$lim int_0^infty frac1n e^-fractn dt ne int_0^infty lim frac1n e^-fractn dt$ in complex as in real?

Clash Royale CLAN TAG#URR8PPP

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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka

This is probably related to a real analysis or even elementary analysis classic counterexample that I forgot. Anyhoo, why doesn't the switch hold?

1. I was about to say that Prop 7.27 doesn't apply because $gamma = G$. Unfortunately, the inclusion is loose.




2. Um, does $G$ have to be a region? If so, then I guess $mathbb R_ge 0$ is not a region? If not, then perhaps




3. $f_n(t)$ is not continuous because the left hand limit at $t=0$ does not exist because $f_n(t)$ is not defined for $t<0$? In that case, what changes if we redefine $f_n: mathbb R to mathbb R$ ?




4. There's also that $f_n$'s codomain isn't C. As with #3, what changes if we redefine $f_n: mathbb R_ge 0 to mathbb C$ ?




5. In re #3 and #4, what changes if we redefine $f_n: mathbb R to mathbb C$ ?

real-analysis complex-analysis uniform-convergence contour-integration

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edited Aug 5 at 11:55

asked Jul 23 at 12:25

BCLC

6,89921973

add a comment | 

up vote
0
down vote

favorite

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka

This is probably related to a real analysis or even elementary analysis classic counterexample that I forgot. Anyhoo, why doesn't the switch hold?

1. I was about to say that Prop 7.27 doesn't apply because $gamma = G$. Unfortunately, the inclusion is loose.




2. Um, does $G$ have to be a region? If so, then I guess $mathbb R_ge 0$ is not a region? If not, then perhaps




3. $f_n(t)$ is not continuous because the left hand limit at $t=0$ does not exist because $f_n(t)$ is not defined for $t<0$? In that case, what changes if we redefine $f_n: mathbb R to mathbb R$ ?




4. There's also that $f_n$'s codomain isn't C. As with #3, what changes if we redefine $f_n: mathbb R_ge 0 to mathbb C$ ?




5. In re #3 and #4, what changes if we redefine $f_n: mathbb R to mathbb C$ ?

real-analysis complex-analysis uniform-convergence contour-integration

share|cite|improve this question

edited Aug 5 at 11:55

asked Jul 23 at 12:25

BCLC

6,89921973

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka

This is probably related to a real analysis or even elementary analysis classic counterexample that I forgot. Anyhoo, why doesn't the switch hold?

1. I was about to say that Prop 7.27 doesn't apply because $gamma = G$. Unfortunately, the inclusion is loose.




2. Um, does $G$ have to be a region? If so, then I guess $mathbb R_ge 0$ is not a region? If not, then perhaps




3. $f_n(t)$ is not continuous because the left hand limit at $t=0$ does not exist because $f_n(t)$ is not defined for $t<0$? In that case, what changes if we redefine $f_n: mathbb R to mathbb R$ ?




4. There's also that $f_n$'s codomain isn't C. As with #3, what changes if we redefine $f_n: mathbb R_ge 0 to mathbb C$ ?




5. In re #3 and #4, what changes if we redefine $f_n: mathbb R to mathbb C$ ?

real-analysis complex-analysis uniform-convergence contour-integration

share|cite|improve this question

edited Aug 5 at 11:55

asked Jul 23 at 12:25

BCLC

6,89921973

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka

This is probably related to a real analysis or even elementary analysis classic counterexample that I forgot. Anyhoo, why doesn't the switch hold?

1. I was about to say that Prop 7.27 doesn't apply because $gamma = G$. Unfortunately, the inclusion is loose.




2. Um, does $G$ have to be a region? If so, then I guess $mathbb R_ge 0$ is not a region? If not, then perhaps




3. $f_n(t)$ is not continuous because the left hand limit at $t=0$ does not exist because $f_n(t)$ is not defined for $t<0$? In that case, what changes if we redefine $f_n: mathbb R to mathbb R$ ?




4. There's also that $f_n$'s codomain isn't C. As with #3, what changes if we redefine $f_n: mathbb R_ge 0 to mathbb C$ ?




5. In re #3 and #4, what changes if we redefine $f_n: mathbb R to mathbb C$ ?

real-analysis complex-analysis uniform-convergence contour-integration

share|cite|improve this question

edited Aug 5 at 11:55

asked Jul 23 at 12:25

BCLC

6,89921973

share|cite|improve this question

share|cite|improve this question

edited Aug 5 at 11:55

edited Aug 5 at 11:55

edited Aug 5 at 11:55

asked Jul 23 at 12:25

BCLC

6,89921973

asked Jul 23 at 12:25

BCLC

6,89921973

asked Jul 23 at 12:25

BCLC

6,89921973

6,89921973

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

We can extend the $f_n$ to functions $tilde f_n:mathbb Ctomathbb R$ which converge uniformly on $mathbb C$, by defining $tilde f_n(z)=f_n(textRe(z))$ for $Re(z)geq0$ and $tilde f_n(z)=frac1n$ otherwise, so the issue isn't with domain of the functions.

Furthermore, changing the codomain to be $mathbb C$ doesn't affect the behavior of the function, as it converges uniformly whenever we consider the codomain to be $mathbb R_>0$, $mathbb R$, $mathbb C$, or any other reasonable subset of $mathbb C$.

The issue is with the "contour" over which we are integrating. The theorem applies to paths, whose images are compact subsets of $mathbb C$, and $mathbb R_geq0$ is not a compact subset of $mathbb C$.

share|cite|improve this answer

edited Jul 23 at 13:20

answered Jul 23 at 12:50

Aweygan

11.9k21437

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Aweygan! Prop 7.27 that path is piecewise smooth. Where does it say in Prop 7.27 anything (implicitly I guess) about compactness?
  – BCLC
  Jul 23 at 13:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am unsure of what you are asking in your second comment.
  – Aweygan
  Jul 23 at 13:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Check the definition of a piecewise smooth curve in the text you link to. They are defined as images of smooth functions over a finite number of disjoint closed (hence compact) intervals, so the images of the curves are compact subsets of $mathbb C$, hence the image of the path is compact in $mathbb C$.
  – Aweygan
  Jul 23 at 13:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If you check your real analysis book, any analogous theorem would probably require that you be integrating over a bounded interval $[a,b]$, not over $[0,infty)$.
  – Aweygan
  Jul 23 at 13:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BCLC You have it now! The integral extends over the entire positive reals and so the referenced theorem does not apply. However, there is a theorem that would permit the interchange of the limit and the integral IF the improper integral converged uniformly. Here, the improper integral does not converge uniformly despite the fact that sequence $f_n(x)$ converges uniformly.
  – Mark Viola
  Jul 23 at 16:23
  
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote

Note that in the proof of given, a key quantity is $textlength(gamma)$. Uniform convergence of $f_n$ is good enough to switch the order when the measure of $G$ is finite:
$$
lim_ntoinftyint_Gf_n(x),mathrmdx=int_Glim_ntoinftyf_n(x),mathrmdx
$$
For the example in the title, the length of the path, $[0,infty)$ is not finite.

share|cite|improve this answer

answered Jul 23 at 15:14

robjohn♦

258k26297612

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good catch and explanation. Thanks!
  – BCLC
  Jul 23 at 15:17
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

We can extend the $f_n$ to functions $tilde f_n:mathbb Ctomathbb R$ which converge uniformly on $mathbb C$, by defining $tilde f_n(z)=f_n(textRe(z))$ for $Re(z)geq0$ and $tilde f_n(z)=frac1n$ otherwise, so the issue isn't with domain of the functions.

Furthermore, changing the codomain to be $mathbb C$ doesn't affect the behavior of the function, as it converges uniformly whenever we consider the codomain to be $mathbb R_>0$, $mathbb R$, $mathbb C$, or any other reasonable subset of $mathbb C$.

The issue is with the "contour" over which we are integrating. The theorem applies to paths, whose images are compact subsets of $mathbb C$, and $mathbb R_geq0$ is not a compact subset of $mathbb C$.

share|cite|improve this answer

edited Jul 23 at 13:20

answered Jul 23 at 12:50

Aweygan

11.9k21437

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Aweygan! Prop 7.27 that path is piecewise smooth. Where does it say in Prop 7.27 anything (implicitly I guess) about compactness?
  – BCLC
  Jul 23 at 13:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am unsure of what you are asking in your second comment.
  – Aweygan
  Jul 23 at 13:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Check the definition of a piecewise smooth curve in the text you link to. They are defined as images of smooth functions over a finite number of disjoint closed (hence compact) intervals, so the images of the curves are compact subsets of $mathbb C$, hence the image of the path is compact in $mathbb C$.
  – Aweygan
  Jul 23 at 13:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If you check your real analysis book, any analogous theorem would probably require that you be integrating over a bounded interval $[a,b]$, not over $[0,infty)$.
  – Aweygan
  Jul 23 at 13:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BCLC You have it now! The integral extends over the entire positive reals and so the referenced theorem does not apply. However, there is a theorem that would permit the interchange of the limit and the integral IF the improper integral converged uniformly. Here, the improper integral does not converge uniformly despite the fact that sequence $f_n(x)$ converges uniformly.
  – Mark Viola
  Jul 23 at 16:23
  
  

  
  
  
  

 | 
show 4 more comments

up vote
2
down vote



accepted

We can extend the $f_n$ to functions $tilde f_n:mathbb Ctomathbb R$ which converge uniformly on $mathbb C$, by defining $tilde f_n(z)=f_n(textRe(z))$ for $Re(z)geq0$ and $tilde f_n(z)=frac1n$ otherwise, so the issue isn't with domain of the functions.

Furthermore, changing the codomain to be $mathbb C$ doesn't affect the behavior of the function, as it converges uniformly whenever we consider the codomain to be $mathbb R_>0$, $mathbb R$, $mathbb C$, or any other reasonable subset of $mathbb C$.

The issue is with the "contour" over which we are integrating. The theorem applies to paths, whose images are compact subsets of $mathbb C$, and $mathbb R_geq0$ is not a compact subset of $mathbb C$.

share|cite|improve this answer

edited Jul 23 at 13:20

answered Jul 23 at 12:50

Aweygan

11.9k21437

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Aweygan! Prop 7.27 that path is piecewise smooth. Where does it say in Prop 7.27 anything (implicitly I guess) about compactness?
  – BCLC
  Jul 23 at 13:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am unsure of what you are asking in your second comment.
  – Aweygan
  Jul 23 at 13:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Check the definition of a piecewise smooth curve in the text you link to. They are defined as images of smooth functions over a finite number of disjoint closed (hence compact) intervals, so the images of the curves are compact subsets of $mathbb C$, hence the image of the path is compact in $mathbb C$.
  – Aweygan
  Jul 23 at 13:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If you check your real analysis book, any analogous theorem would probably require that you be integrating over a bounded interval $[a,b]$, not over $[0,infty)$.
  – Aweygan
  Jul 23 at 13:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BCLC You have it now! The integral extends over the entire positive reals and so the referenced theorem does not apply. However, there is a theorem that would permit the interchange of the limit and the integral IF the improper integral converged uniformly. Here, the improper integral does not converge uniformly despite the fact that sequence $f_n(x)$ converges uniformly.
  – Mark Viola
  Jul 23 at 16:23
  
  

  
  
  
  

 | 
show 4 more comments

up vote
2
down vote



accepted

up vote
2
down vote



accepted

We can extend the $f_n$ to functions $tilde f_n:mathbb Ctomathbb R$ which converge uniformly on $mathbb C$, by defining $tilde f_n(z)=f_n(textRe(z))$ for $Re(z)geq0$ and $tilde f_n(z)=frac1n$ otherwise, so the issue isn't with domain of the functions.

Furthermore, changing the codomain to be $mathbb C$ doesn't affect the behavior of the function, as it converges uniformly whenever we consider the codomain to be $mathbb R_>0$, $mathbb R$, $mathbb C$, or any other reasonable subset of $mathbb C$.

The issue is with the "contour" over which we are integrating. The theorem applies to paths, whose images are compact subsets of $mathbb C$, and $mathbb R_geq0$ is not a compact subset of $mathbb C$.

share|cite|improve this answer

edited Jul 23 at 13:20

answered Jul 23 at 12:50

Aweygan

11.9k21437

We can extend the $f_n$ to functions $tilde f_n:mathbb Ctomathbb R$ which converge uniformly on $mathbb C$, by defining $tilde f_n(z)=f_n(textRe(z))$ for $Re(z)geq0$ and $tilde f_n(z)=frac1n$ otherwise, so the issue isn't with domain of the functions.

Furthermore, changing the codomain to be $mathbb C$ doesn't affect the behavior of the function, as it converges uniformly whenever we consider the codomain to be $mathbb R_>0$, $mathbb R$, $mathbb C$, or any other reasonable subset of $mathbb C$.

The issue is with the "contour" over which we are integrating. The theorem applies to paths, whose images are compact subsets of $mathbb C$, and $mathbb R_geq0$ is not a compact subset of $mathbb C$.

share|cite|improve this answer

edited Jul 23 at 13:20

answered Jul 23 at 12:50

Aweygan

11.9k21437

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 13:20

edited Jul 23 at 13:20

edited Jul 23 at 13:20

answered Jul 23 at 12:50

Aweygan

11.9k21437

answered Jul 23 at 12:50

Aweygan

11.9k21437

answered Jul 23 at 12:50

Aweygan

11.9k21437

11.9k21437

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Aweygan! Prop 7.27 that path is piecewise smooth. Where does it say in Prop 7.27 anything (implicitly I guess) about compactness?
  – BCLC
  Jul 23 at 13:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am unsure of what you are asking in your second comment.
  – Aweygan
  Jul 23 at 13:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Check the definition of a piecewise smooth curve in the text you link to. They are defined as images of smooth functions over a finite number of disjoint closed (hence compact) intervals, so the images of the curves are compact subsets of $mathbb C$, hence the image of the path is compact in $mathbb C$.
  – Aweygan
  Jul 23 at 13:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If you check your real analysis book, any analogous theorem would probably require that you be integrating over a bounded interval $[a,b]$, not over $[0,infty)$.
  – Aweygan
  Jul 23 at 13:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BCLC You have it now! The integral extends over the entire positive reals and so the referenced theorem does not apply. However, there is a theorem that would permit the interchange of the limit and the integral IF the improper integral converged uniformly. Here, the improper integral does not converge uniformly despite the fact that sequence $f_n(x)$ converges uniformly.
  – Mark Viola
  Jul 23 at 16:23
  
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Aweygan! Prop 7.27 that path is piecewise smooth. Where does it say in Prop 7.27 anything (implicitly I guess) about compactness?
  – BCLC
  Jul 23 at 13:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am unsure of what you are asking in your second comment.
  – Aweygan
  Jul 23 at 13:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Check the definition of a piecewise smooth curve in the text you link to. They are defined as images of smooth functions over a finite number of disjoint closed (hence compact) intervals, so the images of the curves are compact subsets of $mathbb C$, hence the image of the path is compact in $mathbb C$.
  – Aweygan
  Jul 23 at 13:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If you check your real analysis book, any analogous theorem would probably require that you be integrating over a bounded interval $[a,b]$, not over $[0,infty)$.
  – Aweygan
  Jul 23 at 13:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BCLC You have it now! The integral extends over the entire positive reals and so the referenced theorem does not apply. However, there is a theorem that would permit the interchange of the limit and the integral IF the improper integral converged uniformly. Here, the improper integral does not converge uniformly despite the fact that sequence $f_n(x)$ converges uniformly.
  – Mark Viola
  Jul 23 at 16:23
  
  

  
  
  
  

Thanks Aweygan! Prop 7.27 that path is piecewise smooth. Where does it say in Prop 7.27 anything (implicitly I guess) about compactness?
– BCLC
Jul 23 at 13:12

Thanks Aweygan! Prop 7.27 that path is piecewise smooth. Where does it say in Prop 7.27 anything (implicitly I guess) about compactness?
– BCLC
Jul 23 at 13:12

1

1

I am unsure of what you are asking in your second comment.
– Aweygan
Jul 23 at 13:21

I am unsure of what you are asking in your second comment.
– Aweygan
Jul 23 at 13:21

1

1

Check the definition of a piecewise smooth curve in the text you link to. They are defined as images of smooth functions over a finite number of disjoint closed (hence compact) intervals, so the images of the curves are compact subsets of $mathbb C$, hence the image of the path is compact in $mathbb C$.
– Aweygan
Jul 23 at 13:26

Check the definition of a piecewise smooth curve in the text you link to. They are defined as images of smooth functions over a finite number of disjoint closed (hence compact) intervals, so the images of the curves are compact subsets of $mathbb C$, hence the image of the path is compact in $mathbb C$.
– Aweygan
Jul 23 at 13:26

1

1

If you check your real analysis book, any analogous theorem would probably require that you be integrating over a bounded interval $[a,b]$, not over $[0,infty)$.
– Aweygan
Jul 23 at 13:27

If you check your real analysis book, any analogous theorem would probably require that you be integrating over a bounded interval $[a,b]$, not over $[0,infty)$.
– Aweygan
Jul 23 at 13:27

1

1

@BCLC You have it now! The integral extends over the entire positive reals and so the referenced theorem does not apply. However, there is a theorem that would permit the interchange of the limit and the integral IF the improper integral converged uniformly. Here, the improper integral does not converge uniformly despite the fact that sequence $f_n(x)$ converges uniformly.
– Mark Viola
Jul 23 at 16:23

@BCLC You have it now! The integral extends over the entire positive reals and so the referenced theorem does not apply. However, there is a theorem that would permit the interchange of the limit and the integral IF the improper integral converged uniformly. Here, the improper integral does not converge uniformly despite the fact that sequence $f_n(x)$ converges uniformly.
– Mark Viola
Jul 23 at 16:23

 | 
show 4 more comments

up vote
1
down vote

Note that in the proof of given, a key quantity is $textlength(gamma)$. Uniform convergence of $f_n$ is good enough to switch the order when the measure of $G$ is finite:
$$
lim_ntoinftyint_Gf_n(x),mathrmdx=int_Glim_ntoinftyf_n(x),mathrmdx
$$
For the example in the title, the length of the path, $[0,infty)$ is not finite.

share|cite|improve this answer

answered Jul 23 at 15:14

robjohn♦

258k26297612

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good catch and explanation. Thanks!
  – BCLC
  Jul 23 at 15:17
  

  
  
  
  

add a comment | 

up vote
1
down vote

Note that in the proof of given, a key quantity is $textlength(gamma)$. Uniform convergence of $f_n$ is good enough to switch the order when the measure of $G$ is finite:
$$
lim_ntoinftyint_Gf_n(x),mathrmdx=int_Glim_ntoinftyf_n(x),mathrmdx
$$
For the example in the title, the length of the path, $[0,infty)$ is not finite.

share|cite|improve this answer

answered Jul 23 at 15:14

robjohn♦

258k26297612

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good catch and explanation. Thanks!
  – BCLC
  Jul 23 at 15:17
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Note that in the proof of given, a key quantity is $textlength(gamma)$. Uniform convergence of $f_n$ is good enough to switch the order when the measure of $G$ is finite:
$$
lim_ntoinftyint_Gf_n(x),mathrmdx=int_Glim_ntoinftyf_n(x),mathrmdx
$$
For the example in the title, the length of the path, $[0,infty)$ is not finite.

share|cite|improve this answer

answered Jul 23 at 15:14

robjohn♦

258k26297612

Note that in the proof of given, a key quantity is $textlength(gamma)$. Uniform convergence of $f_n$ is good enough to switch the order when the measure of $G$ is finite:
$$
lim_ntoinftyint_Gf_n(x),mathrmdx=int_Glim_ntoinftyf_n(x),mathrmdx
$$
For the example in the title, the length of the path, $[0,infty)$ is not finite.

share|cite|improve this answer

answered Jul 23 at 15:14

robjohn♦

258k26297612

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 15:14

robjohn♦

258k26297612

answered Jul 23 at 15:14

robjohn♦

258k26297612

answered Jul 23 at 15:14

robjohn♦

258k26297612

258k26297612

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good catch and explanation. Thanks!
  – BCLC
  Jul 23 at 15:17
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good catch and explanation. Thanks!
  – BCLC
  Jul 23 at 15:17
  

  
  
  
  

Good catch and explanation. Thanks!
– BCLC
Jul 23 at 15:17

Good catch and explanation. Thanks!
– BCLC
Jul 23 at 15:17

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