A weird value obtained by using Cauchy Principal Value on $int_{-infty}^{infty}frac{1}{x^2}dx$
$begingroup$
so I'm trying to evaluate the integral in the title,
$$int_{-infty}^{infty}frac{1}{x^2}dx$$
by using complex plane integration. I've chosen my contour to be a infinte half circle with it's diameter on the real axis. (integration is preformed ccw).
when R tends to infinity, the arch part of the contour yields zero, and so we are left with the part along the real axis, which is the one I'm trying to evalute.
there are no other poles in my contour, only a second order pole at $z=0$ lying on it. the residue of this pole is $0$ so the integral sums up to be zero (by using Cauchy principal value.)
However my function is always positive and greater than $0$, so this doesn't make sense.
Any help would be appreciated
complex-analysis complex-integration
edited Jan 17 at 12:25
Somos
13.7k11235
asked Jan 17 at 11:44
user635635user635635
141
$endgroup$
add a comment |
$begingroup$
so I'm trying to evaluate the integral in the title,
$$int_{-infty}^{infty}frac{1}{x^2}dx$$
by using complex plane integration. I've chosen my contour to be a infinte half circle with it's diameter on the real axis. (integration is preformed ccw).
when R tends to infinity, the arch part of the contour yields zero, and so we are left with the part along the real axis, which is the one I'm trying to evalute.
there are no other poles in my contour, only a second order pole at $z=0$ lying on it. the residue of this pole is $0$ so the integral sums up to be zero (by using Cauchy principal value.)
However my function is always positive and greater than $0$, so this doesn't make sense.
Any help would be appreciated
complex-analysis complex-integration
edited Jan 17 at 12:25
Somos
13.7k11235
asked Jan 17 at 11:44
user635635user635635
141
$endgroup$
$begingroup$
Residue Theorem is not applicable when there is a pole on the contour.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 11:49
$begingroup$
I took an infinitely small half cirlce around that point, so it only contributes half of what is did if it was inside my contour so it contributes ipiRes(z=0). but since the Res is 0 it contributes nothing.
$endgroup$
– user635635
Jan 17 at 11:58
$begingroup$
When there are serious questions about the very existence of the integral you are assuming that it exists and the integral over a line slightly below the real axis tend to integral over the real line. This is way off any rigorous argument.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:07
$begingroup$
Note this is a reply to a comment below - I can't put the reply in its proper place because that answer is deleted: Yes, the integral over the small circle is $0$. But the integral over half of the small circle is not half the integral over the small circle.
$endgroup$
– David C. Ullrich
Jan 17 at 15:16
add a comment |
$begingroup$
so I'm trying to evaluate the integral in the title,
$$int_{-infty}^{infty}frac{1}{x^2}dx$$
by using complex plane integration. I've chosen my contour to be a infinte half circle with it's diameter on the real axis. (integration is preformed ccw).
when R tends to infinity, the arch part of the contour yields zero, and so we are left with the part along the real axis, which is the one I'm trying to evalute.
there are no other poles in my contour, only a second order pole at $z=0$ lying on it. the residue of this pole is $0$ so the integral sums up to be zero (by using Cauchy principal value.)
However my function is always positive and greater than $0$, so this doesn't make sense.
Any help would be appreciated
complex-analysis complex-integration
edited Jan 17 at 12:25
Somos
13.7k11235
asked Jan 17 at 11:44
user635635user635635
141
$endgroup$
so I'm trying to evaluate the integral in the title,
$$int_{-infty}^{infty}frac{1}{x^2}dx$$
by using complex plane integration. I've chosen my contour to be a infinte half circle with it's diameter on the real axis. (integration is preformed ccw).
when R tends to infinity, the arch part of the contour yields zero, and so we are left with the part along the real axis, which is the one I'm trying to evalute.
there are no other poles in my contour, only a second order pole at $z=0$ lying on it. the residue of this pole is $0$ so the integral sums up to be zero (by using Cauchy principal value.)
However my function is always positive and greater than $0$, so this doesn't make sense.
Any help would be appreciated
complex-analysis complex-integration
complex-analysis complex-integration
edited Jan 17 at 12:25
Somos
13.7k11235
asked Jan 17 at 11:44
user635635user635635
141
edited Jan 17 at 12:25
Somos
13.7k11235
asked Jan 17 at 11:44
user635635user635635
141
edited Jan 17 at 12:25
Somos
13.7k11235
edited Jan 17 at 12:25
Somos
13.7k11235
edited Jan 17 at 12:25
Somos
13.7k11235
13.7k11235
asked Jan 17 at 11:44
user635635user635635
141
asked Jan 17 at 11:44
user635635user635635
141
asked Jan 17 at 11:44
user635635user635635
141
141
$begingroup$
Residue Theorem is not applicable when there is a pole on the contour.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 11:49
$begingroup$
I took an infinitely small half cirlce around that point, so it only contributes half of what is did if it was inside my contour so it contributes ipiRes(z=0). but since the Res is 0 it contributes nothing.
$endgroup$
– user635635
Jan 17 at 11:58
$begingroup$
When there are serious questions about the very existence of the integral you are assuming that it exists and the integral over a line slightly below the real axis tend to integral over the real line. This is way off any rigorous argument.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:07
$begingroup$
Note this is a reply to a comment below - I can't put the reply in its proper place because that answer is deleted: Yes, the integral over the small circle is $0$. But the integral over half of the small circle is not half the integral over the small circle.
$endgroup$
– David C. Ullrich
Jan 17 at 15:16
add a comment |
$begingroup$
Residue Theorem is not applicable when there is a pole on the contour.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 11:49
$begingroup$
I took an infinitely small half cirlce around that point, so it only contributes half of what is did if it was inside my contour so it contributes ipiRes(z=0). but since the Res is 0 it contributes nothing.
$endgroup$
– user635635
Jan 17 at 11:58
$begingroup$
When there are serious questions about the very existence of the integral you are assuming that it exists and the integral over a line slightly below the real axis tend to integral over the real line. This is way off any rigorous argument.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:07
$begingroup$
Note this is a reply to a comment below - I can't put the reply in its proper place because that answer is deleted: Yes, the integral over the small circle is $0$. But the integral over half of the small circle is not half the integral over the small circle.
$endgroup$
– David C. Ullrich
Jan 17 at 15:16
$begingroup$
Residue Theorem is not applicable when there is a pole on the contour.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 11:49
$begingroup$
Residue Theorem is not applicable when there is a pole on the contour.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 11:49
$begingroup$
I took an infinitely small half cirlce around that point, so it only contributes half of what is did if it was inside my contour so it contributes ipiRes(z=0). but since the Res is 0 it contributes nothing.
$endgroup$
– user635635
Jan 17 at 11:58
$begingroup$
I took an infinitely small half cirlce around that point, so it only contributes half of what is did if it was inside my contour so it contributes ipiRes(z=0). but since the Res is 0 it contributes nothing.
$endgroup$
– user635635
Jan 17 at 11:58
$begingroup$
When there are serious questions about the very existence of the integral you are assuming that it exists and the integral over a line slightly below the real axis tend to integral over the real line. This is way off any rigorous argument.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:07
$begingroup$
When there are serious questions about the very existence of the integral you are assuming that it exists and the integral over a line slightly below the real axis tend to integral over the real line. This is way off any rigorous argument.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:07
$begingroup$
Note this is a reply to a comment below - I can't put the reply in its proper place because that answer is deleted: Yes, the integral over the small circle is $0$. But the integral over half of the small circle is not half the integral over the small circle.
$endgroup$
– David C. Ullrich
Jan 17 at 15:16
$begingroup$
Note this is a reply to a comment below - I can't put the reply in its proper place because that answer is deleted: Yes, the integral over the small circle is $0$. But the integral over half of the small circle is not half the integral over the small circle.
$endgroup$
– David C. Ullrich
Jan 17 at 15:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The integral diverges even if you chose the principal value. To show this by using residue theorem, first chose a contour C composed by the following:
$$C_R: text{ big semicircle centered at origin, with radius $R$, counterclockwise}$$
$$C_1: text{ straight line from $-R$ to $-r$, towards $+infty$}$$
$$C_r: text{ small semicircle centered at origin, with radius $r$, clockwise}$$
$$C_2: text{ straight line from $R$ to $r$, towards $+infty$}$$
$$int_Cfrac{dz}{z^2}=0$$ by residue theorem.
$$int_{-infty}^inftyfrac{dx}{x^2}=0-lim_{Rtoinfty}int_{C_R}frac{dz}{z^2}-lim_{rto0}int_{C_r}frac{dz}{z^2}\
=0-0-(-infty)=infty$$
The calculation of the limits is left to readers.
answered Jan 17 at 11:58
Kemono ChenKemono Chen
3,0721743
$endgroup$
$begingroup$
perhaps i should clarify that I'm doing this in order to find the integral over real line of (sinc{x})^2 which is pi. Works fine if I take the integral in the title to be zero, but as you said, I believe it should converge.
$endgroup$
– user635635
Jan 17 at 12:06
$begingroup$
@user635635, the integral in your title must diverge I promise. I think, your calculation may be wrong.
$endgroup$
– Kemono Chen
Jan 17 at 12:08
add a comment |
$begingroup$
Usually you'd need this kind of contour if you happen to have poles on the real axis. We take the small radius $rto 0$ and the large radius $Rtoinfty$. The value on the bigger half circle should decay fast enough so the value of the integral would depend on the limit of the smaller half circle.
You can see this example of a similar calculation.
answered Jan 17 at 11:58
BigbearZzzBigbearZzz
8,69121652
$endgroup$
$begingroup$
You should probably mention explicitly that if $f==1/z^2$ then the integral over the smaller half-circle does not tend to $0$ as $rto0$.
$endgroup$
– David C. Ullrich
Jan 17 at 15:13
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076881%2fa-weird-value-obtained-by-using-cauchy-principal-value-on-int-infty-inft%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The integral diverges even if you chose the principal value. To show this by using residue theorem, first chose a contour C composed by the following:
$$C_R: text{ big semicircle centered at origin, with radius $R$, counterclockwise}$$
$$C_1: text{ straight line from $-R$ to $-r$, towards $+infty$}$$
$$C_r: text{ small semicircle centered at origin, with radius $r$, clockwise}$$
$$C_2: text{ straight line from $R$ to $r$, towards $+infty$}$$
$$int_Cfrac{dz}{z^2}=0$$ by residue theorem.
$$int_{-infty}^inftyfrac{dx}{x^2}=0-lim_{Rtoinfty}int_{C_R}frac{dz}{z^2}-lim_{rto0}int_{C_r}frac{dz}{z^2}\
=0-0-(-infty)=infty$$
The calculation of the limits is left to readers.
answered Jan 17 at 11:58
Kemono ChenKemono Chen
3,0721743
$endgroup$
$begingroup$
perhaps i should clarify that I'm doing this in order to find the integral over real line of (sinc{x})^2 which is pi. Works fine if I take the integral in the title to be zero, but as you said, I believe it should converge.
$endgroup$
– user635635
Jan 17 at 12:06
$begingroup$
@user635635, the integral in your title must diverge I promise. I think, your calculation may be wrong.
$endgroup$
– Kemono Chen
Jan 17 at 12:08
add a comment |
$begingroup$
The integral diverges even if you chose the principal value. To show this by using residue theorem, first chose a contour C composed by the following:
$$C_R: text{ big semicircle centered at origin, with radius $R$, counterclockwise}$$
$$C_1: text{ straight line from $-R$ to $-r$, towards $+infty$}$$
$$C_r: text{ small semicircle centered at origin, with radius $r$, clockwise}$$
$$C_2: text{ straight line from $R$ to $r$, towards $+infty$}$$
$$int_Cfrac{dz}{z^2}=0$$ by residue theorem.
$$int_{-infty}^inftyfrac{dx}{x^2}=0-lim_{Rtoinfty}int_{C_R}frac{dz}{z^2}-lim_{rto0}int_{C_r}frac{dz}{z^2}\
=0-0-(-infty)=infty$$
The calculation of the limits is left to readers.
answered Jan 17 at 11:58
Kemono ChenKemono Chen
3,0721743
$endgroup$
$begingroup$
perhaps i should clarify that I'm doing this in order to find the integral over real line of (sinc{x})^2 which is pi. Works fine if I take the integral in the title to be zero, but as you said, I believe it should converge.
$endgroup$
– user635635
Jan 17 at 12:06
$begingroup$
@user635635, the integral in your title must diverge I promise. I think, your calculation may be wrong.
$endgroup$
– Kemono Chen
Jan 17 at 12:08
add a comment |
$begingroup$
The integral diverges even if you chose the principal value. To show this by using residue theorem, first chose a contour C composed by the following:
$$C_R: text{ big semicircle centered at origin, with radius $R$, counterclockwise}$$
$$C_1: text{ straight line from $-R$ to $-r$, towards $+infty$}$$
$$C_r: text{ small semicircle centered at origin, with radius $r$, clockwise}$$
$$C_2: text{ straight line from $R$ to $r$, towards $+infty$}$$
$$int_Cfrac{dz}{z^2}=0$$ by residue theorem.
$$int_{-infty}^inftyfrac{dx}{x^2}=0-lim_{Rtoinfty}int_{C_R}frac{dz}{z^2}-lim_{rto0}int_{C_r}frac{dz}{z^2}\
=0-0-(-infty)=infty$$
The calculation of the limits is left to readers.
answered Jan 17 at 11:58
Kemono ChenKemono Chen
3,0721743
$endgroup$
The integral diverges even if you chose the principal value. To show this by using residue theorem, first chose a contour C composed by the following:
$$C_R: text{ big semicircle centered at origin, with radius $R$, counterclockwise}$$
$$C_1: text{ straight line from $-R$ to $-r$, towards $+infty$}$$
$$C_r: text{ small semicircle centered at origin, with radius $r$, clockwise}$$
$$C_2: text{ straight line from $R$ to $r$, towards $+infty$}$$
$$int_Cfrac{dz}{z^2}=0$$ by residue theorem.
$$int_{-infty}^inftyfrac{dx}{x^2}=0-lim_{Rtoinfty}int_{C_R}frac{dz}{z^2}-lim_{rto0}int_{C_r}frac{dz}{z^2}\
=0-0-(-infty)=infty$$
The calculation of the limits is left to readers.
answered Jan 17 at 11:58
Kemono ChenKemono Chen
3,0721743
answered Jan 17 at 11:58
Kemono ChenKemono Chen
3,0721743
answered Jan 17 at 11:58
Kemono ChenKemono Chen
3,0721743
answered Jan 17 at 11:58
Kemono ChenKemono Chen
3,0721743
3,0721743
$begingroup$
perhaps i should clarify that I'm doing this in order to find the integral over real line of (sinc{x})^2 which is pi. Works fine if I take the integral in the title to be zero, but as you said, I believe it should converge.
$endgroup$
– user635635
Jan 17 at 12:06
$begingroup$
@user635635, the integral in your title must diverge I promise. I think, your calculation may be wrong.
$endgroup$
– Kemono Chen
Jan 17 at 12:08
add a comment |
$begingroup$
perhaps i should clarify that I'm doing this in order to find the integral over real line of (sinc{x})^2 which is pi. Works fine if I take the integral in the title to be zero, but as you said, I believe it should converge.
$endgroup$
– user635635
Jan 17 at 12:06
$begingroup$
@user635635, the integral in your title must diverge I promise. I think, your calculation may be wrong.
$endgroup$
– Kemono Chen
Jan 17 at 12:08
$begingroup$
perhaps i should clarify that I'm doing this in order to find the integral over real line of (sinc{x})^2 which is pi. Works fine if I take the integral in the title to be zero, but as you said, I believe it should converge.
$endgroup$
– user635635
Jan 17 at 12:06
$begingroup$
perhaps i should clarify that I'm doing this in order to find the integral over real line of (sinc{x})^2 which is pi. Works fine if I take the integral in the title to be zero, but as you said, I believe it should converge.
$endgroup$
– user635635
Jan 17 at 12:06
$begingroup$
@user635635, the integral in your title must diverge I promise. I think, your calculation may be wrong.
$endgroup$
– Kemono Chen
Jan 17 at 12:08
$begingroup$
@user635635, the integral in your title must diverge I promise. I think, your calculation may be wrong.
$endgroup$
– Kemono Chen
Jan 17 at 12:08
add a comment |
$begingroup$
Usually you'd need this kind of contour if you happen to have poles on the real axis. We take the small radius $rto 0$ and the large radius $Rtoinfty$. The value on the bigger half circle should decay fast enough so the value of the integral would depend on the limit of the smaller half circle.
You can see this example of a similar calculation.
answered Jan 17 at 11:58
BigbearZzzBigbearZzz
8,69121652
$endgroup$
$begingroup$
You should probably mention explicitly that if $f==1/z^2$ then the integral over the smaller half-circle does not tend to $0$ as $rto0$.
$endgroup$
– David C. Ullrich
Jan 17 at 15:13
add a comment |
$begingroup$
Usually you'd need this kind of contour if you happen to have poles on the real axis. We take the small radius $rto 0$ and the large radius $Rtoinfty$. The value on the bigger half circle should decay fast enough so the value of the integral would depend on the limit of the smaller half circle.
You can see this example of a similar calculation.
answered Jan 17 at 11:58
BigbearZzzBigbearZzz
8,69121652
$endgroup$
$begingroup$
You should probably mention explicitly that if $f==1/z^2$ then the integral over the smaller half-circle does not tend to $0$ as $rto0$.
$endgroup$
– David C. Ullrich
Jan 17 at 15:13
add a comment |
$begingroup$
Usually you'd need this kind of contour if you happen to have poles on the real axis. We take the small radius $rto 0$ and the large radius $Rtoinfty$. The value on the bigger half circle should decay fast enough so the value of the integral would depend on the limit of the smaller half circle.
You can see this example of a similar calculation.
answered Jan 17 at 11:58
BigbearZzzBigbearZzz
8,69121652
$endgroup$
Usually you'd need this kind of contour if you happen to have poles on the real axis. We take the small radius $rto 0$ and the large radius $Rtoinfty$. The value on the bigger half circle should decay fast enough so the value of the integral would depend on the limit of the smaller half circle.
You can see this example of a similar calculation.
answered Jan 17 at 11:58
BigbearZzzBigbearZzz
8,69121652
answered Jan 17 at 11:58
BigbearZzzBigbearZzz
8,69121652
answered Jan 17 at 11:58
BigbearZzzBigbearZzz
8,69121652
answered Jan 17 at 11:58
BigbearZzzBigbearZzz
8,69121652
8,69121652
$begingroup$
You should probably mention explicitly that if $f==1/z^2$ then the integral over the smaller half-circle does not tend to $0$ as $rto0$.
$endgroup$
– David C. Ullrich
Jan 17 at 15:13
add a comment |
$begingroup$
You should probably mention explicitly that if $f==1/z^2$ then the integral over the smaller half-circle does not tend to $0$ as $rto0$.
$endgroup$
– David C. Ullrich
Jan 17 at 15:13
$begingroup$
You should probably mention explicitly that if $f==1/z^2$ then the integral over the smaller half-circle does not tend to $0$ as $rto0$.
$endgroup$
– David C. Ullrich
Jan 17 at 15:13
$begingroup$
You should probably mention explicitly that if $f==1/z^2$ then the integral over the smaller half-circle does not tend to $0$ as $rto0$.
$endgroup$
– David C. Ullrich
Jan 17 at 15:13
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076881%2fa-weird-value-obtained-by-using-cauchy-principal-value-on-int-infty-inft%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Residue Theorem is not applicable when there is a pole on the contour.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 11:49
$begingroup$
I took an infinitely small half cirlce around that point, so it only contributes half of what is did if it was inside my contour so it contributes ipiRes(z=0). but since the Res is 0 it contributes nothing.
$endgroup$
– user635635
Jan 17 at 11:58
$begingroup$
When there are serious questions about the very existence of the integral you are assuming that it exists and the integral over a line slightly below the real axis tend to integral over the real line. This is way off any rigorous argument.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:07
$begingroup$
Note this is a reply to a comment below - I can't put the reply in its proper place because that answer is deleted: Yes, the integral over the small circle is $0$. But the integral over half of the small circle is not half the integral over the small circle.
$endgroup$
– David C. Ullrich
Jan 17 at 15:16