Integrate a function over a contour including infinitely many poles, such as $int_z1/sin(1/z),dz$contour integration: $int_-infty^infty fraccos x + x sin x1+x^2 dx$How many poles have to be inside the contour?Finding a contour to evaluate$int_-infty^inftyfracxsin xx^2+a^2,dx$How to evaluate the contour integral $int_C(0,1) fracz e^z tan^2 zdz$ over the unit circle?how to evaluate this definite integral $int_0^inftyfracsin^2(x)x^2dx$?Complex integration with infinitely many poles on imaginary axishow to solve it by residue theoremintegral of $1 over |z|$ over a circleHow to integrate a complex function which doesn't have any singularities?Integral of a complex function over contour (continued)
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Integrate a function over a contour including infinitely many poles, such as $int_z1/sin(1/z),dz$
contour integration: $int_-infty^infty fraccos x + x sin x1+x^2 dx$How many poles have to be inside the contour?Finding a contour to evaluate$int_-infty^inftyfracxsin xx^2+a^2,dx$How to evaluate the contour integral $int_C(0,1) fracz e^z tan^2 zdz$ over the unit circle?how to evaluate this definite integral $int_0^inftyfracsin^2(x)x^2dx$?Complex integration with infinitely many poles on imaginary axishow to solve it by residue theoremintegral of $1 over |z|$ over a circleHow to integrate a complex function which doesn't have any singularities?Integral of a complex function over contour (continued)
$begingroup$
We can find complex integration of a function over a closed contour by residue theorem if there are only finite many singularity inside the contour. But my question is how to find the integration if there are infinite many singularity inside the contour? Please help me solve this type of problem mention below.
$$int_z frac1sin(frac1z) dz$$
complex-analysis complex-integration
asked Nov 30 '15 at 11:38
neelkanthneelkanth
2,28821129
$endgroup$
|
show 8 more comments
$begingroup$
We can find complex integration of a function over a closed contour by residue theorem if there are only finite many singularity inside the contour. But my question is how to find the integration if there are infinite many singularity inside the contour? Please help me solve this type of problem mention below.
$$int_z frac1sin(frac1z) dz$$
complex-analysis complex-integration
asked Nov 30 '15 at 11:38
neelkanthneelkanth
2,28821129
$endgroup$
1
$begingroup$
It may be helpful that this is $int_0^2pi dz / sin(sinphi - icosphi)$. You might find an antiderivative featuring a Bessel function But that seems very nasty and any residue approach should be preferred.
$endgroup$
– GDumphart
Nov 30 '15 at 11:55
1
$begingroup$
I think that the problem here is that $sin(1/z)$ is not a meromorphic function, so no residue theory can be applied.
$endgroup$
– Crostul
Nov 30 '15 at 12:05
2
$begingroup$
perhaps a new variable u=1/z. we have $dz=-fracduu^2$, the contour is |u|=1 and the integral is more fun. $int_ -fracduu^2sin(u).$
$endgroup$
– Claudeh5
Nov 30 '15 at 12:09
2
$begingroup$
u could use the residue at infinity, so instead of looking at the inside of the unit circle u look at it's exterior. By Cauchy they are the same
$endgroup$
– tired
Nov 30 '15 at 12:13
1
$begingroup$
the result is $i pi/3 $ if i'm not mistaken.
$endgroup$
– tired
Nov 30 '15 at 12:16
|
show 8 more comments
$begingroup$
We can find complex integration of a function over a closed contour by residue theorem if there are only finite many singularity inside the contour. But my question is how to find the integration if there are infinite many singularity inside the contour? Please help me solve this type of problem mention below.
$$int_z frac1sin(frac1z) dz$$
complex-analysis complex-integration
asked Nov 30 '15 at 11:38
neelkanthneelkanth
2,28821129
$endgroup$
We can find complex integration of a function over a closed contour by residue theorem if there are only finite many singularity inside the contour. But my question is how to find the integration if there are infinite many singularity inside the contour? Please help me solve this type of problem mention below.
$$int_z frac1sin(frac1z) dz$$
complex-analysis complex-integration
complex-analysis complex-integration
asked Nov 30 '15 at 11:38
neelkanthneelkanth
2,28821129
asked Nov 30 '15 at 11:38
neelkanthneelkanth
2,28821129
edited Nov 30 '15 at 22:48
user147263
asked Nov 30 '15 at 11:38
neelkanthneelkanth
2,28821129
asked Nov 30 '15 at 11:38
neelkanthneelkanth
2,28821129
asked Nov 30 '15 at 11:38
neelkanthneelkanth
2,28821129
2,28821129
1
$begingroup$
It may be helpful that this is $int_0^2pi dz / sin(sinphi - icosphi)$. You might find an antiderivative featuring a Bessel function But that seems very nasty and any residue approach should be preferred.
$endgroup$
– GDumphart
Nov 30 '15 at 11:55
1
$begingroup$
I think that the problem here is that $sin(1/z)$ is not a meromorphic function, so no residue theory can be applied.
$endgroup$
– Crostul
Nov 30 '15 at 12:05
2
$begingroup$
perhaps a new variable u=1/z. we have $dz=-fracduu^2$, the contour is |u|=1 and the integral is more fun. $int_ -fracduu^2sin(u).$
$endgroup$
– Claudeh5
Nov 30 '15 at 12:09
2
$begingroup$
u could use the residue at infinity, so instead of looking at the inside of the unit circle u look at it's exterior. By Cauchy they are the same
$endgroup$
– tired
Nov 30 '15 at 12:13
1
$begingroup$
the result is $i pi/3 $ if i'm not mistaken.
$endgroup$
– tired
Nov 30 '15 at 12:16
|
show 8 more comments
1
$begingroup$
It may be helpful that this is $int_0^2pi dz / sin(sinphi - icosphi)$. You might find an antiderivative featuring a Bessel function But that seems very nasty and any residue approach should be preferred.
$endgroup$
– GDumphart
Nov 30 '15 at 11:55
1
$begingroup$
I think that the problem here is that $sin(1/z)$ is not a meromorphic function, so no residue theory can be applied.
$endgroup$
– Crostul
Nov 30 '15 at 12:05
2
$begingroup$
perhaps a new variable u=1/z. we have $dz=-fracduu^2$, the contour is |u|=1 and the integral is more fun. $int_ -fracduu^2sin(u).$
$endgroup$
– Claudeh5
Nov 30 '15 at 12:09
2
$begingroup$
u could use the residue at infinity, so instead of looking at the inside of the unit circle u look at it's exterior. By Cauchy they are the same
$endgroup$
– tired
Nov 30 '15 at 12:13
1
$begingroup$
the result is $i pi/3 $ if i'm not mistaken.
$endgroup$
– tired
Nov 30 '15 at 12:16
1
1
$begingroup$
It may be helpful that this is $int_0^2pi dz / sin(sinphi - icosphi)$. You might find an antiderivative featuring a Bessel function But that seems very nasty and any residue approach should be preferred.
$endgroup$
– GDumphart
Nov 30 '15 at 11:55
$begingroup$
It may be helpful that this is $int_0^2pi dz / sin(sinphi - icosphi)$. You might find an antiderivative featuring a Bessel function But that seems very nasty and any residue approach should be preferred.
$endgroup$
– GDumphart
Nov 30 '15 at 11:55
1
1
$begingroup$
I think that the problem here is that $sin(1/z)$ is not a meromorphic function, so no residue theory can be applied.
$endgroup$
– Crostul
Nov 30 '15 at 12:05
$begingroup$
I think that the problem here is that $sin(1/z)$ is not a meromorphic function, so no residue theory can be applied.
$endgroup$
– Crostul
Nov 30 '15 at 12:05
2
2
$begingroup$
perhaps a new variable u=1/z. we have $dz=-fracduu^2$, the contour is |u|=1 and the integral is more fun. $int_ -fracduu^2sin(u).$
$endgroup$
– Claudeh5
Nov 30 '15 at 12:09
$begingroup$
perhaps a new variable u=1/z. we have $dz=-fracduu^2$, the contour is |u|=1 and the integral is more fun. $int_ -fracduu^2sin(u).$
$endgroup$
– Claudeh5
Nov 30 '15 at 12:09
2
2
$begingroup$
u could use the residue at infinity, so instead of looking at the inside of the unit circle u look at it's exterior. By Cauchy they are the same
$endgroup$
– tired
Nov 30 '15 at 12:13
$begingroup$
u could use the residue at infinity, so instead of looking at the inside of the unit circle u look at it's exterior. By Cauchy they are the same
$endgroup$
– tired
Nov 30 '15 at 12:13
1
1
$begingroup$
the result is $i pi/3 $ if i'm not mistaken.
$endgroup$
– tired
Nov 30 '15 at 12:16
$begingroup$
the result is $i pi/3 $ if i'm not mistaken.
$endgroup$
– tired
Nov 30 '15 at 12:16
|
show 8 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Ok this is what i would do (i assume the unit circle is traversed in counter clockwise direction):
Let's denote
$$
f(z)=frac1sin(1/z)
$$
As the OP already noted the integrand behaves quite nasty inside the unit disc. The reason is the infinite number of poles showing up as $zrightarrow 0$.
So what can we do?
One idea would be to look at the outside of the unit disc rather then into its inside because we only have one pole at $z=infty$rather then an uncountable number which is not so bad.
By residue theorem, the countour integral around the pole at infinity is the negative of the integral around the unit circle. We get
$$
oint_zf(z)dz=-2pi itextRes(f(z),z=infty)
$$
the residue at infinity is given by
$$
textRes(f(z),z=infty)=textRes(-f(frac1z)/z^2,z=0)
$$
using $sin(z)=z-z^3/3!+z^5/5!+mathcalO(z^7)$ and expand the fraction as a geometric series we get $textRes(f(z),z=infty)=-1/6$ and
$$
oint_zf(z)dz=fracpi i3
$$
as announced in the comments
edited yesterday
the8thone
1,828932
answered Nov 30 '15 at 12:31
tiredtired
10.6k12043
$endgroup$
$begingroup$
thanking you for your precise time...
$endgroup$
– neelkanth
Nov 30 '15 at 12:38
$begingroup$
Minor comment: The third and fourth summands of the Taylor expansion should be $z^5/5! + mathcalO(z^7)$.
$endgroup$
– GDumphart
Nov 30 '15 at 12:44
$begingroup$
@GDumphart, thanks i corrected it
$endgroup$
– tired
Nov 30 '15 at 13:25
add a comment |
$begingroup$
$I=oint_zf(z)dz=-oint_fracduu^2 sin(u)=fracpi i3$
(the rotation is not in the trigonometric sense, as the residus inside $|z|=1$ are outside |u=1| with $z=1/u$).
The residue of $frac1u^2 sin(u)$ in $u=0$ is $frac16$. This is linked to the theory of residue at infinity.
answered Nov 30 '15 at 17:22
Claudeh5Claudeh5
53028
$endgroup$
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
$begingroup$
Ok this is what i would do (i assume the unit circle is traversed in counter clockwise direction):
Let's denote
$$
f(z)=frac1sin(1/z)
$$
As the OP already noted the integrand behaves quite nasty inside the unit disc. The reason is the infinite number of poles showing up as $zrightarrow 0$.
So what can we do?
One idea would be to look at the outside of the unit disc rather then into its inside because we only have one pole at $z=infty$rather then an uncountable number which is not so bad.
By residue theorem, the countour integral around the pole at infinity is the negative of the integral around the unit circle. We get
$$
oint_zf(z)dz=-2pi itextRes(f(z),z=infty)
$$
the residue at infinity is given by
$$
textRes(f(z),z=infty)=textRes(-f(frac1z)/z^2,z=0)
$$
using $sin(z)=z-z^3/3!+z^5/5!+mathcalO(z^7)$ and expand the fraction as a geometric series we get $textRes(f(z),z=infty)=-1/6$ and
$$
oint_zf(z)dz=fracpi i3
$$
as announced in the comments
edited yesterday
the8thone
1,828932
answered Nov 30 '15 at 12:31
tiredtired
10.6k12043
$endgroup$
$begingroup$
thanking you for your precise time...
$endgroup$
– neelkanth
Nov 30 '15 at 12:38
$begingroup$
Minor comment: The third and fourth summands of the Taylor expansion should be $z^5/5! + mathcalO(z^7)$.
$endgroup$
– GDumphart
Nov 30 '15 at 12:44
$begingroup$
@GDumphart, thanks i corrected it
$endgroup$
– tired
Nov 30 '15 at 13:25
add a comment |
$begingroup$
Ok this is what i would do (i assume the unit circle is traversed in counter clockwise direction):
Let's denote
$$
f(z)=frac1sin(1/z)
$$
As the OP already noted the integrand behaves quite nasty inside the unit disc. The reason is the infinite number of poles showing up as $zrightarrow 0$.
So what can we do?
One idea would be to look at the outside of the unit disc rather then into its inside because we only have one pole at $z=infty$rather then an uncountable number which is not so bad.
By residue theorem, the countour integral around the pole at infinity is the negative of the integral around the unit circle. We get
$$
oint_zf(z)dz=-2pi itextRes(f(z),z=infty)
$$
the residue at infinity is given by
$$
textRes(f(z),z=infty)=textRes(-f(frac1z)/z^2,z=0)
$$
using $sin(z)=z-z^3/3!+z^5/5!+mathcalO(z^7)$ and expand the fraction as a geometric series we get $textRes(f(z),z=infty)=-1/6$ and
$$
oint_zf(z)dz=fracpi i3
$$
as announced in the comments
edited yesterday
the8thone
1,828932
answered Nov 30 '15 at 12:31
tiredtired
10.6k12043
$endgroup$
$begingroup$
thanking you for your precise time...
$endgroup$
– neelkanth
Nov 30 '15 at 12:38
$begingroup$
Minor comment: The third and fourth summands of the Taylor expansion should be $z^5/5! + mathcalO(z^7)$.
$endgroup$
– GDumphart
Nov 30 '15 at 12:44
$begingroup$
@GDumphart, thanks i corrected it
$endgroup$
– tired
Nov 30 '15 at 13:25
add a comment |
$begingroup$
Ok this is what i would do (i assume the unit circle is traversed in counter clockwise direction):
Let's denote
$$
f(z)=frac1sin(1/z)
$$
As the OP already noted the integrand behaves quite nasty inside the unit disc. The reason is the infinite number of poles showing up as $zrightarrow 0$.
So what can we do?
One idea would be to look at the outside of the unit disc rather then into its inside because we only have one pole at $z=infty$rather then an uncountable number which is not so bad.
By residue theorem, the countour integral around the pole at infinity is the negative of the integral around the unit circle. We get
$$
oint_zf(z)dz=-2pi itextRes(f(z),z=infty)
$$
the residue at infinity is given by
$$
textRes(f(z),z=infty)=textRes(-f(frac1z)/z^2,z=0)
$$
using $sin(z)=z-z^3/3!+z^5/5!+mathcalO(z^7)$ and expand the fraction as a geometric series we get $textRes(f(z),z=infty)=-1/6$ and
$$
oint_zf(z)dz=fracpi i3
$$
as announced in the comments
edited yesterday
the8thone
1,828932
answered Nov 30 '15 at 12:31
tiredtired
10.6k12043
$endgroup$
Ok this is what i would do (i assume the unit circle is traversed in counter clockwise direction):
Let's denote
$$
f(z)=frac1sin(1/z)
$$
As the OP already noted the integrand behaves quite nasty inside the unit disc. The reason is the infinite number of poles showing up as $zrightarrow 0$.
So what can we do?
One idea would be to look at the outside of the unit disc rather then into its inside because we only have one pole at $z=infty$rather then an uncountable number which is not so bad.
By residue theorem, the countour integral around the pole at infinity is the negative of the integral around the unit circle. We get
$$
oint_zf(z)dz=-2pi itextRes(f(z),z=infty)
$$
the residue at infinity is given by
$$
textRes(f(z),z=infty)=textRes(-f(frac1z)/z^2,z=0)
$$
using $sin(z)=z-z^3/3!+z^5/5!+mathcalO(z^7)$ and expand the fraction as a geometric series we get $textRes(f(z),z=infty)=-1/6$ and
$$
oint_zf(z)dz=fracpi i3
$$
as announced in the comments
edited yesterday
the8thone
1,828932
answered Nov 30 '15 at 12:31
tiredtired
10.6k12043
edited yesterday
the8thone
1,828932
edited yesterday
the8thone
1,828932
edited yesterday
the8thone
1,828932
1,828932
answered Nov 30 '15 at 12:31
tiredtired
10.6k12043
answered Nov 30 '15 at 12:31
tiredtired
10.6k12043
answered Nov 30 '15 at 12:31
tiredtired
10.6k12043
10.6k12043
$begingroup$
thanking you for your precise time...
$endgroup$
– neelkanth
Nov 30 '15 at 12:38
$begingroup$
Minor comment: The third and fourth summands of the Taylor expansion should be $z^5/5! + mathcalO(z^7)$.
$endgroup$
– GDumphart
Nov 30 '15 at 12:44
$begingroup$
@GDumphart, thanks i corrected it
$endgroup$
– tired
Nov 30 '15 at 13:25
add a comment |
$begingroup$
thanking you for your precise time...
$endgroup$
– neelkanth
Nov 30 '15 at 12:38
$begingroup$
Minor comment: The third and fourth summands of the Taylor expansion should be $z^5/5! + mathcalO(z^7)$.
$endgroup$
– GDumphart
Nov 30 '15 at 12:44
$begingroup$
@GDumphart, thanks i corrected it
$endgroup$
– tired
Nov 30 '15 at 13:25
$begingroup$
thanking you for your precise time...
$endgroup$
– neelkanth
Nov 30 '15 at 12:38
$begingroup$
thanking you for your precise time...
$endgroup$
– neelkanth
Nov 30 '15 at 12:38
$begingroup$
Minor comment: The third and fourth summands of the Taylor expansion should be $z^5/5! + mathcalO(z^7)$.
$endgroup$
– GDumphart
Nov 30 '15 at 12:44
$begingroup$
Minor comment: The third and fourth summands of the Taylor expansion should be $z^5/5! + mathcalO(z^7)$.
$endgroup$
– GDumphart
Nov 30 '15 at 12:44
$begingroup$
@GDumphart, thanks i corrected it
$endgroup$
– tired
Nov 30 '15 at 13:25
$begingroup$
@GDumphart, thanks i corrected it
$endgroup$
– tired
Nov 30 '15 at 13:25
add a comment |
$begingroup$
$I=oint_zf(z)dz=-oint_fracduu^2 sin(u)=fracpi i3$
(the rotation is not in the trigonometric sense, as the residus inside $|z|=1$ are outside |u=1| with $z=1/u$).
The residue of $frac1u^2 sin(u)$ in $u=0$ is $frac16$. This is linked to the theory of residue at infinity.
answered Nov 30 '15 at 17:22
Claudeh5Claudeh5
53028
$endgroup$
add a comment |
$begingroup$
$I=oint_zf(z)dz=-oint_fracduu^2 sin(u)=fracpi i3$
(the rotation is not in the trigonometric sense, as the residus inside $|z|=1$ are outside |u=1| with $z=1/u$).
The residue of $frac1u^2 sin(u)$ in $u=0$ is $frac16$. This is linked to the theory of residue at infinity.
answered Nov 30 '15 at 17:22
Claudeh5Claudeh5
53028
$endgroup$
add a comment |
$begingroup$
$I=oint_zf(z)dz=-oint_fracduu^2 sin(u)=fracpi i3$
(the rotation is not in the trigonometric sense, as the residus inside $|z|=1$ are outside |u=1| with $z=1/u$).
The residue of $frac1u^2 sin(u)$ in $u=0$ is $frac16$. This is linked to the theory of residue at infinity.
answered Nov 30 '15 at 17:22
Claudeh5Claudeh5
53028
$endgroup$
$I=oint_zf(z)dz=-oint_fracduu^2 sin(u)=fracpi i3$
(the rotation is not in the trigonometric sense, as the residus inside $|z|=1$ are outside |u=1| with $z=1/u$).
The residue of $frac1u^2 sin(u)$ in $u=0$ is $frac16$. This is linked to the theory of residue at infinity.
answered Nov 30 '15 at 17:22
Claudeh5Claudeh5
53028
answered Nov 30 '15 at 17:22
Claudeh5Claudeh5
53028
answered Nov 30 '15 at 17:22
Claudeh5Claudeh5
53028
answered Nov 30 '15 at 17:22
Claudeh5Claudeh5
53028
53028
add a comment |
add a comment |
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It may be helpful that this is $int_0^2pi dz / sin(sinphi - icosphi)$. You might find an antiderivative featuring a Bessel function But that seems very nasty and any residue approach should be preferred.
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– GDumphart
Nov 30 '15 at 11:55
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I think that the problem here is that $sin(1/z)$ is not a meromorphic function, so no residue theory can be applied.
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– Crostul
Nov 30 '15 at 12:05
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perhaps a new variable u=1/z. we have $dz=-fracduu^2$, the contour is |u|=1 and the integral is more fun. $int_ -fracduu^2sin(u).$
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– Claudeh5
Nov 30 '15 at 12:09
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u could use the residue at infinity, so instead of looking at the inside of the unit circle u look at it's exterior. By Cauchy they are the same
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– tired
Nov 30 '15 at 12:13
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the result is $i pi/3 $ if i'm not mistaken.
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– tired
Nov 30 '15 at 12:16