Mixing
Evaluate $int frac{1}{z^2+1}$
$begingroup$
Evaluate $$int_{Gamma} frac{1}{z^2+1}$$ where $Gamma={z:|z|=2}$
One way is to use partial fractions and then cauchy integral formula.
The other way is to look at:
$$int_{Gamma} frac{1}{z^2+1}=int_{gamma_1}frac{frac{1}{z-i}}{z+i}+int_{gamma_2}frac{frac{1}{z+i}}{z-i}$$
Where $gamma_1$ is a curve around $i$ and $gamma_2$ is a curve around $-i$ but in which direction? clockwise? as $Gamma$ is anti clockwise?
complex-analysis contour-integration
edited Jan 13 at 17:53
José Carlos Santos
161k22127232
asked Jun 17 '17 at 20:15
gboxgbox
5,41762262
$endgroup$
add a comment |
$begingroup$
Evaluate $$int_{Gamma} frac{1}{z^2+1}$$ where $Gamma={z:|z|=2}$
One way is to use partial fractions and then cauchy integral formula.
The other way is to look at:
$$int_{Gamma} frac{1}{z^2+1}=int_{gamma_1}frac{frac{1}{z-i}}{z+i}+int_{gamma_2}frac{frac{1}{z+i}}{z-i}$$
Where $gamma_1$ is a curve around $i$ and $gamma_2$ is a curve around $-i$ but in which direction? clockwise? as $Gamma$ is anti clockwise?
complex-analysis contour-integration
edited Jan 13 at 17:53
José Carlos Santos
161k22127232
asked Jun 17 '17 at 20:15
gboxgbox
5,41762262
$endgroup$
add a comment |
$begingroup$
Evaluate $$int_{Gamma} frac{1}{z^2+1}$$ where $Gamma={z:|z|=2}$
One way is to use partial fractions and then cauchy integral formula.
The other way is to look at:
$$int_{Gamma} frac{1}{z^2+1}=int_{gamma_1}frac{frac{1}{z-i}}{z+i}+int_{gamma_2}frac{frac{1}{z+i}}{z-i}$$
Where $gamma_1$ is a curve around $i$ and $gamma_2$ is a curve around $-i$ but in which direction? clockwise? as $Gamma$ is anti clockwise?
complex-analysis contour-integration
edited Jan 13 at 17:53
José Carlos Santos
161k22127232
asked Jun 17 '17 at 20:15
gboxgbox
5,41762262
$endgroup$
Evaluate $$int_{Gamma} frac{1}{z^2+1}$$ where $Gamma={z:|z|=2}$
One way is to use partial fractions and then cauchy integral formula.
The other way is to look at:
$$int_{Gamma} frac{1}{z^2+1}=int_{gamma_1}frac{frac{1}{z-i}}{z+i}+int_{gamma_2}frac{frac{1}{z+i}}{z-i}$$
Where $gamma_1$ is a curve around $i$ and $gamma_2$ is a curve around $-i$ but in which direction? clockwise? as $Gamma$ is anti clockwise?
complex-analysis contour-integration
complex-analysis contour-integration
edited Jan 13 at 17:53
José Carlos Santos
161k22127232
asked Jun 17 '17 at 20:15
gboxgbox
5,41762262
edited Jan 13 at 17:53
José Carlos Santos
161k22127232
asked Jun 17 '17 at 20:15
gboxgbox
5,41762262
edited Jan 13 at 17:53
José Carlos Santos
161k22127232
edited Jan 13 at 17:53
José Carlos Santos
161k22127232
edited Jan 13 at 17:53
José Carlos Santos
161k22127232
161k22127232
asked Jun 17 '17 at 20:15
gboxgbox
5,41762262
asked Jun 17 '17 at 20:15
gboxgbox
5,41762262
asked Jun 17 '17 at 20:15
gboxgbox
5,41762262
5,41762262
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, it is not a sum. Usually, the meaning of $int_Gamma$ with $Gamma={zinmathbb{C},:,|z|=r}$ means that we are integrating along the loop $Gammacolon[0,2pi]longrightarrowmathbb C$ defined by $Gamma(t)=re^{it}$.
In this specific case, the value of the integral will be $2pi i$ times the sum of the residues of $frac1{z^2+1}$ at $i$ and $-i$. The first residue is equal to $-frac i2$ and the second one is equal to $frac i2$. Therefore, the integral is equal to $0$.
answered Jun 17 '17 at 20:24
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
add a comment |
$begingroup$
You mean
$$int_Gammafrac{dz}{z^2+1}.$$
You can replace the contour $Gamma$ with the circle of centre $0$ and
radius $r$ for any $r>1$. Letting $rtoinfty$ the integral tends to zero,
so it must have been zero in the first place.
answered Jun 17 '17 at 21:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Can we say that there is antiderivative when where $theta-frac{pi}{2} < Arg(z)leq theta -frac{pi}{2} $
$endgroup$
– gbox
Jun 18 '17 at 13:46
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
$begingroup$
No, it is not a sum. Usually, the meaning of $int_Gamma$ with $Gamma={zinmathbb{C},:,|z|=r}$ means that we are integrating along the loop $Gammacolon[0,2pi]longrightarrowmathbb C$ defined by $Gamma(t)=re^{it}$.
In this specific case, the value of the integral will be $2pi i$ times the sum of the residues of $frac1{z^2+1}$ at $i$ and $-i$. The first residue is equal to $-frac i2$ and the second one is equal to $frac i2$. Therefore, the integral is equal to $0$.
answered Jun 17 '17 at 20:24
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
add a comment |
$begingroup$
No, it is not a sum. Usually, the meaning of $int_Gamma$ with $Gamma={zinmathbb{C},:,|z|=r}$ means that we are integrating along the loop $Gammacolon[0,2pi]longrightarrowmathbb C$ defined by $Gamma(t)=re^{it}$.
In this specific case, the value of the integral will be $2pi i$ times the sum of the residues of $frac1{z^2+1}$ at $i$ and $-i$. The first residue is equal to $-frac i2$ and the second one is equal to $frac i2$. Therefore, the integral is equal to $0$.
answered Jun 17 '17 at 20:24
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
add a comment |
$begingroup$
No, it is not a sum. Usually, the meaning of $int_Gamma$ with $Gamma={zinmathbb{C},:,|z|=r}$ means that we are integrating along the loop $Gammacolon[0,2pi]longrightarrowmathbb C$ defined by $Gamma(t)=re^{it}$.
In this specific case, the value of the integral will be $2pi i$ times the sum of the residues of $frac1{z^2+1}$ at $i$ and $-i$. The first residue is equal to $-frac i2$ and the second one is equal to $frac i2$. Therefore, the integral is equal to $0$.
answered Jun 17 '17 at 20:24
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
No, it is not a sum. Usually, the meaning of $int_Gamma$ with $Gamma={zinmathbb{C},:,|z|=r}$ means that we are integrating along the loop $Gammacolon[0,2pi]longrightarrowmathbb C$ defined by $Gamma(t)=re^{it}$.
In this specific case, the value of the integral will be $2pi i$ times the sum of the residues of $frac1{z^2+1}$ at $i$ and $-i$. The first residue is equal to $-frac i2$ and the second one is equal to $frac i2$. Therefore, the integral is equal to $0$.
answered Jun 17 '17 at 20:24
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jun 17 '17 at 20:24
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jun 17 '17 at 20:24
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jun 17 '17 at 20:24
José Carlos SantosJosé Carlos Santos
161k22127232
161k22127232
add a comment |
add a comment |
$begingroup$
You mean
$$int_Gammafrac{dz}{z^2+1}.$$
You can replace the contour $Gamma$ with the circle of centre $0$ and
radius $r$ for any $r>1$. Letting $rtoinfty$ the integral tends to zero,
so it must have been zero in the first place.
answered Jun 17 '17 at 21:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Can we say that there is antiderivative when where $theta-frac{pi}{2} < Arg(z)leq theta -frac{pi}{2} $
$endgroup$
– gbox
Jun 18 '17 at 13:46
add a comment |
$begingroup$
You mean
$$int_Gammafrac{dz}{z^2+1}.$$
You can replace the contour $Gamma$ with the circle of centre $0$ and
radius $r$ for any $r>1$. Letting $rtoinfty$ the integral tends to zero,
so it must have been zero in the first place.
answered Jun 17 '17 at 21:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Can we say that there is antiderivative when where $theta-frac{pi}{2} < Arg(z)leq theta -frac{pi}{2} $
$endgroup$
– gbox
Jun 18 '17 at 13:46
add a comment |
$begingroup$
You mean
$$int_Gammafrac{dz}{z^2+1}.$$
You can replace the contour $Gamma$ with the circle of centre $0$ and
radius $r$ for any $r>1$. Letting $rtoinfty$ the integral tends to zero,
so it must have been zero in the first place.
answered Jun 17 '17 at 21:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
You mean
$$int_Gammafrac{dz}{z^2+1}.$$
You can replace the contour $Gamma$ with the circle of centre $0$ and
radius $r$ for any $r>1$. Letting $rtoinfty$ the integral tends to zero,
so it must have been zero in the first place.
answered Jun 17 '17 at 21:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jun 17 '17 at 21:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jun 17 '17 at 21:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jun 17 '17 at 21:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
$begingroup$
Can we say that there is antiderivative when where $theta-frac{pi}{2} < Arg(z)leq theta -frac{pi}{2} $
$endgroup$
– gbox
Jun 18 '17 at 13:46
add a comment |
$begingroup$
Can we say that there is antiderivative when where $theta-frac{pi}{2} < Arg(z)leq theta -frac{pi}{2} $
$endgroup$
– gbox
Jun 18 '17 at 13:46
$begingroup$
Can we say that there is antiderivative when where $theta-frac{pi}{2} < Arg(z)leq theta -frac{pi}{2} $
$endgroup$
– gbox
Jun 18 '17 at 13:46
$begingroup$
Can we say that there is antiderivative when where $theta-frac{pi}{2} < Arg(z)leq theta -frac{pi}{2} $
$endgroup$
– gbox
Jun 18 '17 at 13:46
add a comment |
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