Mixing
Help on contour integral on another answer on this site
Clash Royale CLAN TAG#URR8PPP
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Do you mind expanding on the part along the diagonal in the first answer by Robjohn to this question proof? Particularly how to achieve (3).
I am trying use a parameterization for $z=xe^i pi/4$ for x from $0$ to R and I am not getting the correct result.
$$int_diagonale^-z^2=int_0^R e^-(xe^i pi/4)^2e^ipi/4dx=int_0^R e^-x^2e^i pi/2e^ipi/4dx$$
proof-explanation contour-integration
asked Jul 20 at 20:04
MathIsHard
1,122415
add a comment |Â
up vote
1
down vote
favorite
Do you mind expanding on the part along the diagonal in the first answer by Robjohn to this question proof? Particularly how to achieve (3).
I am trying use a parameterization for $z=xe^i pi/4$ for x from $0$ to R and I am not getting the correct result.
$$int_diagonale^-z^2=int_0^R e^-(xe^i pi/4)^2e^ipi/4dx=int_0^R e^-x^2e^i pi/2e^ipi/4dx$$
proof-explanation contour-integration
asked Jul 20 at 20:04
MathIsHard
1,122415
1
$e^ipi/2=i$ and the integral is from $Re^ipi/4$ to 0 so you have to change the sign?
– Bob
Jul 20 at 20:14
Oh you might be right. Let me take a look at that in a few. Thanks. :)
– MathIsHard
Jul 20 at 20:39
Thank you. I see it now. I appreciate the help.
– MathIsHard
Jul 20 at 20:50
1
One of you should post that as an answer so that the question doesn't remain unanswered.
– joriki
Jul 20 at 23:06
Yes, please post that Bob or let me know if you want me to do it. Thanks!
– MathIsHard
Jul 20 at 23:34
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Do you mind expanding on the part along the diagonal in the first answer by Robjohn to this question proof? Particularly how to achieve (3).
I am trying use a parameterization for $z=xe^i pi/4$ for x from $0$ to R and I am not getting the correct result.
$$int_diagonale^-z^2=int_0^R e^-(xe^i pi/4)^2e^ipi/4dx=int_0^R e^-x^2e^i pi/2e^ipi/4dx$$
proof-explanation contour-integration
asked Jul 20 at 20:04
MathIsHard
1,122415
Do you mind expanding on the part along the diagonal in the first answer by Robjohn to this question proof? Particularly how to achieve (3).
I am trying use a parameterization for $z=xe^i pi/4$ for x from $0$ to R and I am not getting the correct result.
$$int_diagonale^-z^2=int_0^R e^-(xe^i pi/4)^2e^ipi/4dx=int_0^R e^-x^2e^i pi/2e^ipi/4dx$$
proof-explanation contour-integration
asked Jul 20 at 20:04
MathIsHard
1,122415
edited Jul 20 at 22:45
asked Jul 20 at 20:04
MathIsHard
1,122415
asked Jul 20 at 20:04
MathIsHard
1,122415
asked Jul 20 at 20:04
MathIsHard
1,122415
1,122415
1
$e^ipi/2=i$ and the integral is from $Re^ipi/4$ to 0 so you have to change the sign?
– Bob
Jul 20 at 20:14
Oh you might be right. Let me take a look at that in a few. Thanks. :)
– MathIsHard
Jul 20 at 20:39
Thank you. I see it now. I appreciate the help.
– MathIsHard
Jul 20 at 20:50
1
One of you should post that as an answer so that the question doesn't remain unanswered.
– joriki
Jul 20 at 23:06
Yes, please post that Bob or let me know if you want me to do it. Thanks!
– MathIsHard
Jul 20 at 23:34
add a comment |Â
1
$e^ipi/2=i$ and the integral is from $Re^ipi/4$ to 0 so you have to change the sign?
– Bob
Jul 20 at 20:14
Oh you might be right. Let me take a look at that in a few. Thanks. :)
– MathIsHard
Jul 20 at 20:39
Thank you. I see it now. I appreciate the help.
– MathIsHard
Jul 20 at 20:50
1
One of you should post that as an answer so that the question doesn't remain unanswered.
– joriki
Jul 20 at 23:06
Yes, please post that Bob or let me know if you want me to do it. Thanks!
– MathIsHard
Jul 20 at 23:34
1
1
$e^ipi/2=i$ and the integral is from $Re^ipi/4$ to 0 so you have to change the sign?
– Bob
Jul 20 at 20:14
$e^ipi/2=i$ and the integral is from $Re^ipi/4$ to 0 so you have to change the sign?
– Bob
Jul 20 at 20:14
Oh you might be right. Let me take a look at that in a few. Thanks. :)
– MathIsHard
Jul 20 at 20:39
Oh you might be right. Let me take a look at that in a few. Thanks. :)
– MathIsHard
Jul 20 at 20:39
Thank you. I see it now. I appreciate the help.
– MathIsHard
Jul 20 at 20:50
Thank you. I see it now. I appreciate the help.
– MathIsHard
Jul 20 at 20:50
1
1
One of you should post that as an answer so that the question doesn't remain unanswered.
– joriki
Jul 20 at 23:06
One of you should post that as an answer so that the question doesn't remain unanswered.
– joriki
Jul 20 at 23:06
Yes, please post that Bob or let me know if you want me to do it. Thanks!
– MathIsHard
Jul 20 at 23:34
Yes, please post that Bob or let me know if you want me to do it. Thanks!
– MathIsHard
Jul 20 at 23:34
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
Parametrize the segment from $0$ to $R e^ipi / 4$ by:
$$gamma :[0,R]rightarrow mathbbC, tmapsto te^i(pi/4).$$
Since you want to integrate from $Re^ipi/4$ to $0$ and not from $0$ to $Re^ipi/4$, you have to switch the orientation of this curve, with the result that the integral changes sign, so the value of the integral you're looking for is
$$-int_gamma e^-z^2dz= -int_0 ^ R e^-gamma(t)^2gamma'(t)dt = -int_0 ^R e^-(te^ipi/4)^2e^ipi/4dt=\-e^ipi/4int_0 ^R e^-t^2e^ipi/2dt=-e^ipi/4int_0 ^R e^-it^2dt,$$
where we used the well known fact that $e^ipi/2=i.$ Now, letting $Rrightarrowinfty$ you get the result claimed in the linked answer.
answered Jul 21 at 3:39
Bob
1,467522
Thanks Bob :) appreciate it.
– MathIsHard
Jul 21 at 5:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Parametrize the segment from $0$ to $R e^ipi / 4$ by:
$$gamma :[0,R]rightarrow mathbbC, tmapsto te^i(pi/4).$$
Since you want to integrate from $Re^ipi/4$ to $0$ and not from $0$ to $Re^ipi/4$, you have to switch the orientation of this curve, with the result that the integral changes sign, so the value of the integral you're looking for is
$$-int_gamma e^-z^2dz= -int_0 ^ R e^-gamma(t)^2gamma'(t)dt = -int_0 ^R e^-(te^ipi/4)^2e^ipi/4dt=\-e^ipi/4int_0 ^R e^-t^2e^ipi/2dt=-e^ipi/4int_0 ^R e^-it^2dt,$$
where we used the well known fact that $e^ipi/2=i.$ Now, letting $Rrightarrowinfty$ you get the result claimed in the linked answer.
answered Jul 21 at 3:39
Bob
1,467522
Thanks Bob :) appreciate it.
– MathIsHard
Jul 21 at 5:03
add a comment |Â
up vote
1
down vote
accepted
Parametrize the segment from $0$ to $R e^ipi / 4$ by:
$$gamma :[0,R]rightarrow mathbbC, tmapsto te^i(pi/4).$$
Since you want to integrate from $Re^ipi/4$ to $0$ and not from $0$ to $Re^ipi/4$, you have to switch the orientation of this curve, with the result that the integral changes sign, so the value of the integral you're looking for is
$$-int_gamma e^-z^2dz= -int_0 ^ R e^-gamma(t)^2gamma'(t)dt = -int_0 ^R e^-(te^ipi/4)^2e^ipi/4dt=\-e^ipi/4int_0 ^R e^-t^2e^ipi/2dt=-e^ipi/4int_0 ^R e^-it^2dt,$$
where we used the well known fact that $e^ipi/2=i.$ Now, letting $Rrightarrowinfty$ you get the result claimed in the linked answer.
answered Jul 21 at 3:39
Bob
1,467522
Thanks Bob :) appreciate it.
– MathIsHard
Jul 21 at 5:03
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Parametrize the segment from $0$ to $R e^ipi / 4$ by:
$$gamma :[0,R]rightarrow mathbbC, tmapsto te^i(pi/4).$$
Since you want to integrate from $Re^ipi/4$ to $0$ and not from $0$ to $Re^ipi/4$, you have to switch the orientation of this curve, with the result that the integral changes sign, so the value of the integral you're looking for is
$$-int_gamma e^-z^2dz= -int_0 ^ R e^-gamma(t)^2gamma'(t)dt = -int_0 ^R e^-(te^ipi/4)^2e^ipi/4dt=\-e^ipi/4int_0 ^R e^-t^2e^ipi/2dt=-e^ipi/4int_0 ^R e^-it^2dt,$$
where we used the well known fact that $e^ipi/2=i.$ Now, letting $Rrightarrowinfty$ you get the result claimed in the linked answer.
answered Jul 21 at 3:39
Bob
1,467522
Parametrize the segment from $0$ to $R e^ipi / 4$ by:
$$gamma :[0,R]rightarrow mathbbC, tmapsto te^i(pi/4).$$
Since you want to integrate from $Re^ipi/4$ to $0$ and not from $0$ to $Re^ipi/4$, you have to switch the orientation of this curve, with the result that the integral changes sign, so the value of the integral you're looking for is
$$-int_gamma e^-z^2dz= -int_0 ^ R e^-gamma(t)^2gamma'(t)dt = -int_0 ^R e^-(te^ipi/4)^2e^ipi/4dt=\-e^ipi/4int_0 ^R e^-t^2e^ipi/2dt=-e^ipi/4int_0 ^R e^-it^2dt,$$
where we used the well known fact that $e^ipi/2=i.$ Now, letting $Rrightarrowinfty$ you get the result claimed in the linked answer.
answered Jul 21 at 3:39
Bob
1,467522
answered Jul 21 at 3:39
Bob
1,467522
answered Jul 21 at 3:39
Bob
1,467522
answered Jul 21 at 3:39
Bob
1,467522
1,467522
Thanks Bob :) appreciate it.
– MathIsHard
Jul 21 at 5:03
add a comment |Â
Thanks Bob :) appreciate it.
– MathIsHard
Jul 21 at 5:03
Thanks Bob :) appreciate it.
– MathIsHard
Jul 21 at 5:03
Thanks Bob :) appreciate it.
– MathIsHard
Jul 21 at 5:03
add a comment |Â
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1
$e^ipi/2=i$ and the integral is from $Re^ipi/4$ to 0 so you have to change the sign?
– Bob
Jul 20 at 20:14
Oh you might be right. Let me take a look at that in a few. Thanks. :)
– MathIsHard
Jul 20 at 20:39
Thank you. I see it now. I appreciate the help.
– MathIsHard
Jul 20 at 20:50
1
One of you should post that as an answer so that the question doesn't remain unanswered.
– joriki
Jul 20 at 23:06
Yes, please post that Bob or let me know if you want me to do it. Thanks!
– MathIsHard
Jul 20 at 23:34