Real integrals with two poles in the complex plane
I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.
Consider an integral over the function $f: mathbb{R} to mathbb{C}$
$$
I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
$$
This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as
$$
I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
= 2pi i frac{e^{-1}}{2i} =pi e^{-1}
quad ,
$$
where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.
Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give
$$
I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
= - 2pi i frac{e^{+1}}{-2i}
=pi e^{+1}
quad .
$$
But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?
complex-analysis contour-integration cauchy-integral-formula
asked Dec 28 '18 at 11:22
CharlieB
1233
add a comment |
I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.
Consider an integral over the function $f: mathbb{R} to mathbb{C}$
$$
I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
$$
This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as
$$
I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
= 2pi i frac{e^{-1}}{2i} =pi e^{-1}
quad ,
$$
where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.
Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give
$$
I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
= - 2pi i frac{e^{+1}}{-2i}
=pi e^{+1}
quad .
$$
But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?
complex-analysis contour-integration cauchy-integral-formula
asked Dec 28 '18 at 11:22
CharlieB
1233
add a comment |
I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.
Consider an integral over the function $f: mathbb{R} to mathbb{C}$
$$
I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
$$
This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as
$$
I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
= 2pi i frac{e^{-1}}{2i} =pi e^{-1}
quad ,
$$
where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.
Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give
$$
I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
= - 2pi i frac{e^{+1}}{-2i}
=pi e^{+1}
quad .
$$
But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?
complex-analysis contour-integration cauchy-integral-formula
asked Dec 28 '18 at 11:22
CharlieB
1233
I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.
Consider an integral over the function $f: mathbb{R} to mathbb{C}$
$$
I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
$$
This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as
$$
I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
= 2pi i frac{e^{-1}}{2i} =pi e^{-1}
quad ,
$$
where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.
Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give
$$
I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
= - 2pi i frac{e^{+1}}{-2i}
=pi e^{+1}
quad .
$$
But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?
complex-analysis contour-integration cauchy-integral-formula
complex-analysis contour-integration cauchy-integral-formula
asked Dec 28 '18 at 11:22
CharlieB
1233
asked Dec 28 '18 at 11:22
CharlieB
1233
asked Dec 28 '18 at 11:22
CharlieB
1233
asked Dec 28 '18 at 11:22
CharlieB
1233
asked Dec 28 '18 at 11:22
CharlieB
1233
1233
add a comment |
add a comment |
2 Answers
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oldest
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No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.
answered Dec 28 '18 at 11:32
José Carlos Santos
150k22121221
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
Dec 28 '18 at 12:03
add a comment |
If you choose the contour in the upper half-plane, the integral over the
semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
for the absolute value of the integral, when $R>1$.
But on the lower half-plane, the integral over the semicircle does not tend to
zero as $Rtoinfty$.
answered Dec 28 '18 at 11:29
Lord Shark the Unknown
101k958132
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
Dec 28 '18 at 12:04
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
No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.
answered Dec 28 '18 at 11:32
José Carlos Santos
150k22121221
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
Dec 28 '18 at 12:03
add a comment |
No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.
answered Dec 28 '18 at 11:32
José Carlos Santos
150k22121221
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
Dec 28 '18 at 12:03
add a comment |
No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.
answered Dec 28 '18 at 11:32
José Carlos Santos
150k22121221
No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.
answered Dec 28 '18 at 11:32
José Carlos Santos
150k22121221
answered Dec 28 '18 at 11:32
José Carlos Santos
150k22121221
answered Dec 28 '18 at 11:32
José Carlos Santos
150k22121221
answered Dec 28 '18 at 11:32
José Carlos Santos
150k22121221
150k22121221
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
Dec 28 '18 at 12:03
add a comment |
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
Dec 28 '18 at 12:03
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
Dec 28 '18 at 12:03
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
Dec 28 '18 at 12:03
add a comment |
If you choose the contour in the upper half-plane, the integral over the
semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
for the absolute value of the integral, when $R>1$.
But on the lower half-plane, the integral over the semicircle does not tend to
zero as $Rtoinfty$.
answered Dec 28 '18 at 11:29
Lord Shark the Unknown
101k958132
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
Dec 28 '18 at 12:04
add a comment |
If you choose the contour in the upper half-plane, the integral over the
semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
for the absolute value of the integral, when $R>1$.
But on the lower half-plane, the integral over the semicircle does not tend to
zero as $Rtoinfty$.
answered Dec 28 '18 at 11:29
Lord Shark the Unknown
101k958132
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
Dec 28 '18 at 12:04
add a comment |
If you choose the contour in the upper half-plane, the integral over the
semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
for the absolute value of the integral, when $R>1$.
But on the lower half-plane, the integral over the semicircle does not tend to
zero as $Rtoinfty$.
answered Dec 28 '18 at 11:29
Lord Shark the Unknown
101k958132
If you choose the contour in the upper half-plane, the integral over the
semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
for the absolute value of the integral, when $R>1$.
But on the lower half-plane, the integral over the semicircle does not tend to
zero as $Rtoinfty$.
answered Dec 28 '18 at 11:29
Lord Shark the Unknown
101k958132
answered Dec 28 '18 at 11:29
Lord Shark the Unknown
101k958132
answered Dec 28 '18 at 11:29
Lord Shark the Unknown
101k958132
answered Dec 28 '18 at 11:29
Lord Shark the Unknown
101k958132
101k958132
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
Dec 28 '18 at 12:04
add a comment |
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
Dec 28 '18 at 12:04
1
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
Dec 28 '18 at 12:04
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
Dec 28 '18 at 12:04
add a comment |
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