A weird value obtained by using Cauchy Principal Value on $int_{-infty}^{infty}frac{1}{x^2}dx$












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2


0



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 12:25









Somos

15k11337




15k11337










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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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





















0












$begingroup$

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.



fig 1






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 11:58









Kemono ChenKemono Chen

3,2461844




3,2461844












  • $begingroup$
    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










  • $begingroup$
    @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













0












$begingroup$

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.



fig 1






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















0












$begingroup$

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.



fig 1






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

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.



fig 1






share|cite|improve this answer









$endgroup$



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.



fig 1







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 11:58









BigbearZzzBigbearZzz

9,07421653




9,07421653












  • $begingroup$
    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






$begingroup$
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




















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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2
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