How can I prove that $int _{-1}^{1} frac{1}{x} dx =0 $
According to WolframAlpha $int _{-1}^{1} frac{1}{x} dx =0 $. But how can this be proved rigorously? I know that the function is odd, but it's unbounded on $[-1;1]$. Leibniz–Newton formula cannot be applied here either. I would really appreciate some help with this matter.
real-analysis calculus integration
edited 2 days ago
Asaf Karagila♦
302k32427757
asked Jan 6 at 3:36
Est MayhemEst Mayhem
463
New contributor
Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
According to WolframAlpha $int _{-1}^{1} frac{1}{x} dx =0 $. But how can this be proved rigorously? I know that the function is odd, but it's unbounded on $[-1;1]$. Leibniz–Newton formula cannot be applied here either. I would really appreciate some help with this matter.
real-analysis calculus integration
edited 2 days ago
Asaf Karagila♦
302k32427757
asked Jan 6 at 3:36
Est MayhemEst Mayhem
463
New contributor
Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
What is your background?
– Will Jagy
Jan 6 at 3:38
5
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
Jan 6 at 3:39
1
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
Jan 6 at 3:40
@Tyberius thank you! That's gonna help.
– Est Mayhem
Jan 6 at 3:45
add a comment |
According to WolframAlpha $int _{-1}^{1} frac{1}{x} dx =0 $. But how can this be proved rigorously? I know that the function is odd, but it's unbounded on $[-1;1]$. Leibniz–Newton formula cannot be applied here either. I would really appreciate some help with this matter.
real-analysis calculus integration
edited 2 days ago
Asaf Karagila♦
302k32427757
asked Jan 6 at 3:36
Est MayhemEst Mayhem
463
New contributor
Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
According to WolframAlpha $int _{-1}^{1} frac{1}{x} dx =0 $. But how can this be proved rigorously? I know that the function is odd, but it's unbounded on $[-1;1]$. Leibniz–Newton formula cannot be applied here either. I would really appreciate some help with this matter.
real-analysis calculus integration
real-analysis calculus integration
edited 2 days ago
Asaf Karagila♦
302k32427757
asked Jan 6 at 3:36
Est MayhemEst Mayhem
463
New contributor
Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Asaf Karagila♦
302k32427757
asked Jan 6 at 3:36
Est MayhemEst Mayhem
463
New contributor
Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Asaf Karagila♦
302k32427757
edited 2 days ago
Asaf Karagila♦
302k32427757
edited 2 days ago
Asaf Karagila♦
302k32427757
302k32427757
asked Jan 6 at 3:36
Est MayhemEst Mayhem
463
New contributor
Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 6 at 3:36
Est MayhemEst Mayhem
463
asked Jan 6 at 3:36
Est MayhemEst Mayhem
463
463
New contributor
Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
What is your background?
– Will Jagy
Jan 6 at 3:38
5
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
Jan 6 at 3:39
1
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
Jan 6 at 3:40
@Tyberius thank you! That's gonna help.
– Est Mayhem
Jan 6 at 3:45
add a comment |
1
What is your background?
– Will Jagy
Jan 6 at 3:38
5
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
Jan 6 at 3:39
1
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
Jan 6 at 3:40
@Tyberius thank you! That's gonna help.
– Est Mayhem
Jan 6 at 3:45
1
1
What is your background?
– Will Jagy
Jan 6 at 3:38
What is your background?
– Will Jagy
Jan 6 at 3:38
5
5
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
Jan 6 at 3:39
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
Jan 6 at 3:39
1
1
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
Jan 6 at 3:40
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
Jan 6 at 3:40
@Tyberius thank you! That's gonna help.
– Est Mayhem
Jan 6 at 3:45
@Tyberius thank you! That's gonna help.
– Est Mayhem
Jan 6 at 3:45
add a comment |
2 Answers
2
active
oldest
votes
The integral is divergent due to $f(x)=frac 1x$ being undefined at $x=0$. However, if we consider the principal value of the integral, then the answer is zero. I'm also assuming this is what Wolfram Alpha gives.
To prove this, split the integral up over its positive and negative parts and replace the zero with a variable and have it tend towards zero.$$begin{align*}intlimits_{-1}^1frac {mathrm dx}x & =limlimits_{varepsilonto0}left[intlimits_{-1}^{-varepsilon}frac {mathrm dx}x+intlimits_{varepsilon}^1frac {mathrm dx}xright]\ & =limlimits_{varepsilonto0}Bigr[log(-varepsilon)-log(-1)+log 1-log(varepsilon)Bigr]\ & =limlimits_{varepsilonto0}logleft(frac {-varepsilon}{-1}frac 1{varepsilon}right)\ & =log 1\ & =0end{align*}$$
answered Jan 6 at 3:49
Frank W.Frank W.
3,1891321
2
The first sentence is not correct. The reason the integral is divergent is not because $1/x$ is undefined at $x=0$, it's because a computation shows that it's divergent.
– Hans Lundmark
2 days ago
Also, you need to take $epsilon to 0^+$ and use $log|x|$ as your antiderivative instead of $log(x)$. Your computation with $log(-1)$ is clearly nonsense.
– Hans Lundmark
2 days ago
@HansLundmark: $log x$ as antiderivative of $1/x$ is correct for $x$ in the complex plane. The subtraction $log(-varepsilon)-log(-1)$, is real, although each of the two logs is complex. This way of doing it may be inadvisable for beginners, but it is not "clearly nonsense".
– GEdgar
2 days ago
@GEdgar: I know that. But in the context of trying to explain things about real integrals to someone whose background you don't know (a calculus student perhaps?), I don't think it makes any sense at all. And the log laws ($log(zw)=log z + log w$, etc.) are not valid without restrictions in the complex domain, so the step that follows definitely needs to be carefully justfied. At the very least, one should specify what branch of the complex log that is used. Just writing out a formal computation like that without any explanation doesn't prove anything.
– Hans Lundmark
2 days ago
@GEdgar: And if one uses the complex logarithm in an answer like this, at least one ought to say so! With no mention of that, it's natural to assume that it's the usual real logarithm, in which case "log(-1)" is indeed nonsense.
– Hans Lundmark
2 days ago
add a comment |
This cannot be proven rigorously because it is not technically true.
The Cauchy principal value of your integral is $0$, but if you go back to the definition of the improper Riemann integral you will find that this quantity is undefined.
answered Jan 6 at 3:38
ItsJustLogicBroItsJustLogicBro
2263
New contributor
ItsJustLogicBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What if this a Lebesgue integral?
– Est Mayhem
Jan 6 at 3:45
I got it. Thanks for the reply!
– Est Mayhem
Jan 6 at 3:49
@EstMayhem Well I don't know who replied, but you are welcome? The answer is that the function $1/x$ is not Lebesgue integrable on this interval.
– ItsJustLogicBro
Jan 6 at 3:51
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
Jan 6 at 4:36
The Lebesgue integral also does not exist.
– GEdgar
2 days ago
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
});
}
});
Est Mayhem is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3063452%2fhow-can-i-prove-that-int-11-frac1x-dx-0%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
The integral is divergent due to $f(x)=frac 1x$ being undefined at $x=0$. However, if we consider the principal value of the integral, then the answer is zero. I'm also assuming this is what Wolfram Alpha gives.
To prove this, split the integral up over its positive and negative parts and replace the zero with a variable and have it tend towards zero.$$begin{align*}intlimits_{-1}^1frac {mathrm dx}x & =limlimits_{varepsilonto0}left[intlimits_{-1}^{-varepsilon}frac {mathrm dx}x+intlimits_{varepsilon}^1frac {mathrm dx}xright]\ & =limlimits_{varepsilonto0}Bigr[log(-varepsilon)-log(-1)+log 1-log(varepsilon)Bigr]\ & =limlimits_{varepsilonto0}logleft(frac {-varepsilon}{-1}frac 1{varepsilon}right)\ & =log 1\ & =0end{align*}$$
answered Jan 6 at 3:49
Frank W.Frank W.
3,1891321
2
The first sentence is not correct. The reason the integral is divergent is not because $1/x$ is undefined at $x=0$, it's because a computation shows that it's divergent.
– Hans Lundmark
2 days ago
Also, you need to take $epsilon to 0^+$ and use $log|x|$ as your antiderivative instead of $log(x)$. Your computation with $log(-1)$ is clearly nonsense.
– Hans Lundmark
2 days ago
@HansLundmark: $log x$ as antiderivative of $1/x$ is correct for $x$ in the complex plane. The subtraction $log(-varepsilon)-log(-1)$, is real, although each of the two logs is complex. This way of doing it may be inadvisable for beginners, but it is not "clearly nonsense".
– GEdgar
2 days ago
@GEdgar: I know that. But in the context of trying to explain things about real integrals to someone whose background you don't know (a calculus student perhaps?), I don't think it makes any sense at all. And the log laws ($log(zw)=log z + log w$, etc.) are not valid without restrictions in the complex domain, so the step that follows definitely needs to be carefully justfied. At the very least, one should specify what branch of the complex log that is used. Just writing out a formal computation like that without any explanation doesn't prove anything.
– Hans Lundmark
2 days ago
@GEdgar: And if one uses the complex logarithm in an answer like this, at least one ought to say so! With no mention of that, it's natural to assume that it's the usual real logarithm, in which case "log(-1)" is indeed nonsense.
– Hans Lundmark
2 days ago
add a comment |
The integral is divergent due to $f(x)=frac 1x$ being undefined at $x=0$. However, if we consider the principal value of the integral, then the answer is zero. I'm also assuming this is what Wolfram Alpha gives.
To prove this, split the integral up over its positive and negative parts and replace the zero with a variable and have it tend towards zero.$$begin{align*}intlimits_{-1}^1frac {mathrm dx}x & =limlimits_{varepsilonto0}left[intlimits_{-1}^{-varepsilon}frac {mathrm dx}x+intlimits_{varepsilon}^1frac {mathrm dx}xright]\ & =limlimits_{varepsilonto0}Bigr[log(-varepsilon)-log(-1)+log 1-log(varepsilon)Bigr]\ & =limlimits_{varepsilonto0}logleft(frac {-varepsilon}{-1}frac 1{varepsilon}right)\ & =log 1\ & =0end{align*}$$
answered Jan 6 at 3:49
Frank W.Frank W.
3,1891321
2
The first sentence is not correct. The reason the integral is divergent is not because $1/x$ is undefined at $x=0$, it's because a computation shows that it's divergent.
– Hans Lundmark
2 days ago
Also, you need to take $epsilon to 0^+$ and use $log|x|$ as your antiderivative instead of $log(x)$. Your computation with $log(-1)$ is clearly nonsense.
– Hans Lundmark
2 days ago
@HansLundmark: $log x$ as antiderivative of $1/x$ is correct for $x$ in the complex plane. The subtraction $log(-varepsilon)-log(-1)$, is real, although each of the two logs is complex. This way of doing it may be inadvisable for beginners, but it is not "clearly nonsense".
– GEdgar
2 days ago
@GEdgar: I know that. But in the context of trying to explain things about real integrals to someone whose background you don't know (a calculus student perhaps?), I don't think it makes any sense at all. And the log laws ($log(zw)=log z + log w$, etc.) are not valid without restrictions in the complex domain, so the step that follows definitely needs to be carefully justfied. At the very least, one should specify what branch of the complex log that is used. Just writing out a formal computation like that without any explanation doesn't prove anything.
– Hans Lundmark
2 days ago
@GEdgar: And if one uses the complex logarithm in an answer like this, at least one ought to say so! With no mention of that, it's natural to assume that it's the usual real logarithm, in which case "log(-1)" is indeed nonsense.
– Hans Lundmark
2 days ago
add a comment |
The integral is divergent due to $f(x)=frac 1x$ being undefined at $x=0$. However, if we consider the principal value of the integral, then the answer is zero. I'm also assuming this is what Wolfram Alpha gives.
To prove this, split the integral up over its positive and negative parts and replace the zero with a variable and have it tend towards zero.$$begin{align*}intlimits_{-1}^1frac {mathrm dx}x & =limlimits_{varepsilonto0}left[intlimits_{-1}^{-varepsilon}frac {mathrm dx}x+intlimits_{varepsilon}^1frac {mathrm dx}xright]\ & =limlimits_{varepsilonto0}Bigr[log(-varepsilon)-log(-1)+log 1-log(varepsilon)Bigr]\ & =limlimits_{varepsilonto0}logleft(frac {-varepsilon}{-1}frac 1{varepsilon}right)\ & =log 1\ & =0end{align*}$$
answered Jan 6 at 3:49
Frank W.Frank W.
3,1891321
The integral is divergent due to $f(x)=frac 1x$ being undefined at $x=0$. However, if we consider the principal value of the integral, then the answer is zero. I'm also assuming this is what Wolfram Alpha gives.
To prove this, split the integral up over its positive and negative parts and replace the zero with a variable and have it tend towards zero.$$begin{align*}intlimits_{-1}^1frac {mathrm dx}x & =limlimits_{varepsilonto0}left[intlimits_{-1}^{-varepsilon}frac {mathrm dx}x+intlimits_{varepsilon}^1frac {mathrm dx}xright]\ & =limlimits_{varepsilonto0}Bigr[log(-varepsilon)-log(-1)+log 1-log(varepsilon)Bigr]\ & =limlimits_{varepsilonto0}logleft(frac {-varepsilon}{-1}frac 1{varepsilon}right)\ & =log 1\ & =0end{align*}$$
answered Jan 6 at 3:49
Frank W.Frank W.
3,1891321
answered Jan 6 at 3:49
Frank W.Frank W.
3,1891321
answered Jan 6 at 3:49
Frank W.Frank W.
3,1891321
answered Jan 6 at 3:49
Frank W.Frank W.
3,1891321
3,1891321
2
The first sentence is not correct. The reason the integral is divergent is not because $1/x$ is undefined at $x=0$, it's because a computation shows that it's divergent.
– Hans Lundmark
2 days ago
Also, you need to take $epsilon to 0^+$ and use $log|x|$ as your antiderivative instead of $log(x)$. Your computation with $log(-1)$ is clearly nonsense.
– Hans Lundmark
2 days ago
@HansLundmark: $log x$ as antiderivative of $1/x$ is correct for $x$ in the complex plane. The subtraction $log(-varepsilon)-log(-1)$, is real, although each of the two logs is complex. This way of doing it may be inadvisable for beginners, but it is not "clearly nonsense".
– GEdgar
2 days ago
@GEdgar: I know that. But in the context of trying to explain things about real integrals to someone whose background you don't know (a calculus student perhaps?), I don't think it makes any sense at all. And the log laws ($log(zw)=log z + log w$, etc.) are not valid without restrictions in the complex domain, so the step that follows definitely needs to be carefully justfied. At the very least, one should specify what branch of the complex log that is used. Just writing out a formal computation like that without any explanation doesn't prove anything.
– Hans Lundmark
2 days ago
@GEdgar: And if one uses the complex logarithm in an answer like this, at least one ought to say so! With no mention of that, it's natural to assume that it's the usual real logarithm, in which case "log(-1)" is indeed nonsense.
– Hans Lundmark
2 days ago
add a comment |
2
The first sentence is not correct. The reason the integral is divergent is not because $1/x$ is undefined at $x=0$, it's because a computation shows that it's divergent.
– Hans Lundmark
2 days ago
Also, you need to take $epsilon to 0^+$ and use $log|x|$ as your antiderivative instead of $log(x)$. Your computation with $log(-1)$ is clearly nonsense.
– Hans Lundmark
2 days ago
@HansLundmark: $log x$ as antiderivative of $1/x$ is correct for $x$ in the complex plane. The subtraction $log(-varepsilon)-log(-1)$, is real, although each of the two logs is complex. This way of doing it may be inadvisable for beginners, but it is not "clearly nonsense".
– GEdgar
2 days ago
@GEdgar: I know that. But in the context of trying to explain things about real integrals to someone whose background you don't know (a calculus student perhaps?), I don't think it makes any sense at all. And the log laws ($log(zw)=log z + log w$, etc.) are not valid without restrictions in the complex domain, so the step that follows definitely needs to be carefully justfied. At the very least, one should specify what branch of the complex log that is used. Just writing out a formal computation like that without any explanation doesn't prove anything.
– Hans Lundmark
2 days ago
@GEdgar: And if one uses the complex logarithm in an answer like this, at least one ought to say so! With no mention of that, it's natural to assume that it's the usual real logarithm, in which case "log(-1)" is indeed nonsense.
– Hans Lundmark
2 days ago
2
2
The first sentence is not correct. The reason the integral is divergent is not because $1/x$ is undefined at $x=0$, it's because a computation shows that it's divergent.
– Hans Lundmark
2 days ago
The first sentence is not correct. The reason the integral is divergent is not because $1/x$ is undefined at $x=0$, it's because a computation shows that it's divergent.
– Hans Lundmark
2 days ago
Also, you need to take $epsilon to 0^+$ and use $log|x|$ as your antiderivative instead of $log(x)$. Your computation with $log(-1)$ is clearly nonsense.
– Hans Lundmark
2 days ago
Also, you need to take $epsilon to 0^+$ and use $log|x|$ as your antiderivative instead of $log(x)$. Your computation with $log(-1)$ is clearly nonsense.
– Hans Lundmark
2 days ago
@HansLundmark: $log x$ as antiderivative of $1/x$ is correct for $x$ in the complex plane. The subtraction $log(-varepsilon)-log(-1)$, is real, although each of the two logs is complex. This way of doing it may be inadvisable for beginners, but it is not "clearly nonsense".
– GEdgar
2 days ago
@HansLundmark: $log x$ as antiderivative of $1/x$ is correct for $x$ in the complex plane. The subtraction $log(-varepsilon)-log(-1)$, is real, although each of the two logs is complex. This way of doing it may be inadvisable for beginners, but it is not "clearly nonsense".
– GEdgar
2 days ago
@GEdgar: I know that. But in the context of trying to explain things about real integrals to someone whose background you don't know (a calculus student perhaps?), I don't think it makes any sense at all. And the log laws ($log(zw)=log z + log w$, etc.) are not valid without restrictions in the complex domain, so the step that follows definitely needs to be carefully justfied. At the very least, one should specify what branch of the complex log that is used. Just writing out a formal computation like that without any explanation doesn't prove anything.
– Hans Lundmark
2 days ago
@GEdgar: I know that. But in the context of trying to explain things about real integrals to someone whose background you don't know (a calculus student perhaps?), I don't think it makes any sense at all. And the log laws ($log(zw)=log z + log w$, etc.) are not valid without restrictions in the complex domain, so the step that follows definitely needs to be carefully justfied. At the very least, one should specify what branch of the complex log that is used. Just writing out a formal computation like that without any explanation doesn't prove anything.
– Hans Lundmark
2 days ago
@GEdgar: And if one uses the complex logarithm in an answer like this, at least one ought to say so! With no mention of that, it's natural to assume that it's the usual real logarithm, in which case "log(-1)" is indeed nonsense.
– Hans Lundmark
2 days ago
@GEdgar: And if one uses the complex logarithm in an answer like this, at least one ought to say so! With no mention of that, it's natural to assume that it's the usual real logarithm, in which case "log(-1)" is indeed nonsense.
– Hans Lundmark
2 days ago
add a comment |
This cannot be proven rigorously because it is not technically true.
The Cauchy principal value of your integral is $0$, but if you go back to the definition of the improper Riemann integral you will find that this quantity is undefined.
answered Jan 6 at 3:38
ItsJustLogicBroItsJustLogicBro
2263
New contributor
ItsJustLogicBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What if this a Lebesgue integral?
– Est Mayhem
Jan 6 at 3:45
I got it. Thanks for the reply!
– Est Mayhem
Jan 6 at 3:49
@EstMayhem Well I don't know who replied, but you are welcome? The answer is that the function $1/x$ is not Lebesgue integrable on this interval.
– ItsJustLogicBro
Jan 6 at 3:51
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
Jan 6 at 4:36
The Lebesgue integral also does not exist.
– GEdgar
2 days ago
add a comment |
This cannot be proven rigorously because it is not technically true.
The Cauchy principal value of your integral is $0$, but if you go back to the definition of the improper Riemann integral you will find that this quantity is undefined.
answered Jan 6 at 3:38
ItsJustLogicBroItsJustLogicBro
2263
New contributor
ItsJustLogicBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What if this a Lebesgue integral?
– Est Mayhem
Jan 6 at 3:45
I got it. Thanks for the reply!
– Est Mayhem
Jan 6 at 3:49
@EstMayhem Well I don't know who replied, but you are welcome? The answer is that the function $1/x$ is not Lebesgue integrable on this interval.
– ItsJustLogicBro
Jan 6 at 3:51
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
Jan 6 at 4:36
The Lebesgue integral also does not exist.
– GEdgar
2 days ago
add a comment |
This cannot be proven rigorously because it is not technically true.
The Cauchy principal value of your integral is $0$, but if you go back to the definition of the improper Riemann integral you will find that this quantity is undefined.
answered Jan 6 at 3:38
ItsJustLogicBroItsJustLogicBro
2263
New contributor
ItsJustLogicBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This cannot be proven rigorously because it is not technically true.
The Cauchy principal value of your integral is $0$, but if you go back to the definition of the improper Riemann integral you will find that this quantity is undefined.
answered Jan 6 at 3:38
ItsJustLogicBroItsJustLogicBro
2263
New contributor
ItsJustLogicBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 6 at 3:38
ItsJustLogicBroItsJustLogicBro
2263
New contributor
ItsJustLogicBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 6 at 3:38
ItsJustLogicBroItsJustLogicBro
2263
answered Jan 6 at 3:38
ItsJustLogicBroItsJustLogicBro
2263
2263
New contributor
ItsJustLogicBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ItsJustLogicBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ItsJustLogicBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What if this a Lebesgue integral?
– Est Mayhem
Jan 6 at 3:45
I got it. Thanks for the reply!
– Est Mayhem
Jan 6 at 3:49
@EstMayhem Well I don't know who replied, but you are welcome? The answer is that the function $1/x$ is not Lebesgue integrable on this interval.
– ItsJustLogicBro
Jan 6 at 3:51
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
Jan 6 at 4:36
The Lebesgue integral also does not exist.
– GEdgar
2 days ago
add a comment |
What if this a Lebesgue integral?
– Est Mayhem
Jan 6 at 3:45
I got it. Thanks for the reply!
– Est Mayhem
Jan 6 at 3:49
@EstMayhem Well I don't know who replied, but you are welcome? The answer is that the function $1/x$ is not Lebesgue integrable on this interval.
– ItsJustLogicBro
Jan 6 at 3:51
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
Jan 6 at 4:36
The Lebesgue integral also does not exist.
– GEdgar
2 days ago
What if this a Lebesgue integral?
– Est Mayhem
Jan 6 at 3:45
What if this a Lebesgue integral?
– Est Mayhem
Jan 6 at 3:45
I got it. Thanks for the reply!
– Est Mayhem
Jan 6 at 3:49
I got it. Thanks for the reply!
– Est Mayhem
Jan 6 at 3:49
@EstMayhem Well I don't know who replied, but you are welcome? The answer is that the function $1/x$ is not Lebesgue integrable on this interval.
– ItsJustLogicBro
Jan 6 at 3:51
@EstMayhem Well I don't know who replied, but you are welcome? The answer is that the function $1/x$ is not Lebesgue integrable on this interval.
– ItsJustLogicBro
Jan 6 at 3:51
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
Jan 6 at 4:36
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
Jan 6 at 4:36
The Lebesgue integral also does not exist.
– GEdgar
2 days ago
The Lebesgue integral also does not exist.
– GEdgar
2 days ago
add a comment |
Est Mayhem is a new contributor. Be nice, and check out our Code of Conduct.
Est Mayhem is a new contributor. Be nice, and check out our Code of Conduct.
Est Mayhem is a new contributor. Be nice, and check out our Code of Conduct.
Est Mayhem is a new contributor. Be nice, and check out our Code of Conduct.
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3063452%2fhow-can-i-prove-that-int-11-frac1x-dx-0%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
1
What is your background?
– Will Jagy
Jan 6 at 3:38
5
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
Jan 6 at 3:39
1
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
Jan 6 at 3:40
@Tyberius thank you! That's gonna help.
– Est Mayhem
Jan 6 at 3:45