Is it possible to evaulate $I = frac{2i}{pi} int_gamma ln(z) dz$ explicitly even though $ln(z)$ isn't...
$begingroup$
Is it possible to calculate the following integral explicitly:
$$
I = frac{2i}{pi} int_gamma ln(z) dz,
$$
where $gamma$ can be a disk at the origin of $mathbb{C}$? Unfortunately $ln$ isn't holomorphic at $0$ so we can't use Cauchy's integral theorem.
But it seems like it should be possible as if we identify $mathbb{C}$ with $mathbb{R^2}$ then this would be a considered a weakly singular integral in potential theory in $mathbb{R}^2$ which I know is integrable.
So can this integral be evaluated explicitly if the contour is a disk at the origin?
integration complex-analysis cauchy-integral-formula potential-theory singular-integrals
asked Jan 11 at 21:19
eurocodereurocoder
1,126315
$endgroup$
add a comment |
$begingroup$
Is it possible to calculate the following integral explicitly:
$$
I = frac{2i}{pi} int_gamma ln(z) dz,
$$
where $gamma$ can be a disk at the origin of $mathbb{C}$? Unfortunately $ln$ isn't holomorphic at $0$ so we can't use Cauchy's integral theorem.
But it seems like it should be possible as if we identify $mathbb{C}$ with $mathbb{R^2}$ then this would be a considered a weakly singular integral in potential theory in $mathbb{R}^2$ which I know is integrable.
So can this integral be evaluated explicitly if the contour is a disk at the origin?
integration complex-analysis cauchy-integral-formula potential-theory singular-integrals
asked Jan 11 at 21:19
eurocodereurocoder
1,126315
$endgroup$
$begingroup$
How do even define the logarithm? There is no branch in $mathbb{C}setminus{0}$. It isn't a function with a singularity at the origin. It is a function which can be defined only on a specific branch.
$endgroup$
– Mark
Jan 11 at 21:27
$begingroup$
Yes there is a singularity at the origin..and it has a branch cut. I would like to be able to evaluate it explicitly if possible.
$endgroup$
– eurocoder
Jan 11 at 21:34
1
$begingroup$
If $gamma$ is a curve in $mathbb{C}^*$ and the principal branch of $log(z)$ is continued analytically along $gamma$ then $int_gamma log (z) dz= F(gamma(1))-F(gamma(0))$ where $F(z) = zlog z-z$ and $F(gamma(t))$ is continued continuously in $t in [0,1]$. With the unit circle $gamma(t) = e^{2i pi t}, t in [0,1] $ then $gamma(0) = gamma(1)=1, F(gamma(1) ) = 2ipi - 1, F(gamma(0)) = -1$.
$endgroup$
– reuns
Jan 11 at 21:54
add a comment |
$begingroup$
Is it possible to calculate the following integral explicitly:
$$
I = frac{2i}{pi} int_gamma ln(z) dz,
$$
where $gamma$ can be a disk at the origin of $mathbb{C}$? Unfortunately $ln$ isn't holomorphic at $0$ so we can't use Cauchy's integral theorem.
But it seems like it should be possible as if we identify $mathbb{C}$ with $mathbb{R^2}$ then this would be a considered a weakly singular integral in potential theory in $mathbb{R}^2$ which I know is integrable.
So can this integral be evaluated explicitly if the contour is a disk at the origin?
integration complex-analysis cauchy-integral-formula potential-theory singular-integrals
asked Jan 11 at 21:19
eurocodereurocoder
1,126315
$endgroup$
Is it possible to calculate the following integral explicitly:
$$
I = frac{2i}{pi} int_gamma ln(z) dz,
$$
where $gamma$ can be a disk at the origin of $mathbb{C}$? Unfortunately $ln$ isn't holomorphic at $0$ so we can't use Cauchy's integral theorem.
But it seems like it should be possible as if we identify $mathbb{C}$ with $mathbb{R^2}$ then this would be a considered a weakly singular integral in potential theory in $mathbb{R}^2$ which I know is integrable.
So can this integral be evaluated explicitly if the contour is a disk at the origin?
integration complex-analysis cauchy-integral-formula potential-theory singular-integrals
integration complex-analysis cauchy-integral-formula potential-theory singular-integrals
asked Jan 11 at 21:19
eurocodereurocoder
1,126315
asked Jan 11 at 21:19
eurocodereurocoder
1,126315
edited Jan 11 at 21:35
eurocoder
asked Jan 11 at 21:19
eurocodereurocoder
1,126315
asked Jan 11 at 21:19
eurocodereurocoder
1,126315
asked Jan 11 at 21:19
eurocodereurocoder
1,126315
1,126315
$begingroup$
How do even define the logarithm? There is no branch in $mathbb{C}setminus{0}$. It isn't a function with a singularity at the origin. It is a function which can be defined only on a specific branch.
$endgroup$
– Mark
Jan 11 at 21:27
$begingroup$
Yes there is a singularity at the origin..and it has a branch cut. I would like to be able to evaluate it explicitly if possible.
$endgroup$
– eurocoder
Jan 11 at 21:34
1
$begingroup$
If $gamma$ is a curve in $mathbb{C}^*$ and the principal branch of $log(z)$ is continued analytically along $gamma$ then $int_gamma log (z) dz= F(gamma(1))-F(gamma(0))$ where $F(z) = zlog z-z$ and $F(gamma(t))$ is continued continuously in $t in [0,1]$. With the unit circle $gamma(t) = e^{2i pi t}, t in [0,1] $ then $gamma(0) = gamma(1)=1, F(gamma(1) ) = 2ipi - 1, F(gamma(0)) = -1$.
$endgroup$
– reuns
Jan 11 at 21:54
add a comment |
$begingroup$
How do even define the logarithm? There is no branch in $mathbb{C}setminus{0}$. It isn't a function with a singularity at the origin. It is a function which can be defined only on a specific branch.
$endgroup$
– Mark
Jan 11 at 21:27
$begingroup$
Yes there is a singularity at the origin..and it has a branch cut. I would like to be able to evaluate it explicitly if possible.
$endgroup$
– eurocoder
Jan 11 at 21:34
1
$begingroup$
If $gamma$ is a curve in $mathbb{C}^*$ and the principal branch of $log(z)$ is continued analytically along $gamma$ then $int_gamma log (z) dz= F(gamma(1))-F(gamma(0))$ where $F(z) = zlog z-z$ and $F(gamma(t))$ is continued continuously in $t in [0,1]$. With the unit circle $gamma(t) = e^{2i pi t}, t in [0,1] $ then $gamma(0) = gamma(1)=1, F(gamma(1) ) = 2ipi - 1, F(gamma(0)) = -1$.
$endgroup$
– reuns
Jan 11 at 21:54
$begingroup$
How do even define the logarithm? There is no branch in $mathbb{C}setminus{0}$. It isn't a function with a singularity at the origin. It is a function which can be defined only on a specific branch.
$endgroup$
– Mark
Jan 11 at 21:27
$begingroup$
How do even define the logarithm? There is no branch in $mathbb{C}setminus{0}$. It isn't a function with a singularity at the origin. It is a function which can be defined only on a specific branch.
$endgroup$
– Mark
Jan 11 at 21:27
$begingroup$
Yes there is a singularity at the origin..and it has a branch cut. I would like to be able to evaluate it explicitly if possible.
$endgroup$
– eurocoder
Jan 11 at 21:34
$begingroup$
Yes there is a singularity at the origin..and it has a branch cut. I would like to be able to evaluate it explicitly if possible.
$endgroup$
– eurocoder
Jan 11 at 21:34
1
1
$begingroup$
If $gamma$ is a curve in $mathbb{C}^*$ and the principal branch of $log(z)$ is continued analytically along $gamma$ then $int_gamma log (z) dz= F(gamma(1))-F(gamma(0))$ where $F(z) = zlog z-z$ and $F(gamma(t))$ is continued continuously in $t in [0,1]$. With the unit circle $gamma(t) = e^{2i pi t}, t in [0,1] $ then $gamma(0) = gamma(1)=1, F(gamma(1) ) = 2ipi - 1, F(gamma(0)) = -1$.
$endgroup$
– reuns
Jan 11 at 21:54
$begingroup$
If $gamma$ is a curve in $mathbb{C}^*$ and the principal branch of $log(z)$ is continued analytically along $gamma$ then $int_gamma log (z) dz= F(gamma(1))-F(gamma(0))$ where $F(z) = zlog z-z$ and $F(gamma(t))$ is continued continuously in $t in [0,1]$. With the unit circle $gamma(t) = e^{2i pi t}, t in [0,1] $ then $gamma(0) = gamma(1)=1, F(gamma(1) ) = 2ipi - 1, F(gamma(0)) = -1$.
$endgroup$
– reuns
Jan 11 at 21:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is a difficult question because there is some monodromy.
You can integrate this directly. If $z=rmathrm e^{mathrm itheta}$, then we can circle the origin many different ways, e.g. $0 < theta < 2pi$, or $-pi < theta < pi$, or $5pi < theta < 7pi$. In general, $t < theta < t+2pi$ would give a loop around the origin. As Mark Viola points out: we need the branch cut to pass through $|z|=r$ and $theta = t$.
If $z=rmathrm e^{mathrm i theta}$, then $mathrm dz=mathrm irmathrm e^{mathrm itheta}~mathrm dtheta$, and so
begin{eqnarray*}
int_{gamma} ln z~mathrm dz &=& int_t^{t+2pi}lnleft(rmathrm e^{mathrm itheta}right)~mathrm irmathrm e^{mathrm itheta}~mathrm dtheta \ \
&=& int_t^{t+2pi}left[ln r + mathrm ithetaright]~mathrm irmathrm e^{mathrm itheta}~mathrm dtheta \ \
&=& mathrm i rln rint_t^{t+2pi} mathrm e^{mathrm i theta}~mathrm dtheta - rint_t^{t+2pi}thetamathrm e^{mathrm i theta}~mathrm dtheta \ \
&=& 0 - rleft[ (1-mathrm itheta)mathrm e^{mathrm itheta}right]_t^{t+2pi} \ \
&=& 2pimathrm irmathrm e^{mathrm it}
end{eqnarray*}
answered Jan 11 at 22:46
Fly by NightFly by Night
25.8k32978
$endgroup$
1
$begingroup$
I assume that the branch cut passes through $|z|=r$ at $theta =t$. Is that what you have in mind here?
$endgroup$
– Mark Viola
Jan 11 at 22:51
$begingroup$
That's right. Thank you Mark :-)
$endgroup$
– Fly by Night
Jan 11 at 22:54
add a comment |
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1 Answer
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1 Answer
1
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oldest
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active
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$begingroup$
This is a difficult question because there is some monodromy.
You can integrate this directly. If $z=rmathrm e^{mathrm itheta}$, then we can circle the origin many different ways, e.g. $0 < theta < 2pi$, or $-pi < theta < pi$, or $5pi < theta < 7pi$. In general, $t < theta < t+2pi$ would give a loop around the origin. As Mark Viola points out: we need the branch cut to pass through $|z|=r$ and $theta = t$.
If $z=rmathrm e^{mathrm i theta}$, then $mathrm dz=mathrm irmathrm e^{mathrm itheta}~mathrm dtheta$, and so
begin{eqnarray*}
int_{gamma} ln z~mathrm dz &=& int_t^{t+2pi}lnleft(rmathrm e^{mathrm itheta}right)~mathrm irmathrm e^{mathrm itheta}~mathrm dtheta \ \
&=& int_t^{t+2pi}left[ln r + mathrm ithetaright]~mathrm irmathrm e^{mathrm itheta}~mathrm dtheta \ \
&=& mathrm i rln rint_t^{t+2pi} mathrm e^{mathrm i theta}~mathrm dtheta - rint_t^{t+2pi}thetamathrm e^{mathrm i theta}~mathrm dtheta \ \
&=& 0 - rleft[ (1-mathrm itheta)mathrm e^{mathrm itheta}right]_t^{t+2pi} \ \
&=& 2pimathrm irmathrm e^{mathrm it}
end{eqnarray*}
answered Jan 11 at 22:46
Fly by NightFly by Night
25.8k32978
$endgroup$
1
$begingroup$
I assume that the branch cut passes through $|z|=r$ at $theta =t$. Is that what you have in mind here?
$endgroup$
– Mark Viola
Jan 11 at 22:51
$begingroup$
That's right. Thank you Mark :-)
$endgroup$
– Fly by Night
Jan 11 at 22:54
add a comment |
$begingroup$
This is a difficult question because there is some monodromy.
You can integrate this directly. If $z=rmathrm e^{mathrm itheta}$, then we can circle the origin many different ways, e.g. $0 < theta < 2pi$, or $-pi < theta < pi$, or $5pi < theta < 7pi$. In general, $t < theta < t+2pi$ would give a loop around the origin. As Mark Viola points out: we need the branch cut to pass through $|z|=r$ and $theta = t$.
If $z=rmathrm e^{mathrm i theta}$, then $mathrm dz=mathrm irmathrm e^{mathrm itheta}~mathrm dtheta$, and so
begin{eqnarray*}
int_{gamma} ln z~mathrm dz &=& int_t^{t+2pi}lnleft(rmathrm e^{mathrm itheta}right)~mathrm irmathrm e^{mathrm itheta}~mathrm dtheta \ \
&=& int_t^{t+2pi}left[ln r + mathrm ithetaright]~mathrm irmathrm e^{mathrm itheta}~mathrm dtheta \ \
&=& mathrm i rln rint_t^{t+2pi} mathrm e^{mathrm i theta}~mathrm dtheta - rint_t^{t+2pi}thetamathrm e^{mathrm i theta}~mathrm dtheta \ \
&=& 0 - rleft[ (1-mathrm itheta)mathrm e^{mathrm itheta}right]_t^{t+2pi} \ \
&=& 2pimathrm irmathrm e^{mathrm it}
end{eqnarray*}
answered Jan 11 at 22:46
Fly by NightFly by Night
25.8k32978
$endgroup$
1
$begingroup$
I assume that the branch cut passes through $|z|=r$ at $theta =t$. Is that what you have in mind here?
$endgroup$
– Mark Viola
Jan 11 at 22:51
$begingroup$
That's right. Thank you Mark :-)
$endgroup$
– Fly by Night
Jan 11 at 22:54
add a comment |
$begingroup$
This is a difficult question because there is some monodromy.
You can integrate this directly. If $z=rmathrm e^{mathrm itheta}$, then we can circle the origin many different ways, e.g. $0 < theta < 2pi$, or $-pi < theta < pi$, or $5pi < theta < 7pi$. In general, $t < theta < t+2pi$ would give a loop around the origin. As Mark Viola points out: we need the branch cut to pass through $|z|=r$ and $theta = t$.
If $z=rmathrm e^{mathrm i theta}$, then $mathrm dz=mathrm irmathrm e^{mathrm itheta}~mathrm dtheta$, and so
begin{eqnarray*}
int_{gamma} ln z~mathrm dz &=& int_t^{t+2pi}lnleft(rmathrm e^{mathrm itheta}right)~mathrm irmathrm e^{mathrm itheta}~mathrm dtheta \ \
&=& int_t^{t+2pi}left[ln r + mathrm ithetaright]~mathrm irmathrm e^{mathrm itheta}~mathrm dtheta \ \
&=& mathrm i rln rint_t^{t+2pi} mathrm e^{mathrm i theta}~mathrm dtheta - rint_t^{t+2pi}thetamathrm e^{mathrm i theta}~mathrm dtheta \ \
&=& 0 - rleft[ (1-mathrm itheta)mathrm e^{mathrm itheta}right]_t^{t+2pi} \ \
&=& 2pimathrm irmathrm e^{mathrm it}
end{eqnarray*}
answered Jan 11 at 22:46
Fly by NightFly by Night
25.8k32978
$endgroup$
This is a difficult question because there is some monodromy.
You can integrate this directly. If $z=rmathrm e^{mathrm itheta}$, then we can circle the origin many different ways, e.g. $0 < theta < 2pi$, or $-pi < theta < pi$, or $5pi < theta < 7pi$. In general, $t < theta < t+2pi$ would give a loop around the origin. As Mark Viola points out: we need the branch cut to pass through $|z|=r$ and $theta = t$.
If $z=rmathrm e^{mathrm i theta}$, then $mathrm dz=mathrm irmathrm e^{mathrm itheta}~mathrm dtheta$, and so
begin{eqnarray*}
int_{gamma} ln z~mathrm dz &=& int_t^{t+2pi}lnleft(rmathrm e^{mathrm itheta}right)~mathrm irmathrm e^{mathrm itheta}~mathrm dtheta \ \
&=& int_t^{t+2pi}left[ln r + mathrm ithetaright]~mathrm irmathrm e^{mathrm itheta}~mathrm dtheta \ \
&=& mathrm i rln rint_t^{t+2pi} mathrm e^{mathrm i theta}~mathrm dtheta - rint_t^{t+2pi}thetamathrm e^{mathrm i theta}~mathrm dtheta \ \
&=& 0 - rleft[ (1-mathrm itheta)mathrm e^{mathrm itheta}right]_t^{t+2pi} \ \
&=& 2pimathrm irmathrm e^{mathrm it}
end{eqnarray*}
answered Jan 11 at 22:46
Fly by NightFly by Night
25.8k32978
edited Jan 11 at 22:59
answered Jan 11 at 22:46
Fly by NightFly by Night
25.8k32978
answered Jan 11 at 22:46
Fly by NightFly by Night
25.8k32978
answered Jan 11 at 22:46
Fly by NightFly by Night
25.8k32978
25.8k32978
1
$begingroup$
I assume that the branch cut passes through $|z|=r$ at $theta =t$. Is that what you have in mind here?
$endgroup$
– Mark Viola
Jan 11 at 22:51
$begingroup$
That's right. Thank you Mark :-)
$endgroup$
– Fly by Night
Jan 11 at 22:54
add a comment |
1
$begingroup$
I assume that the branch cut passes through $|z|=r$ at $theta =t$. Is that what you have in mind here?
$endgroup$
– Mark Viola
Jan 11 at 22:51
$begingroup$
That's right. Thank you Mark :-)
$endgroup$
– Fly by Night
Jan 11 at 22:54
1
1
$begingroup$
I assume that the branch cut passes through $|z|=r$ at $theta =t$. Is that what you have in mind here?
$endgroup$
– Mark Viola
Jan 11 at 22:51
$begingroup$
I assume that the branch cut passes through $|z|=r$ at $theta =t$. Is that what you have in mind here?
$endgroup$
– Mark Viola
Jan 11 at 22:51
$begingroup$
That's right. Thank you Mark :-)
$endgroup$
– Fly by Night
Jan 11 at 22:54
$begingroup$
That's right. Thank you Mark :-)
$endgroup$
– Fly by Night
Jan 11 at 22:54
add a comment |
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$begingroup$
How do even define the logarithm? There is no branch in $mathbb{C}setminus{0}$. It isn't a function with a singularity at the origin. It is a function which can be defined only on a specific branch.
$endgroup$
– Mark
Jan 11 at 21:27
$begingroup$
Yes there is a singularity at the origin..and it has a branch cut. I would like to be able to evaluate it explicitly if possible.
$endgroup$
– eurocoder
Jan 11 at 21:34
1
$begingroup$
If $gamma$ is a curve in $mathbb{C}^*$ and the principal branch of $log(z)$ is continued analytically along $gamma$ then $int_gamma log (z) dz= F(gamma(1))-F(gamma(0))$ where $F(z) = zlog z-z$ and $F(gamma(t))$ is continued continuously in $t in [0,1]$. With the unit circle $gamma(t) = e^{2i pi t}, t in [0,1] $ then $gamma(0) = gamma(1)=1, F(gamma(1) ) = 2ipi - 1, F(gamma(0)) = -1$.
$endgroup$
– reuns
Jan 11 at 21:54