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 yesterday
Asaf Karagila♦
302k32426756
asked yesterday
Est Mayhem
463
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Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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 yesterday
Asaf Karagila♦
302k32426756
asked yesterday
Est 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
yesterday
5
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
yesterday
1
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
yesterday
@Tyberius thank you! That's gonna help.
– Est Mayhem
yesterday
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 yesterday
Asaf Karagila♦
302k32426756
asked yesterday
Est 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 yesterday
Asaf Karagila♦
302k32426756
asked yesterday
Est 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 yesterday
Asaf Karagila♦
302k32426756
asked yesterday
Est 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 yesterday
Asaf Karagila♦
302k32426756
edited yesterday
Asaf Karagila♦
302k32426756
edited yesterday
Asaf Karagila♦
302k32426756
302k32426756
asked yesterday
Est 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 yesterday
Est Mayhem
463
asked yesterday
Est 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
yesterday
5
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
yesterday
1
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
yesterday
@Tyberius thank you! That's gonna help.
– Est Mayhem
yesterday
add a comment |
1
What is your background?
– Will Jagy
yesterday
5
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
yesterday
1
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
yesterday
@Tyberius thank you! That's gonna help.
– Est Mayhem
yesterday
1
1
What is your background?
– Will Jagy
yesterday
What is your background?
– Will Jagy
yesterday
5
5
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
yesterday
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
yesterday
1
1
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
yesterday
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
yesterday
@Tyberius thank you! That's gonna help.
– Est Mayhem
yesterday
@Tyberius thank you! That's gonna help.
– Est Mayhem
yesterday
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 yesterday
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
yesterday
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
yesterday
@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
22 hours 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
22 hours 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
22 hours 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 yesterday
ItsJustLogicBro
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
yesterday
I got it. Thanks for the reply!
– Est Mayhem
yesterday
@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
yesterday
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
yesterday
The Lebesgue integral also does not exist.
– GEdgar
22 hours ago
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
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 yesterday
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
yesterday
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
yesterday
@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
22 hours 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
22 hours 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
22 hours 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 yesterday
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
yesterday
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
yesterday
@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
22 hours 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
22 hours 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
22 hours 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 yesterday
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 yesterday
Frank W.
3,1891321
answered yesterday
Frank W.
3,1891321
answered yesterday
Frank W.
3,1891321
answered yesterday
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
yesterday
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
yesterday
@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
22 hours 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
22 hours 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
22 hours 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
yesterday
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
yesterday
@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
22 hours 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
22 hours 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
22 hours 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
yesterday
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
yesterday
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
yesterday
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
yesterday
@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
22 hours 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
22 hours 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
22 hours 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
22 hours 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
22 hours 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
22 hours 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 yesterday
ItsJustLogicBro
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
yesterday
I got it. Thanks for the reply!
– Est Mayhem
yesterday
@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
yesterday
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
yesterday
The Lebesgue integral also does not exist.
– GEdgar
22 hours 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 yesterday
ItsJustLogicBro
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
yesterday
I got it. Thanks for the reply!
– Est Mayhem
yesterday
@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
yesterday
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
yesterday
The Lebesgue integral also does not exist.
– GEdgar
22 hours 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 yesterday
ItsJustLogicBro
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 yesterday
ItsJustLogicBro
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 yesterday
ItsJustLogicBro
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 yesterday
ItsJustLogicBro
2263
answered yesterday
ItsJustLogicBro
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
yesterday
I got it. Thanks for the reply!
– Est Mayhem
yesterday
@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
yesterday
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
yesterday
The Lebesgue integral also does not exist.
– GEdgar
22 hours ago
add a comment |
What if this a Lebesgue integral?
– Est Mayhem
yesterday
I got it. Thanks for the reply!
– Est Mayhem
yesterday
@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
yesterday
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
yesterday
The Lebesgue integral also does not exist.
– GEdgar
22 hours ago
What if this a Lebesgue integral?
– Est Mayhem
yesterday
What if this a Lebesgue integral?
– Est Mayhem
yesterday
I got it. Thanks for the reply!
– Est Mayhem
yesterday
I got it. Thanks for the reply!
– Est Mayhem
yesterday
@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
yesterday
@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
yesterday
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
yesterday
I meant your initial reply for the question itself :) Thanks.
– Est Mayhem
yesterday
The Lebesgue integral also does not exist.
– GEdgar
22 hours ago
The Lebesgue integral also does not exist.
– GEdgar
22 hours 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.
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What is your background?
– Will Jagy
yesterday
5
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
yesterday
1
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
yesterday
@Tyberius thank you! That's gonna help.
– Est Mayhem
yesterday