Placement of singularities in the residue theorem
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Why do the singularities in Cauchy's residue theorem have to be within the contour, and why do they still count if they're not on the path of integration, like I'd suspect for real integrals? Sorry if this is basic but I haven't found it anywhere else.
complex-integration
asked Jan 24 at 21:19
Benjamin ThoburnBenjamin Thoburn
352213
$endgroup$
add a comment |
$begingroup$
Why do the singularities in Cauchy's residue theorem have to be within the contour, and why do they still count if they're not on the path of integration, like I'd suspect for real integrals? Sorry if this is basic but I haven't found it anywhere else.
complex-integration
asked Jan 24 at 21:19
Benjamin ThoburnBenjamin Thoburn
352213
$endgroup$
2
$begingroup$
Are you familiar with Cauchy's Integral Theorem? If $f$ is analytic in and on a rectifiable curve, then the contour integral on that path is $0$. Now, deform the contour to exclude any singularities and proceed.
$endgroup$
– Mark Viola
Jan 24 at 21:24
add a comment |
$begingroup$
Why do the singularities in Cauchy's residue theorem have to be within the contour, and why do they still count if they're not on the path of integration, like I'd suspect for real integrals? Sorry if this is basic but I haven't found it anywhere else.
complex-integration
asked Jan 24 at 21:19
Benjamin ThoburnBenjamin Thoburn
352213
$endgroup$
Why do the singularities in Cauchy's residue theorem have to be within the contour, and why do they still count if they're not on the path of integration, like I'd suspect for real integrals? Sorry if this is basic but I haven't found it anywhere else.
complex-integration
complex-integration
asked Jan 24 at 21:19
Benjamin ThoburnBenjamin Thoburn
352213
asked Jan 24 at 21:19
Benjamin ThoburnBenjamin Thoburn
352213
asked Jan 24 at 21:19
Benjamin ThoburnBenjamin Thoburn
352213
asked Jan 24 at 21:19
Benjamin ThoburnBenjamin Thoburn
352213
asked Jan 24 at 21:19
Benjamin ThoburnBenjamin Thoburn
352213
352213
2
$begingroup$
Are you familiar with Cauchy's Integral Theorem? If $f$ is analytic in and on a rectifiable curve, then the contour integral on that path is $0$. Now, deform the contour to exclude any singularities and proceed.
$endgroup$
– Mark Viola
Jan 24 at 21:24
add a comment |
2
$begingroup$
Are you familiar with Cauchy's Integral Theorem? If $f$ is analytic in and on a rectifiable curve, then the contour integral on that path is $0$. Now, deform the contour to exclude any singularities and proceed.
$endgroup$
– Mark Viola
Jan 24 at 21:24
2
2
$begingroup$
Are you familiar with Cauchy's Integral Theorem? If $f$ is analytic in and on a rectifiable curve, then the contour integral on that path is $0$. Now, deform the contour to exclude any singularities and proceed.
$endgroup$
– Mark Viola
Jan 24 at 21:24
$begingroup$
Are you familiar with Cauchy's Integral Theorem? If $f$ is analytic in and on a rectifiable curve, then the contour integral on that path is $0$. Now, deform the contour to exclude any singularities and proceed.
$endgroup$
– Mark Viola
Jan 24 at 21:24
add a comment |
1 Answer
1
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oldest
votes
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I find homotopy to the most intuitive explanation: suppose you have two closed contours $gamma_1$ and $gamma_2$ and a function $f(z).$ If you can smoothly deform $gamma_1$ into $gamma_2$ without "crossing over" a singularity of $f,$ then
$$oint_{gamma_1}f(z),dz=oint_{gamma_2}f(z),dz.$$
So, for example, if you can smoothly (continuously) deform a closed contour down to a single point (not a singularity), then it would make sense the integral would be zero, right? So in that sense, whatever singularities start out inside your contour have to stay inside that contour, or you change the value of the integral.
answered Jan 24 at 21:35
Adrian KeisterAdrian Keister
5,27371933
$endgroup$
$begingroup$
Nice, why does homotopy apply? I've thought about it before.
$endgroup$
– Benjamin Thoburn
Jan 24 at 21:42
$begingroup$
Well, within the limits of the assumptions of homotopy, it always applies! Homotopy doesn't get you everywhere you need to go, but it is quite intuitive.
$endgroup$
– Adrian Keister
Jan 25 at 18:24
$begingroup$
I think I get it, it works because the integral is linear, you can write it as a sum of the integral along other paths, then each of those be be changed thus you can change the whole closed contour... I’ll think more about the gritty details of singularities.
$endgroup$
– Benjamin Thoburn
Jan 25 at 18:44
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
I find homotopy to the most intuitive explanation: suppose you have two closed contours $gamma_1$ and $gamma_2$ and a function $f(z).$ If you can smoothly deform $gamma_1$ into $gamma_2$ without "crossing over" a singularity of $f,$ then
$$oint_{gamma_1}f(z),dz=oint_{gamma_2}f(z),dz.$$
So, for example, if you can smoothly (continuously) deform a closed contour down to a single point (not a singularity), then it would make sense the integral would be zero, right? So in that sense, whatever singularities start out inside your contour have to stay inside that contour, or you change the value of the integral.
answered Jan 24 at 21:35
Adrian KeisterAdrian Keister
5,27371933
$endgroup$
$begingroup$
Nice, why does homotopy apply? I've thought about it before.
$endgroup$
– Benjamin Thoburn
Jan 24 at 21:42
$begingroup$
Well, within the limits of the assumptions of homotopy, it always applies! Homotopy doesn't get you everywhere you need to go, but it is quite intuitive.
$endgroup$
– Adrian Keister
Jan 25 at 18:24
$begingroup$
I think I get it, it works because the integral is linear, you can write it as a sum of the integral along other paths, then each of those be be changed thus you can change the whole closed contour... I’ll think more about the gritty details of singularities.
$endgroup$
– Benjamin Thoburn
Jan 25 at 18:44
add a comment |
$begingroup$
I find homotopy to the most intuitive explanation: suppose you have two closed contours $gamma_1$ and $gamma_2$ and a function $f(z).$ If you can smoothly deform $gamma_1$ into $gamma_2$ without "crossing over" a singularity of $f,$ then
$$oint_{gamma_1}f(z),dz=oint_{gamma_2}f(z),dz.$$
So, for example, if you can smoothly (continuously) deform a closed contour down to a single point (not a singularity), then it would make sense the integral would be zero, right? So in that sense, whatever singularities start out inside your contour have to stay inside that contour, or you change the value of the integral.
answered Jan 24 at 21:35
Adrian KeisterAdrian Keister
5,27371933
$endgroup$
$begingroup$
Nice, why does homotopy apply? I've thought about it before.
$endgroup$
– Benjamin Thoburn
Jan 24 at 21:42
$begingroup$
Well, within the limits of the assumptions of homotopy, it always applies! Homotopy doesn't get you everywhere you need to go, but it is quite intuitive.
$endgroup$
– Adrian Keister
Jan 25 at 18:24
$begingroup$
I think I get it, it works because the integral is linear, you can write it as a sum of the integral along other paths, then each of those be be changed thus you can change the whole closed contour... I’ll think more about the gritty details of singularities.
$endgroup$
– Benjamin Thoburn
Jan 25 at 18:44
add a comment |
$begingroup$
I find homotopy to the most intuitive explanation: suppose you have two closed contours $gamma_1$ and $gamma_2$ and a function $f(z).$ If you can smoothly deform $gamma_1$ into $gamma_2$ without "crossing over" a singularity of $f,$ then
$$oint_{gamma_1}f(z),dz=oint_{gamma_2}f(z),dz.$$
So, for example, if you can smoothly (continuously) deform a closed contour down to a single point (not a singularity), then it would make sense the integral would be zero, right? So in that sense, whatever singularities start out inside your contour have to stay inside that contour, or you change the value of the integral.
answered Jan 24 at 21:35
Adrian KeisterAdrian Keister
5,27371933
$endgroup$
I find homotopy to the most intuitive explanation: suppose you have two closed contours $gamma_1$ and $gamma_2$ and a function $f(z).$ If you can smoothly deform $gamma_1$ into $gamma_2$ without "crossing over" a singularity of $f,$ then
$$oint_{gamma_1}f(z),dz=oint_{gamma_2}f(z),dz.$$
So, for example, if you can smoothly (continuously) deform a closed contour down to a single point (not a singularity), then it would make sense the integral would be zero, right? So in that sense, whatever singularities start out inside your contour have to stay inside that contour, or you change the value of the integral.
answered Jan 24 at 21:35
Adrian KeisterAdrian Keister
5,27371933
answered Jan 24 at 21:35
Adrian KeisterAdrian Keister
5,27371933
answered Jan 24 at 21:35
Adrian KeisterAdrian Keister
5,27371933
answered Jan 24 at 21:35
Adrian KeisterAdrian Keister
5,27371933
5,27371933
$begingroup$
Nice, why does homotopy apply? I've thought about it before.
$endgroup$
– Benjamin Thoburn
Jan 24 at 21:42
$begingroup$
Well, within the limits of the assumptions of homotopy, it always applies! Homotopy doesn't get you everywhere you need to go, but it is quite intuitive.
$endgroup$
– Adrian Keister
Jan 25 at 18:24
$begingroup$
I think I get it, it works because the integral is linear, you can write it as a sum of the integral along other paths, then each of those be be changed thus you can change the whole closed contour... I’ll think more about the gritty details of singularities.
$endgroup$
– Benjamin Thoburn
Jan 25 at 18:44
add a comment |
$begingroup$
Nice, why does homotopy apply? I've thought about it before.
$endgroup$
– Benjamin Thoburn
Jan 24 at 21:42
$begingroup$
Well, within the limits of the assumptions of homotopy, it always applies! Homotopy doesn't get you everywhere you need to go, but it is quite intuitive.
$endgroup$
– Adrian Keister
Jan 25 at 18:24
$begingroup$
I think I get it, it works because the integral is linear, you can write it as a sum of the integral along other paths, then each of those be be changed thus you can change the whole closed contour... I’ll think more about the gritty details of singularities.
$endgroup$
– Benjamin Thoburn
Jan 25 at 18:44
$begingroup$
Nice, why does homotopy apply? I've thought about it before.
$endgroup$
– Benjamin Thoburn
Jan 24 at 21:42
$begingroup$
Nice, why does homotopy apply? I've thought about it before.
$endgroup$
– Benjamin Thoburn
Jan 24 at 21:42
$begingroup$
Well, within the limits of the assumptions of homotopy, it always applies! Homotopy doesn't get you everywhere you need to go, but it is quite intuitive.
$endgroup$
– Adrian Keister
Jan 25 at 18:24
$begingroup$
Well, within the limits of the assumptions of homotopy, it always applies! Homotopy doesn't get you everywhere you need to go, but it is quite intuitive.
$endgroup$
– Adrian Keister
Jan 25 at 18:24
$begingroup$
I think I get it, it works because the integral is linear, you can write it as a sum of the integral along other paths, then each of those be be changed thus you can change the whole closed contour... I’ll think more about the gritty details of singularities.
$endgroup$
– Benjamin Thoburn
Jan 25 at 18:44
$begingroup$
I think I get it, it works because the integral is linear, you can write it as a sum of the integral along other paths, then each of those be be changed thus you can change the whole closed contour... I’ll think more about the gritty details of singularities.
$endgroup$
– Benjamin Thoburn
Jan 25 at 18:44
add a comment |
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$begingroup$
Are you familiar with Cauchy's Integral Theorem? If $f$ is analytic in and on a rectifiable curve, then the contour integral on that path is $0$. Now, deform the contour to exclude any singularities and proceed.
$endgroup$
– Mark Viola
Jan 24 at 21:24