Mixing
Residue at infinity calculating integrals
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I have the following problem which I want to evaluate at infinity:
$$oint dfrac(z+2)(z^2+9)dz$$
I approach this problem by saying that $z=dfrac1t$ and $dz=dfrac-1t^2dt$. And I plug them inside my integral and obtain:
$$oint dfrac-(2t+1)(t+9t^3)dz=-2pi i Res(0)=-2pi i$$
Yet this result is not in accordance with the usual integration using residues which yield $2pi i$. I was wondering where am I doing a mistake of minus.
complex-analysis residue-calculus
asked Aug 3 at 21:59
GGphys
876
add a comment |Â
up vote
3
down vote
favorite
I have the following problem which I want to evaluate at infinity:
$$oint dfrac(z+2)(z^2+9)dz$$
I approach this problem by saying that $z=dfrac1t$ and $dz=dfrac-1t^2dt$. And I plug them inside my integral and obtain:
$$oint dfrac-(2t+1)(t+9t^3)dz=-2pi i Res(0)=-2pi i$$
Yet this result is not in accordance with the usual integration using residues which yield $2pi i$. I was wondering where am I doing a mistake of minus.
complex-analysis residue-calculus
asked Aug 3 at 21:59
GGphys
876
1
What do you mean by "evaluate at infinity"? $oint$ generally means an integral around a contour. What contour?
– Robert Israel
Aug 3 at 22:04
I was trying to imply when z goes to infinity.
– GGphys
Aug 3 at 22:05
Note that 0 is not the only point where your new function has poles, in fact it has poles at $1/3i$ and $-1/3i$ which is worse than the original, so I don't understand the point of the change of variables. Also by "going to infinity" do you mean evaluating the contour integral for a circle whose radius goes to infinity?
– Quantaliinuxite
Aug 3 at 22:15
The residue at $infty$ being $-1$ is the correct result. You can also check with this: the sum of residues should be $0$. The residues at $pm 3texti$ are $frac12mp frac13texti$, which sum to $1$. Hence, the residue at $infty$ is $-1$.
– Batominovski
Aug 3 at 22:20
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have the following problem which I want to evaluate at infinity:
$$oint dfrac(z+2)(z^2+9)dz$$
I approach this problem by saying that $z=dfrac1t$ and $dz=dfrac-1t^2dt$. And I plug them inside my integral and obtain:
$$oint dfrac-(2t+1)(t+9t^3)dz=-2pi i Res(0)=-2pi i$$
Yet this result is not in accordance with the usual integration using residues which yield $2pi i$. I was wondering where am I doing a mistake of minus.
complex-analysis residue-calculus
asked Aug 3 at 21:59
GGphys
876
I have the following problem which I want to evaluate at infinity:
$$oint dfrac(z+2)(z^2+9)dz$$
I approach this problem by saying that $z=dfrac1t$ and $dz=dfrac-1t^2dt$. And I plug them inside my integral and obtain:
$$oint dfrac-(2t+1)(t+9t^3)dz=-2pi i Res(0)=-2pi i$$
Yet this result is not in accordance with the usual integration using residues which yield $2pi i$. I was wondering where am I doing a mistake of minus.
complex-analysis residue-calculus
asked Aug 3 at 21:59
GGphys
876
asked Aug 3 at 21:59
GGphys
876
asked Aug 3 at 21:59
GGphys
876
asked Aug 3 at 21:59
GGphys
876
876
1
What do you mean by "evaluate at infinity"? $oint$ generally means an integral around a contour. What contour?
– Robert Israel
Aug 3 at 22:04
I was trying to imply when z goes to infinity.
– GGphys
Aug 3 at 22:05
Note that 0 is not the only point where your new function has poles, in fact it has poles at $1/3i$ and $-1/3i$ which is worse than the original, so I don't understand the point of the change of variables. Also by "going to infinity" do you mean evaluating the contour integral for a circle whose radius goes to infinity?
– Quantaliinuxite
Aug 3 at 22:15
The residue at $infty$ being $-1$ is the correct result. You can also check with this: the sum of residues should be $0$. The residues at $pm 3texti$ are $frac12mp frac13texti$, which sum to $1$. Hence, the residue at $infty$ is $-1$.
– Batominovski
Aug 3 at 22:20
add a comment |Â
1
What do you mean by "evaluate at infinity"? $oint$ generally means an integral around a contour. What contour?
– Robert Israel
Aug 3 at 22:04
I was trying to imply when z goes to infinity.
– GGphys
Aug 3 at 22:05
Note that 0 is not the only point where your new function has poles, in fact it has poles at $1/3i$ and $-1/3i$ which is worse than the original, so I don't understand the point of the change of variables. Also by "going to infinity" do you mean evaluating the contour integral for a circle whose radius goes to infinity?
– Quantaliinuxite
Aug 3 at 22:15
The residue at $infty$ being $-1$ is the correct result. You can also check with this: the sum of residues should be $0$. The residues at $pm 3texti$ are $frac12mp frac13texti$, which sum to $1$. Hence, the residue at $infty$ is $-1$.
– Batominovski
Aug 3 at 22:20
1
1
What do you mean by "evaluate at infinity"? $oint$ generally means an integral around a contour. What contour?
– Robert Israel
Aug 3 at 22:04
What do you mean by "evaluate at infinity"? $oint$ generally means an integral around a contour. What contour?
– Robert Israel
Aug 3 at 22:04
I was trying to imply when z goes to infinity.
– GGphys
Aug 3 at 22:05
I was trying to imply when z goes to infinity.
– GGphys
Aug 3 at 22:05
Note that 0 is not the only point where your new function has poles, in fact it has poles at $1/3i$ and $-1/3i$ which is worse than the original, so I don't understand the point of the change of variables. Also by "going to infinity" do you mean evaluating the contour integral for a circle whose radius goes to infinity?
– Quantaliinuxite
Aug 3 at 22:15
Note that 0 is not the only point where your new function has poles, in fact it has poles at $1/3i$ and $-1/3i$ which is worse than the original, so I don't understand the point of the change of variables. Also by "going to infinity" do you mean evaluating the contour integral for a circle whose radius goes to infinity?
– Quantaliinuxite
Aug 3 at 22:15
The residue at $infty$ being $-1$ is the correct result. You can also check with this: the sum of residues should be $0$. The residues at $pm 3texti$ are $frac12mp frac13texti$, which sum to $1$. Hence, the residue at $infty$ is $-1$.
– Batominovski
Aug 3 at 22:20
The residue at $infty$ being $-1$ is the correct result. You can also check with this: the sum of residues should be $0$. The residues at $pm 3texti$ are $frac12mp frac13texti$, which sum to $1$. Hence, the residue at $infty$ is $-1$.
– Batominovski
Aug 3 at 22:20
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
If you are trying to evaluate the integral,$oint_=Rfracz+2z^2+9,dz$, after making the transformation $zmapsto 1/z$, we can proceed as follows.
For $R>3$, we have for $zmapsto 1/z$
$$beginalign
oint_=Rfracz+2z^2+9,dz&=oint_frac1+2zz(1+9z^2),dz\\
&=2pi i textResleft(frac1+2zz(1+9z^2),z=0 right)\\
&=2pi i
endalign$$
as expected.
Recall that under the transformation, $dzmapsto -frac1z^2,dz$ and the contour is traversed clockwise.
answered Aug 3 at 22:39
Mark Viola
126k1171166
I was interested why the contour is trasvered clockwise actually. I assume because $z=e^ialpha$ and $t=e^-ibeta$ so it causes the negative sign.
– GGphys
2 days ago
1
Indeed. You have it. Well done.
– Mark Viola
2 days ago
add a comment |Â
up vote
1
down vote
If you mean you want $displaystyle oint_Gamma dfracz+2z^2+9 ; dz$ where $Gamma$ is a positively oriented contour near $infty$ (i.e. outside the poles at $z = pm 3 i$), the simplest thing to do is consider the start of the Laurent series near $infty$:
$$ dfracz+2z^2+9 = dfracz+2z^2 (1 - 9/z^2 + ldots) = frac1z -frac7z^2 + ldots $$
so the result is $2pi i$.
answered Aug 3 at 22:11
Robert Israel
303k22199438
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you are trying to evaluate the integral,$oint_=Rfracz+2z^2+9,dz$, after making the transformation $zmapsto 1/z$, we can proceed as follows.
For $R>3$, we have for $zmapsto 1/z$
$$beginalign
oint_=Rfracz+2z^2+9,dz&=oint_frac1+2zz(1+9z^2),dz\\
&=2pi i textResleft(frac1+2zz(1+9z^2),z=0 right)\\
&=2pi i
endalign$$
as expected.
Recall that under the transformation, $dzmapsto -frac1z^2,dz$ and the contour is traversed clockwise.
answered Aug 3 at 22:39
Mark Viola
126k1171166
I was interested why the contour is trasvered clockwise actually. I assume because $z=e^ialpha$ and $t=e^-ibeta$ so it causes the negative sign.
– GGphys
2 days ago
1
Indeed. You have it. Well done.
– Mark Viola
2 days ago
add a comment |Â
up vote
1
down vote
accepted
If you are trying to evaluate the integral,$oint_=Rfracz+2z^2+9,dz$, after making the transformation $zmapsto 1/z$, we can proceed as follows.
For $R>3$, we have for $zmapsto 1/z$
$$beginalign
oint_=Rfracz+2z^2+9,dz&=oint_frac1+2zz(1+9z^2),dz\\
&=2pi i textResleft(frac1+2zz(1+9z^2),z=0 right)\\
&=2pi i
endalign$$
as expected.
Recall that under the transformation, $dzmapsto -frac1z^2,dz$ and the contour is traversed clockwise.
answered Aug 3 at 22:39
Mark Viola
126k1171166
I was interested why the contour is trasvered clockwise actually. I assume because $z=e^ialpha$ and $t=e^-ibeta$ so it causes the negative sign.
– GGphys
2 days ago
1
Indeed. You have it. Well done.
– Mark Viola
2 days ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you are trying to evaluate the integral,$oint_=Rfracz+2z^2+9,dz$, after making the transformation $zmapsto 1/z$, we can proceed as follows.
For $R>3$, we have for $zmapsto 1/z$
$$beginalign
oint_=Rfracz+2z^2+9,dz&=oint_frac1+2zz(1+9z^2),dz\\
&=2pi i textResleft(frac1+2zz(1+9z^2),z=0 right)\\
&=2pi i
endalign$$
as expected.
Recall that under the transformation, $dzmapsto -frac1z^2,dz$ and the contour is traversed clockwise.
answered Aug 3 at 22:39
Mark Viola
126k1171166
If you are trying to evaluate the integral,$oint_=Rfracz+2z^2+9,dz$, after making the transformation $zmapsto 1/z$, we can proceed as follows.
For $R>3$, we have for $zmapsto 1/z$
$$beginalign
oint_=Rfracz+2z^2+9,dz&=oint_frac1+2zz(1+9z^2),dz\\
&=2pi i textResleft(frac1+2zz(1+9z^2),z=0 right)\\
&=2pi i
endalign$$
as expected.
Recall that under the transformation, $dzmapsto -frac1z^2,dz$ and the contour is traversed clockwise.
answered Aug 3 at 22:39
Mark Viola
126k1171166
edited Aug 3 at 23:31
answered Aug 3 at 22:39
Mark Viola
126k1171166
answered Aug 3 at 22:39
Mark Viola
126k1171166
answered Aug 3 at 22:39
Mark Viola
126k1171166
126k1171166
I was interested why the contour is trasvered clockwise actually. I assume because $z=e^ialpha$ and $t=e^-ibeta$ so it causes the negative sign.
– GGphys
2 days ago
1
Indeed. You have it. Well done.
– Mark Viola
2 days ago
add a comment |Â
I was interested why the contour is trasvered clockwise actually. I assume because $z=e^ialpha$ and $t=e^-ibeta$ so it causes the negative sign.
– GGphys
2 days ago
1
Indeed. You have it. Well done.
– Mark Viola
2 days ago
I was interested why the contour is trasvered clockwise actually. I assume because $z=e^ialpha$ and $t=e^-ibeta$ so it causes the negative sign.
– GGphys
2 days ago
I was interested why the contour is trasvered clockwise actually. I assume because $z=e^ialpha$ and $t=e^-ibeta$ so it causes the negative sign.
– GGphys
2 days ago
1
1
Indeed. You have it. Well done.
– Mark Viola
2 days ago
Indeed. You have it. Well done.
– Mark Viola
2 days ago
add a comment |Â
up vote
1
down vote
If you mean you want $displaystyle oint_Gamma dfracz+2z^2+9 ; dz$ where $Gamma$ is a positively oriented contour near $infty$ (i.e. outside the poles at $z = pm 3 i$), the simplest thing to do is consider the start of the Laurent series near $infty$:
$$ dfracz+2z^2+9 = dfracz+2z^2 (1 - 9/z^2 + ldots) = frac1z -frac7z^2 + ldots $$
so the result is $2pi i$.
answered Aug 3 at 22:11
Robert Israel
303k22199438
add a comment |Â
up vote
1
down vote
If you mean you want $displaystyle oint_Gamma dfracz+2z^2+9 ; dz$ where $Gamma$ is a positively oriented contour near $infty$ (i.e. outside the poles at $z = pm 3 i$), the simplest thing to do is consider the start of the Laurent series near $infty$:
$$ dfracz+2z^2+9 = dfracz+2z^2 (1 - 9/z^2 + ldots) = frac1z -frac7z^2 + ldots $$
so the result is $2pi i$.
answered Aug 3 at 22:11
Robert Israel
303k22199438
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you mean you want $displaystyle oint_Gamma dfracz+2z^2+9 ; dz$ where $Gamma$ is a positively oriented contour near $infty$ (i.e. outside the poles at $z = pm 3 i$), the simplest thing to do is consider the start of the Laurent series near $infty$:
$$ dfracz+2z^2+9 = dfracz+2z^2 (1 - 9/z^2 + ldots) = frac1z -frac7z^2 + ldots $$
so the result is $2pi i$.
answered Aug 3 at 22:11
Robert Israel
303k22199438
If you mean you want $displaystyle oint_Gamma dfracz+2z^2+9 ; dz$ where $Gamma$ is a positively oriented contour near $infty$ (i.e. outside the poles at $z = pm 3 i$), the simplest thing to do is consider the start of the Laurent series near $infty$:
$$ dfracz+2z^2+9 = dfracz+2z^2 (1 - 9/z^2 + ldots) = frac1z -frac7z^2 + ldots $$
so the result is $2pi i$.
answered Aug 3 at 22:11
Robert Israel
303k22199438
edited Aug 3 at 22:20
answered Aug 3 at 22:11
Robert Israel
303k22199438
answered Aug 3 at 22:11
Robert Israel
303k22199438
answered Aug 3 at 22:11
Robert Israel
303k22199438
303k22199438
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871535%2fresidue-at-infinity-calculating-integrals%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
What do you mean by "evaluate at infinity"? $oint$ generally means an integral around a contour. What contour?
– Robert Israel
Aug 3 at 22:04
I was trying to imply when z goes to infinity.
– GGphys
Aug 3 at 22:05
Note that 0 is not the only point where your new function has poles, in fact it has poles at $1/3i$ and $-1/3i$ which is worse than the original, so I don't understand the point of the change of variables. Also by "going to infinity" do you mean evaluating the contour integral for a circle whose radius goes to infinity?
– Quantaliinuxite
Aug 3 at 22:15
The residue at $infty$ being $-1$ is the correct result. You can also check with this: the sum of residues should be $0$. The residues at $pm 3texti$ are $frac12mp frac13texti$, which sum to $1$. Hence, the residue at $infty$ is $-1$.
– Batominovski
Aug 3 at 22:20