Analytic function need not have a primitiveconvergence of analytic functionsHow to understand the Identity Theorem in complex analysis, from the point of view of power series expansionsHolomorphic function on a domain has a primitiveLimit of nonnegative analytic functionsIf analytic $f_n rightrightarrows f$ on a closed disk and $f$ is injective, then almost all $f_n$ are injectiveIf $f$ is analytic then $|f|$ is not constant unless $f$ is constantCondition for $f-fracAz-a-fracBz-b-fracCz-c$ to have primitiveCauchy integral theorem in a Jordan domain?Confusion about the definition of analytic and singularity.A function on a domain $D$ has a primitive on $D$ but is not analytic on $D$?
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Analytic function need not have a primitive
convergence of analytic functionsHow to understand the Identity Theorem in complex analysis, from the point of view of power series expansionsHolomorphic function on a domain has a primitiveLimit of nonnegative analytic functionsIf analytic $f_n rightrightarrows f$ on a closed disk and $f$ is injective, then almost all $f_n$ are injectiveIf $f$ is analytic then $|f|$ is not constant unless $f$ is constantCondition for $f-fracAz-a-fracBz-b-fracCz-c$ to have primitiveCauchy integral theorem in a Jordan domain?Confusion about the definition of analytic and singularity.A function on a domain $D$ has a primitive on $D$ but is not analytic on $D$?
$begingroup$
I am told that an analytic function need not possess a primitive in its domain of analyticity. However, if $f$ is analytic in the disk $textB(a,r)$, then it has a primitive in that disk. Suppose $f$ is analytic in an open set $G$. Then for each $ain G$, $exists r_a>0$ such that the disk $textB(a,r_a)subset G$. In each of these disks $f$ has a primitive. Why can't we always "stitch" these "local primitives" to obtain a primitive for $f$ in the set $G$? When can we do such a thing?
complex-analysis soft-question analytic-functions
asked Mar 21 at 6:40
Hrit RoyHrit Roy
896215
$endgroup$
add a comment |
$begingroup$
I am told that an analytic function need not possess a primitive in its domain of analyticity. However, if $f$ is analytic in the disk $textB(a,r)$, then it has a primitive in that disk. Suppose $f$ is analytic in an open set $G$. Then for each $ain G$, $exists r_a>0$ such that the disk $textB(a,r_a)subset G$. In each of these disks $f$ has a primitive. Why can't we always "stitch" these "local primitives" to obtain a primitive for $f$ in the set $G$? When can we do such a thing?
complex-analysis soft-question analytic-functions
asked Mar 21 at 6:40
Hrit RoyHrit Roy
896215
$endgroup$
add a comment |
$begingroup$
I am told that an analytic function need not possess a primitive in its domain of analyticity. However, if $f$ is analytic in the disk $textB(a,r)$, then it has a primitive in that disk. Suppose $f$ is analytic in an open set $G$. Then for each $ain G$, $exists r_a>0$ such that the disk $textB(a,r_a)subset G$. In each of these disks $f$ has a primitive. Why can't we always "stitch" these "local primitives" to obtain a primitive for $f$ in the set $G$? When can we do such a thing?
complex-analysis soft-question analytic-functions
asked Mar 21 at 6:40
Hrit RoyHrit Roy
896215
$endgroup$
I am told that an analytic function need not possess a primitive in its domain of analyticity. However, if $f$ is analytic in the disk $textB(a,r)$, then it has a primitive in that disk. Suppose $f$ is analytic in an open set $G$. Then for each $ain G$, $exists r_a>0$ such that the disk $textB(a,r_a)subset G$. In each of these disks $f$ has a primitive. Why can't we always "stitch" these "local primitives" to obtain a primitive for $f$ in the set $G$? When can we do such a thing?
complex-analysis soft-question analytic-functions
complex-analysis soft-question analytic-functions
asked Mar 21 at 6:40
Hrit RoyHrit Roy
896215
asked Mar 21 at 6:40
Hrit RoyHrit Roy
896215
asked Mar 21 at 6:40
Hrit RoyHrit Roy
896215
asked Mar 21 at 6:40
Hrit RoyHrit Roy
896215
asked Mar 21 at 6:40
Hrit RoyHrit Roy
896215
896215
add a comment |
add a comment |
1 Answer
1
active
oldest
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If $Omega$ is simply connected, then every holomorphic function on $Omega$ has an anti-derivative. This is a standard theorem proven in almost every textbook.
Conversely, if $Omega$ is not simply connected, you can always find an example of a holomorphic function on $Omega$ that doesn't admit an anti-derivative. The canonical example is $f(z) = 1/z$ on the punctured disk (or punctured plane), and this example is easy to adapt to any non-simply connected domain.
Note that if $f$ has an anti-derivative $F$, then
$$
int_gamma f(z),dz = F(b)-F(a)
$$
if $gamma$ is a curve from $a$ to $b$. (This is the complex analogue of the fundamental theorem of calculus.) In particular, the integral of $f$ along any closed curve in $Omega$ is $0$, whereas e.g.
$$
int_=1 fracdzz = 2pi i.
$$
answered Mar 21 at 12:40
mrfmrf
37.6k64786
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add a comment |
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1 Answer
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$begingroup$
If $Omega$ is simply connected, then every holomorphic function on $Omega$ has an anti-derivative. This is a standard theorem proven in almost every textbook.
Conversely, if $Omega$ is not simply connected, you can always find an example of a holomorphic function on $Omega$ that doesn't admit an anti-derivative. The canonical example is $f(z) = 1/z$ on the punctured disk (or punctured plane), and this example is easy to adapt to any non-simply connected domain.
Note that if $f$ has an anti-derivative $F$, then
$$
int_gamma f(z),dz = F(b)-F(a)
$$
if $gamma$ is a curve from $a$ to $b$. (This is the complex analogue of the fundamental theorem of calculus.) In particular, the integral of $f$ along any closed curve in $Omega$ is $0$, whereas e.g.
$$
int_=1 fracdzz = 2pi i.
$$
answered Mar 21 at 12:40
mrfmrf
37.6k64786
$endgroup$
add a comment |
$begingroup$
If $Omega$ is simply connected, then every holomorphic function on $Omega$ has an anti-derivative. This is a standard theorem proven in almost every textbook.
Conversely, if $Omega$ is not simply connected, you can always find an example of a holomorphic function on $Omega$ that doesn't admit an anti-derivative. The canonical example is $f(z) = 1/z$ on the punctured disk (or punctured plane), and this example is easy to adapt to any non-simply connected domain.
Note that if $f$ has an anti-derivative $F$, then
$$
int_gamma f(z),dz = F(b)-F(a)
$$
if $gamma$ is a curve from $a$ to $b$. (This is the complex analogue of the fundamental theorem of calculus.) In particular, the integral of $f$ along any closed curve in $Omega$ is $0$, whereas e.g.
$$
int_=1 fracdzz = 2pi i.
$$
answered Mar 21 at 12:40
mrfmrf
37.6k64786
$endgroup$
add a comment |
$begingroup$
If $Omega$ is simply connected, then every holomorphic function on $Omega$ has an anti-derivative. This is a standard theorem proven in almost every textbook.
Conversely, if $Omega$ is not simply connected, you can always find an example of a holomorphic function on $Omega$ that doesn't admit an anti-derivative. The canonical example is $f(z) = 1/z$ on the punctured disk (or punctured plane), and this example is easy to adapt to any non-simply connected domain.
Note that if $f$ has an anti-derivative $F$, then
$$
int_gamma f(z),dz = F(b)-F(a)
$$
if $gamma$ is a curve from $a$ to $b$. (This is the complex analogue of the fundamental theorem of calculus.) In particular, the integral of $f$ along any closed curve in $Omega$ is $0$, whereas e.g.
$$
int_=1 fracdzz = 2pi i.
$$
answered Mar 21 at 12:40
mrfmrf
37.6k64786
$endgroup$
If $Omega$ is simply connected, then every holomorphic function on $Omega$ has an anti-derivative. This is a standard theorem proven in almost every textbook.
Conversely, if $Omega$ is not simply connected, you can always find an example of a holomorphic function on $Omega$ that doesn't admit an anti-derivative. The canonical example is $f(z) = 1/z$ on the punctured disk (or punctured plane), and this example is easy to adapt to any non-simply connected domain.
Note that if $f$ has an anti-derivative $F$, then
$$
int_gamma f(z),dz = F(b)-F(a)
$$
if $gamma$ is a curve from $a$ to $b$. (This is the complex analogue of the fundamental theorem of calculus.) In particular, the integral of $f$ along any closed curve in $Omega$ is $0$, whereas e.g.
$$
int_=1 fracdzz = 2pi i.
$$
answered Mar 21 at 12:40
mrfmrf
37.6k64786
answered Mar 21 at 12:40
mrfmrf
37.6k64786
answered Mar 21 at 12:40
mrfmrf
37.6k64786
answered Mar 21 at 12:40
mrfmrf
37.6k64786
37.6k64786
add a comment |
add a comment |
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