int_0^1 rm dx , e^ipi(2n+1)x = frac2ipi(2n+1) , .
$$
So the result of the integral is multi-valued, depending on which specific Riemann-Sheet I look at.
But still one is tempted to ask, what is the intuitively correct result? The one for the principal branch, or is each result equally correct?
It is also hard to incorporate this into the picture of an integral measuring some area underneath the curve.
So does anyone have some insights or has thought about it more thoroughly?
EDIT2: Apparently it is "unclear" how to interpret a complex integral as the area under a curve. Use some imagination!
Here is an example:
How would you calculate the area under a curve in a 2d case? I guess something like
$$
int_gamma f(x,y) , rm ds
$$
would qualify where $rm ds = |dotgamma(t)| rm dt$ is the arc-length of the curve, or? Consider $f(x,y)=x$ along the curve $gamma:[0,1] rightarrow mathbb R^2, t mapsto (t,t)$. Then
$$
int_gamma x , rm ds = sqrt2 int_0^1 t , rm dt = frac1sqrt2 , .
$$
What would one have guessed??? The slope in $x$-direction is $1$ and in $y$-direction $0$. Therefore the directional derivative along the bisector gives $(1,0)cdot (1/sqrt2,1/sqrt2)^t = frac1sqrt2$ and $f(s)$ is expected to be $fracssqrt2$.
In fact the arc-length is $s=sqrt2,t$, so likewise $f(t,t)=t=fracssqrt2=f(s)$ is a 1d function with a proper interpretation of area under it
$$
int_0^sqrt2 f(s) , rm ds = frac1sqrt2 int_0^sqrt2 s , rm ds = frac1sqrt2 , .
$$
Now let's go to complex analysis. It's basically the same thing.
Let $f(z)=z=x+iy$ along the curve $gamma:[0,1] rightarrow mathbb C, t mapsto t+it$. The integrals over $rm Re(f)$ and $rm Im(f)$ can be considered in the same way the example above illustrated. So lets start with $rm Re(f)$
$$
int_gamma rm Re(f) , rm dz = int_0^1 t , (1+i) , rm dt = frac1+i2
$$
whose absolute value is precisely $frac1sqrt2$. Due to symmetry the result for $rm Im(f)$ is the same, so adding together we have
$$
frac1+i2 + i , frac1+i2 = i
$$
with absolute value $1$.
As mentioned hitherto the area of a function is interpreted as in the example from $mathbb R^2$ above.
$rm Re(f)$ and $rm Im(f)$ are thus giving areas each which have to be added vectorially.
In fact this is just what happens for 2 functions $f(x,y)=x$ and $g(x,y)=y$ in $mathbb R^2$ whose areas were determined by $frac1sqrt2$. Adding vectorially leads to
$$
sqrtfrac12 + frac12 = 1
$$
which is exactly the result from the complex function.
complex-analysis definite-integrals exponential-function
edited Jul 25 at 0:41
quid♦
36.2k85090
asked Jul 24 at 11:42
Diger
52029
I guess everyone has thought about this integral and I just want to discuss what to think of the implications. So,
$$int_0^1 rm dx , (-1)^x = int_0^1 rm dx , left(e^ipi(2n+1)right)^x$$
where $nin mathbb Z$. Now I guess the next step implies deciding for a specific Riemann-Sheet, thus
$$
int_0^1 rm dx , e^ipi(2n+1)x = frac2ipi(2n+1) , .
$$
So the result of the integral is multi-valued, depending on which specific Riemann-Sheet I look at.
But still one is tempted to ask, what is the intuitively correct result? The one for the principal branch, or is each result equally correct?
It is also hard to incorporate this into the picture of an integral measuring some area underneath the curve.
So does anyone have some insights or has thought about it more thoroughly?
EDIT2: Apparently it is "unclear" how to interpret a complex integral as the area under a curve. Use some imagination!
Here is an example:
How would you calculate the area under a curve in a 2d case? I guess something like
$$
int_gamma f(x,y) , rm ds
$$
would qualify where $rm ds = |dotgamma(t)| rm dt$ is the arc-length of the curve, or? Consider $f(x,y)=x$ along the curve $gamma:[0,1] rightarrow mathbb R^2, t mapsto (t,t)$. Then
$$
int_gamma x , rm ds = sqrt2 int_0^1 t , rm dt = frac1sqrt2 , .
$$
What would one have guessed??? The slope in $x$-direction is $1$ and in $y$-direction $0$. Therefore the directional derivative along the bisector gives $(1,0)cdot (1/sqrt2,1/sqrt2)^t = frac1sqrt2$ and $f(s)$ is expected to be $fracssqrt2$.
In fact the arc-length is $s=sqrt2,t$, so likewise $f(t,t)=t=fracssqrt2=f(s)$ is a 1d function with a proper interpretation of area under it
$$
int_0^sqrt2 f(s) , rm ds = frac1sqrt2 int_0^sqrt2 s , rm ds = frac1sqrt2 , .
$$
Now let's go to complex analysis. It's basically the same thing.
Let $f(z)=z=x+iy$ along the curve $gamma:[0,1] rightarrow mathbb C, t mapsto t+it$. The integrals over $rm Re(f)$ and $rm Im(f)$ can be considered in the same way the example above illustrated. So lets start with $rm Re(f)$
$$
int_gamma rm Re(f) , rm dz = int_0^1 t , (1+i) , rm dt = frac1+i2
$$
whose absolute value is precisely $frac1sqrt2$. Due to symmetry the result for $rm Im(f)$ is the same, so adding together we have
$$
frac1+i2 + i , frac1+i2 = i
$$
with absolute value $1$.
As mentioned hitherto the area of a function is interpreted as in the example from $mathbb R^2$ above.
$rm Re(f)$ and $rm Im(f)$ are thus giving areas each which have to be added vectorially.
In fact this is just what happens for 2 functions $f(x,y)=x$ and $g(x,y)=y$ in $mathbb R^2$ whose areas were determined by $frac1sqrt2$. Adding vectorially leads to
$$
sqrtfrac12 + frac12 = 1
$$
which is exactly the result from the complex function.
complex-analysis definite-integrals exponential-function
edited Jul 25 at 0:41
quid♦
36.2k85090
asked Jul 24 at 11:42
Diger
52029
edited Jul 25 at 0:41
quid♦
36.2k85090
edited Jul 25 at 0:41
quid♦
36.2k85090
edited Jul 25 at 0:41
quid♦
36.2k85090
36.2k85090
asked Jul 24 at 11:42
Diger
52029
asked Jul 24 at 11:42
Diger
52029
asked Jul 24 at 11:42
Diger
52029
52029
closed as primarily opinion-based by Did, amWhy, Leucippus, Parcly Taxel, Claude Leibovici Jul 25 at 5:55
Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as primarily opinion-based by Did, amWhy, Leucippus, Parcly Taxel, Claude Leibovici Jul 25 at 5:55
Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.
2
Relevant: youtu.be/pYJQZDMayeI
– aidangallagher4
Jul 24 at 11:47
It's unclear how "area under a curve" applies to any complex integral (except when it's actually just a real integral after all).
– David K
Jul 24 at 20:44
Clearer now????
– Diger
Jul 24 at 23:12
add a comment |Â
2
Relevant: youtu.be/pYJQZDMayeI
– aidangallagher4
Jul 24 at 11:47
It's unclear how "area under a curve" applies to any complex integral (except when it's actually just a real integral after all).
– David K
Jul 24 at 20:44
Clearer now????
– Diger
Jul 24 at 23:12
2
2
Relevant: youtu.be/pYJQZDMayeI
– aidangallagher4
Jul 24 at 11:47
Relevant: youtu.be/pYJQZDMayeI
– aidangallagher4
Jul 24 at 11:47
It's unclear how "area under a curve" applies to any complex integral (except when it's actually just a real integral after all).
– David K
Jul 24 at 20:44
It's unclear how "area under a curve" applies to any complex integral (except when it's actually just a real integral after all).
– David K
Jul 24 at 20:44
Clearer now????
– Diger
Jul 24 at 23:12
Clearer now????
– Diger
Jul 24 at 23:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
Before asking what the value of the integral is you have to decide how you are going to define $(-1)^x=e^xlog (-1)$. For each choice of $log (-1)$ you will get one value for the integral. There is no such thing as the right value of the logarithm.
answered Jul 24 at 11:49
Kavi Rama Murthy
20.2k2829
You are basically just rephrasing what I formulated as "deciding for a specific Riemann-Sheet".
– Diger
Jul 24 at 11:58
True, but the OP thinks that there is a specific Riemann sheet that should be preferred. There is no such thing I just tried to explain that.
– Kavi Rama Murthy
Jul 24 at 12:00
So you are saying the first integral in essence doesn't make sense? The result is just playing around for the enjoyment (as the video guy above puts it: "This is so cool, isn't it?" - apart from the big mess on the board) without any further depth one might remotely be able to interpret into it?
– Diger
Jul 24 at 12:05
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Before asking what the value of the integral is you have to decide how you are going to define $(-1)^x=e^xlog (-1)$. For each choice of $log (-1)$ you will get one value for the integral. There is no such thing as the right value of the logarithm.
answered Jul 24 at 11:49
Kavi Rama Murthy
20.2k2829
You are basically just rephrasing what I formulated as "deciding for a specific Riemann-Sheet".
– Diger
Jul 24 at 11:58
True, but the OP thinks that there is a specific Riemann sheet that should be preferred. There is no such thing I just tried to explain that.
– Kavi Rama Murthy
Jul 24 at 12:00
So you are saying the first integral in essence doesn't make sense? The result is just playing around for the enjoyment (as the video guy above puts it: "This is so cool, isn't it?" - apart from the big mess on the board) without any further depth one might remotely be able to interpret into it?
– Diger
Jul 24 at 12:05
add a comment |Â
up vote
4
down vote
Before asking what the value of the integral is you have to decide how you are going to define $(-1)^x=e^xlog (-1)$. For each choice of $log (-1)$ you will get one value for the integral. There is no such thing as the right value of the logarithm.
answered Jul 24 at 11:49
Kavi Rama Murthy
20.2k2829
You are basically just rephrasing what I formulated as "deciding for a specific Riemann-Sheet".
– Diger
Jul 24 at 11:58
True, but the OP thinks that there is a specific Riemann sheet that should be preferred. There is no such thing I just tried to explain that.
– Kavi Rama Murthy
Jul 24 at 12:00
So you are saying the first integral in essence doesn't make sense? The result is just playing around for the enjoyment (as the video guy above puts it: "This is so cool, isn't it?" - apart from the big mess on the board) without any further depth one might remotely be able to interpret into it?
– Diger
Jul 24 at 12:05
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Before asking what the value of the integral is you have to decide how you are going to define $(-1)^x=e^xlog (-1)$. For each choice of $log (-1)$ you will get one value for the integral. There is no such thing as the right value of the logarithm.
answered Jul 24 at 11:49
Kavi Rama Murthy
20.2k2829
Before asking what the value of the integral is you have to decide how you are going to define $(-1)^x=e^xlog (-1)$. For each choice of $log (-1)$ you will get one value for the integral. There is no such thing as the right value of the logarithm.
answered Jul 24 at 11:49
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 11:49
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 11:49
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 11:49
Kavi Rama Murthy
20.2k2829
20.2k2829
You are basically just rephrasing what I formulated as "deciding for a specific Riemann-Sheet".
– Diger
Jul 24 at 11:58
True, but the OP thinks that there is a specific Riemann sheet that should be preferred. There is no such thing I just tried to explain that.
– Kavi Rama Murthy
Jul 24 at 12:00
So you are saying the first integral in essence doesn't make sense? The result is just playing around for the enjoyment (as the video guy above puts it: "This is so cool, isn't it?" - apart from the big mess on the board) without any further depth one might remotely be able to interpret into it?
– Diger
Jul 24 at 12:05
add a comment |Â
You are basically just rephrasing what I formulated as "deciding for a specific Riemann-Sheet".
– Diger
Jul 24 at 11:58
True, but the OP thinks that there is a specific Riemann sheet that should be preferred. There is no such thing I just tried to explain that.
– Kavi Rama Murthy
Jul 24 at 12:00
So you are saying the first integral in essence doesn't make sense? The result is just playing around for the enjoyment (as the video guy above puts it: "This is so cool, isn't it?" - apart from the big mess on the board) without any further depth one might remotely be able to interpret into it?
– Diger
Jul 24 at 12:05
You are basically just rephrasing what I formulated as "deciding for a specific Riemann-Sheet".
– Diger
Jul 24 at 11:58
You are basically just rephrasing what I formulated as "deciding for a specific Riemann-Sheet".
– Diger
Jul 24 at 11:58
True, but the OP thinks that there is a specific Riemann sheet that should be preferred. There is no such thing I just tried to explain that.
– Kavi Rama Murthy
Jul 24 at 12:00
True, but the OP thinks that there is a specific Riemann sheet that should be preferred. There is no such thing I just tried to explain that.
– Kavi Rama Murthy
Jul 24 at 12:00
So you are saying the first integral in essence doesn't make sense? The result is just playing around for the enjoyment (as the video guy above puts it: "This is so cool, isn't it?" - apart from the big mess on the board) without any further depth one might remotely be able to interpret into it?
– Diger
Jul 24 at 12:05
So you are saying the first integral in essence doesn't make sense? The result is just playing around for the enjoyment (as the video guy above puts it: "This is so cool, isn't it?" - apart from the big mess on the board) without any further depth one might remotely be able to interpret into it?
– Diger
Jul 24 at 12:05
add a comment |Â
2
Relevant: youtu.be/pYJQZDMayeI
– aidangallagher4
Jul 24 at 11:47
It's unclear how "area under a curve" applies to any complex integral (except when it's actually just a real integral after all).
– David K
Jul 24 at 20:44
Clearer now????
– Diger
Jul 24 at 23:12