Evaluate $int_gammafrac{1}{z}dz$ over a circle outside the origin











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I am trying to do the evaluate this integral:



$$int_{gamma} frac{1}{z} dz$$



where $gamma$ is a circle that does not contain the origin on itself either on its inside, I mean, $gamma$ is a circle center at $z_0 neq 0$ with radious $r < |z_0|$



My attempt:



Since the idea of the exercise is to evaluate the integral to get an insight into later results (Cauchy's theorem) I made a parametrization:
$z(t) = z_0 + re^{it}$ with $ 0 leq t leq 2pi$. So I had



$$int_{gamma} frac{1}{z} dz = int_{0}^{2pi} frac{1}{z(t)}z'(t)dt = int_0^{2pi}frac{ire^{it}}{z_0 + re^{it}}dt$$



Here what I did to evaluate this integral is a change of variable:
call $u = z_0 + re^{it}$ then $du = ire^{it}dt$ and when $t = 0$, $u = z_0 + r$ and when $t = 2pi$, $u = z_0 + r$. So now I had:



$$int_{gamma} f(z) dz = int_{z_0 + r}^{z_0 + r} frac{1}{u}du = 0$$



Where the last integral is due to fact that the limits of the integral are equal.



I am not quite confident with my change of variable because:




  • What I get after my change of variable is totally trivially and it totally forget the integrant function ($1/u$).


  • The notes that I was provided take a totally different approach: they use the fact that $frac{1}{z}$ is analytic in $gamma$ and the result that you can interchange the limit and the integral sign (in fact. they use some tricky steps to justify this) to come to the same answer ( exercise three, part b) )



So, will you please say me if what I did is correct or not¡? Thanks in advance!










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  • When you define $u$ that way you don't get an integral over a single point, you get an integral over a closed curve.
    – Umberto P.
    Nov 17 at 1:41










  • This seems OK. You have made use of that $r<mid z_0 mid$ in last step. Think about why this is not working for the otherwise case.
    – AdditIdent
    Nov 17 at 1:46















up vote
1
down vote

favorite












I am trying to do the evaluate this integral:



$$int_{gamma} frac{1}{z} dz$$



where $gamma$ is a circle that does not contain the origin on itself either on its inside, I mean, $gamma$ is a circle center at $z_0 neq 0$ with radious $r < |z_0|$



My attempt:



Since the idea of the exercise is to evaluate the integral to get an insight into later results (Cauchy's theorem) I made a parametrization:
$z(t) = z_0 + re^{it}$ with $ 0 leq t leq 2pi$. So I had



$$int_{gamma} frac{1}{z} dz = int_{0}^{2pi} frac{1}{z(t)}z'(t)dt = int_0^{2pi}frac{ire^{it}}{z_0 + re^{it}}dt$$



Here what I did to evaluate this integral is a change of variable:
call $u = z_0 + re^{it}$ then $du = ire^{it}dt$ and when $t = 0$, $u = z_0 + r$ and when $t = 2pi$, $u = z_0 + r$. So now I had:



$$int_{gamma} f(z) dz = int_{z_0 + r}^{z_0 + r} frac{1}{u}du = 0$$



Where the last integral is due to fact that the limits of the integral are equal.



I am not quite confident with my change of variable because:




  • What I get after my change of variable is totally trivially and it totally forget the integrant function ($1/u$).


  • The notes that I was provided take a totally different approach: they use the fact that $frac{1}{z}$ is analytic in $gamma$ and the result that you can interchange the limit and the integral sign (in fact. they use some tricky steps to justify this) to come to the same answer ( exercise three, part b) )



So, will you please say me if what I did is correct or not¡? Thanks in advance!










share|cite|improve this question
























  • When you define $u$ that way you don't get an integral over a single point, you get an integral over a closed curve.
    – Umberto P.
    Nov 17 at 1:41










  • This seems OK. You have made use of that $r<mid z_0 mid$ in last step. Think about why this is not working for the otherwise case.
    – AdditIdent
    Nov 17 at 1:46













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I am trying to do the evaluate this integral:



$$int_{gamma} frac{1}{z} dz$$



where $gamma$ is a circle that does not contain the origin on itself either on its inside, I mean, $gamma$ is a circle center at $z_0 neq 0$ with radious $r < |z_0|$



My attempt:



Since the idea of the exercise is to evaluate the integral to get an insight into later results (Cauchy's theorem) I made a parametrization:
$z(t) = z_0 + re^{it}$ with $ 0 leq t leq 2pi$. So I had



$$int_{gamma} frac{1}{z} dz = int_{0}^{2pi} frac{1}{z(t)}z'(t)dt = int_0^{2pi}frac{ire^{it}}{z_0 + re^{it}}dt$$



Here what I did to evaluate this integral is a change of variable:
call $u = z_0 + re^{it}$ then $du = ire^{it}dt$ and when $t = 0$, $u = z_0 + r$ and when $t = 2pi$, $u = z_0 + r$. So now I had:



$$int_{gamma} f(z) dz = int_{z_0 + r}^{z_0 + r} frac{1}{u}du = 0$$



Where the last integral is due to fact that the limits of the integral are equal.



I am not quite confident with my change of variable because:




  • What I get after my change of variable is totally trivially and it totally forget the integrant function ($1/u$).


  • The notes that I was provided take a totally different approach: they use the fact that $frac{1}{z}$ is analytic in $gamma$ and the result that you can interchange the limit and the integral sign (in fact. they use some tricky steps to justify this) to come to the same answer ( exercise three, part b) )



So, will you please say me if what I did is correct or not¡? Thanks in advance!










share|cite|improve this question















I am trying to do the evaluate this integral:



$$int_{gamma} frac{1}{z} dz$$



where $gamma$ is a circle that does not contain the origin on itself either on its inside, I mean, $gamma$ is a circle center at $z_0 neq 0$ with radious $r < |z_0|$



My attempt:



Since the idea of the exercise is to evaluate the integral to get an insight into later results (Cauchy's theorem) I made a parametrization:
$z(t) = z_0 + re^{it}$ with $ 0 leq t leq 2pi$. So I had



$$int_{gamma} frac{1}{z} dz = int_{0}^{2pi} frac{1}{z(t)}z'(t)dt = int_0^{2pi}frac{ire^{it}}{z_0 + re^{it}}dt$$



Here what I did to evaluate this integral is a change of variable:
call $u = z_0 + re^{it}$ then $du = ire^{it}dt$ and when $t = 0$, $u = z_0 + r$ and when $t = 2pi$, $u = z_0 + r$. So now I had:



$$int_{gamma} f(z) dz = int_{z_0 + r}^{z_0 + r} frac{1}{u}du = 0$$



Where the last integral is due to fact that the limits of the integral are equal.



I am not quite confident with my change of variable because:




  • What I get after my change of variable is totally trivially and it totally forget the integrant function ($1/u$).


  • The notes that I was provided take a totally different approach: they use the fact that $frac{1}{z}$ is analytic in $gamma$ and the result that you can interchange the limit and the integral sign (in fact. they use some tricky steps to justify this) to come to the same answer ( exercise three, part b) )



So, will you please say me if what I did is correct or not¡? Thanks in advance!







complex-analysis contour-integration






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edited Nov 17 at 1:17









Tianlalu

2,8811835




2,8811835










asked Nov 17 at 1:04









Aldebaran

284




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  • When you define $u$ that way you don't get an integral over a single point, you get an integral over a closed curve.
    – Umberto P.
    Nov 17 at 1:41










  • This seems OK. You have made use of that $r<mid z_0 mid$ in last step. Think about why this is not working for the otherwise case.
    – AdditIdent
    Nov 17 at 1:46


















  • When you define $u$ that way you don't get an integral over a single point, you get an integral over a closed curve.
    – Umberto P.
    Nov 17 at 1:41










  • This seems OK. You have made use of that $r<mid z_0 mid$ in last step. Think about why this is not working for the otherwise case.
    – AdditIdent
    Nov 17 at 1:46
















When you define $u$ that way you don't get an integral over a single point, you get an integral over a closed curve.
– Umberto P.
Nov 17 at 1:41




When you define $u$ that way you don't get an integral over a single point, you get an integral over a closed curve.
– Umberto P.
Nov 17 at 1:41












This seems OK. You have made use of that $r<mid z_0 mid$ in last step. Think about why this is not working for the otherwise case.
– AdditIdent
Nov 17 at 1:46




This seems OK. You have made use of that $r<mid z_0 mid$ in last step. Think about why this is not working for the otherwise case.
– AdditIdent
Nov 17 at 1:46










1 Answer
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1
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The integral $int_a^b f(x),dx$ only has a precise meaning when $a$ and $b$ are real. If $a$ and $b$ are complex, you could define it to mean the integral along a straight-line path in the complex plane from $a$ to $b$, but line integrals are not always path independent. The technique of $u$-substitution is derived from the fundamental theorem of calculus, a theorem about real-valued functions on the reals. This theorem (and therefore the $u$-substitution technique) doesn’t necessarily work if you use complex numbers where real numbers appear in the theorem. (Lots of things from real analysis don’t generalize. For example, the definition of $f'(z)$ as a limit where $hto0$ is wrong, because this is only the derivative in a particular direction in the complex plane.) A full understanding of these issues is a good chunk of a course in complex analysis.



Short answer: $u$-substitution (or “change of variable” for a definite integral) is a technique for functions of real numbers.



(Worth noting: Even for functions of real numbers, the “change the limits” approach for changing variables in a definite integral $int_a^b f(x),dx$ can fail if the function $u(x)$ used for the substitution doesn’t map the interval $[a,b]$ to $[u(a),u(b)]$ in one-to-one fashion. Unfortunately, this issue is often ignored in elementary calculus courses.)






share|cite|improve this answer





















  • wao!! than you so much... Now everything is clearer... I was almost sure that what I did was wrong but now I know why... Just one more favor. In about three weeks I will be on vacations and I really want to learn much more about complex analysis, so... could you recommend me some great books¡?¡
    – Aldebaran
    Nov 17 at 4:13










  • There are good suggestions here: math.stackexchange.com/questions/2107445/complex-analysis-books. You could also search Math StackExchange for other discussions of your question.
    – Steve Kass
    Nov 17 at 19:01











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes








up vote
1
down vote













The integral $int_a^b f(x),dx$ only has a precise meaning when $a$ and $b$ are real. If $a$ and $b$ are complex, you could define it to mean the integral along a straight-line path in the complex plane from $a$ to $b$, but line integrals are not always path independent. The technique of $u$-substitution is derived from the fundamental theorem of calculus, a theorem about real-valued functions on the reals. This theorem (and therefore the $u$-substitution technique) doesn’t necessarily work if you use complex numbers where real numbers appear in the theorem. (Lots of things from real analysis don’t generalize. For example, the definition of $f'(z)$ as a limit where $hto0$ is wrong, because this is only the derivative in a particular direction in the complex plane.) A full understanding of these issues is a good chunk of a course in complex analysis.



Short answer: $u$-substitution (or “change of variable” for a definite integral) is a technique for functions of real numbers.



(Worth noting: Even for functions of real numbers, the “change the limits” approach for changing variables in a definite integral $int_a^b f(x),dx$ can fail if the function $u(x)$ used for the substitution doesn’t map the interval $[a,b]$ to $[u(a),u(b)]$ in one-to-one fashion. Unfortunately, this issue is often ignored in elementary calculus courses.)






share|cite|improve this answer





















  • wao!! than you so much... Now everything is clearer... I was almost sure that what I did was wrong but now I know why... Just one more favor. In about three weeks I will be on vacations and I really want to learn much more about complex analysis, so... could you recommend me some great books¡?¡
    – Aldebaran
    Nov 17 at 4:13










  • There are good suggestions here: math.stackexchange.com/questions/2107445/complex-analysis-books. You could also search Math StackExchange for other discussions of your question.
    – Steve Kass
    Nov 17 at 19:01















up vote
1
down vote













The integral $int_a^b f(x),dx$ only has a precise meaning when $a$ and $b$ are real. If $a$ and $b$ are complex, you could define it to mean the integral along a straight-line path in the complex plane from $a$ to $b$, but line integrals are not always path independent. The technique of $u$-substitution is derived from the fundamental theorem of calculus, a theorem about real-valued functions on the reals. This theorem (and therefore the $u$-substitution technique) doesn’t necessarily work if you use complex numbers where real numbers appear in the theorem. (Lots of things from real analysis don’t generalize. For example, the definition of $f'(z)$ as a limit where $hto0$ is wrong, because this is only the derivative in a particular direction in the complex plane.) A full understanding of these issues is a good chunk of a course in complex analysis.



Short answer: $u$-substitution (or “change of variable” for a definite integral) is a technique for functions of real numbers.



(Worth noting: Even for functions of real numbers, the “change the limits” approach for changing variables in a definite integral $int_a^b f(x),dx$ can fail if the function $u(x)$ used for the substitution doesn’t map the interval $[a,b]$ to $[u(a),u(b)]$ in one-to-one fashion. Unfortunately, this issue is often ignored in elementary calculus courses.)






share|cite|improve this answer





















  • wao!! than you so much... Now everything is clearer... I was almost sure that what I did was wrong but now I know why... Just one more favor. In about three weeks I will be on vacations and I really want to learn much more about complex analysis, so... could you recommend me some great books¡?¡
    – Aldebaran
    Nov 17 at 4:13










  • There are good suggestions here: math.stackexchange.com/questions/2107445/complex-analysis-books. You could also search Math StackExchange for other discussions of your question.
    – Steve Kass
    Nov 17 at 19:01













up vote
1
down vote










up vote
1
down vote









The integral $int_a^b f(x),dx$ only has a precise meaning when $a$ and $b$ are real. If $a$ and $b$ are complex, you could define it to mean the integral along a straight-line path in the complex plane from $a$ to $b$, but line integrals are not always path independent. The technique of $u$-substitution is derived from the fundamental theorem of calculus, a theorem about real-valued functions on the reals. This theorem (and therefore the $u$-substitution technique) doesn’t necessarily work if you use complex numbers where real numbers appear in the theorem. (Lots of things from real analysis don’t generalize. For example, the definition of $f'(z)$ as a limit where $hto0$ is wrong, because this is only the derivative in a particular direction in the complex plane.) A full understanding of these issues is a good chunk of a course in complex analysis.



Short answer: $u$-substitution (or “change of variable” for a definite integral) is a technique for functions of real numbers.



(Worth noting: Even for functions of real numbers, the “change the limits” approach for changing variables in a definite integral $int_a^b f(x),dx$ can fail if the function $u(x)$ used for the substitution doesn’t map the interval $[a,b]$ to $[u(a),u(b)]$ in one-to-one fashion. Unfortunately, this issue is often ignored in elementary calculus courses.)






share|cite|improve this answer












The integral $int_a^b f(x),dx$ only has a precise meaning when $a$ and $b$ are real. If $a$ and $b$ are complex, you could define it to mean the integral along a straight-line path in the complex plane from $a$ to $b$, but line integrals are not always path independent. The technique of $u$-substitution is derived from the fundamental theorem of calculus, a theorem about real-valued functions on the reals. This theorem (and therefore the $u$-substitution technique) doesn’t necessarily work if you use complex numbers where real numbers appear in the theorem. (Lots of things from real analysis don’t generalize. For example, the definition of $f'(z)$ as a limit where $hto0$ is wrong, because this is only the derivative in a particular direction in the complex plane.) A full understanding of these issues is a good chunk of a course in complex analysis.



Short answer: $u$-substitution (or “change of variable” for a definite integral) is a technique for functions of real numbers.



(Worth noting: Even for functions of real numbers, the “change the limits” approach for changing variables in a definite integral $int_a^b f(x),dx$ can fail if the function $u(x)$ used for the substitution doesn’t map the interval $[a,b]$ to $[u(a),u(b)]$ in one-to-one fashion. Unfortunately, this issue is often ignored in elementary calculus courses.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 17 at 2:15









Steve Kass

10.9k11429




10.9k11429












  • wao!! than you so much... Now everything is clearer... I was almost sure that what I did was wrong but now I know why... Just one more favor. In about three weeks I will be on vacations and I really want to learn much more about complex analysis, so... could you recommend me some great books¡?¡
    – Aldebaran
    Nov 17 at 4:13










  • There are good suggestions here: math.stackexchange.com/questions/2107445/complex-analysis-books. You could also search Math StackExchange for other discussions of your question.
    – Steve Kass
    Nov 17 at 19:01


















  • wao!! than you so much... Now everything is clearer... I was almost sure that what I did was wrong but now I know why... Just one more favor. In about three weeks I will be on vacations and I really want to learn much more about complex analysis, so... could you recommend me some great books¡?¡
    – Aldebaran
    Nov 17 at 4:13










  • There are good suggestions here: math.stackexchange.com/questions/2107445/complex-analysis-books. You could also search Math StackExchange for other discussions of your question.
    – Steve Kass
    Nov 17 at 19:01
















wao!! than you so much... Now everything is clearer... I was almost sure that what I did was wrong but now I know why... Just one more favor. In about three weeks I will be on vacations and I really want to learn much more about complex analysis, so... could you recommend me some great books¡?¡
– Aldebaran
Nov 17 at 4:13




wao!! than you so much... Now everything is clearer... I was almost sure that what I did was wrong but now I know why... Just one more favor. In about three weeks I will be on vacations and I really want to learn much more about complex analysis, so... could you recommend me some great books¡?¡
– Aldebaran
Nov 17 at 4:13












There are good suggestions here: math.stackexchange.com/questions/2107445/complex-analysis-books. You could also search Math StackExchange for other discussions of your question.
– Steve Kass
Nov 17 at 19:01




There are good suggestions here: math.stackexchange.com/questions/2107445/complex-analysis-books. You could also search Math StackExchange for other discussions of your question.
– Steve Kass
Nov 17 at 19:01


















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