How can I prove that integral is zero precisely?
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
asked 34 mins ago
Est Mayhem
111
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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
asked 34 mins ago
Est Mayhem
111
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.
What is your background?
– Will Jagy
32 mins ago
3
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
31 mins ago
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
30 mins ago
@Tyberius thank you! That's gonna help.
– Est Mayhem
25 mins ago
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
asked 34 mins ago
Est Mayhem
111
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
asked 34 mins ago
Est Mayhem
111
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 34 mins ago
Est Mayhem
111
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 34 mins ago
Est Mayhem
111
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 34 mins ago
Est Mayhem
111
asked 34 mins ago
Est Mayhem
111
111
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.
What is your background?
– Will Jagy
32 mins ago
3
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
31 mins ago
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
30 mins ago
@Tyberius thank you! That's gonna help.
– Est Mayhem
25 mins ago
add a comment |
What is your background?
– Will Jagy
32 mins ago
3
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
31 mins ago
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
30 mins ago
@Tyberius thank you! That's gonna help.
– Est Mayhem
25 mins ago
What is your background?
– Will Jagy
32 mins ago
What is your background?
– Will Jagy
32 mins ago
3
3
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
31 mins ago
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
31 mins ago
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
30 mins ago
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
30 mins ago
@Tyberius thank you! That's gonna help.
– Est Mayhem
25 mins ago
@Tyberius thank you! That's gonna help.
– Est Mayhem
25 mins ago
add a comment |
2 Answers
2
active
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votes
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 31 mins ago
ItsJustLogicBro
1461
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ItsJustLogicBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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What if this a Lebesgue integral?
– Est Mayhem
25 mins ago
I got it. Thanks for the reply!
– Est Mayhem
21 mins ago
@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
19 mins 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 21 mins ago
Frank W.
3,0461320
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 31 mins ago
ItsJustLogicBro
1461
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
25 mins ago
I got it. Thanks for the reply!
– Est Mayhem
21 mins ago
@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
19 mins 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 31 mins ago
ItsJustLogicBro
1461
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
25 mins ago
I got it. Thanks for the reply!
– Est Mayhem
21 mins ago
@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
19 mins 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 31 mins ago
ItsJustLogicBro
1461
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 31 mins ago
ItsJustLogicBro
1461
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 31 mins ago
ItsJustLogicBro
1461
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 31 mins ago
ItsJustLogicBro
1461
answered 31 mins ago
ItsJustLogicBro
1461
1461
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.
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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
25 mins ago
I got it. Thanks for the reply!
– Est Mayhem
21 mins ago
@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
19 mins ago
add a comment |
What if this a Lebesgue integral?
– Est Mayhem
25 mins ago
I got it. Thanks for the reply!
– Est Mayhem
21 mins ago
@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
19 mins ago
What if this a Lebesgue integral?
– Est Mayhem
25 mins ago
What if this a Lebesgue integral?
– Est Mayhem
25 mins ago
I got it. Thanks for the reply!
– Est Mayhem
21 mins ago
I got it. Thanks for the reply!
– Est Mayhem
21 mins ago
@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
19 mins ago
@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
19 mins 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 21 mins ago
Frank W.
3,0461320
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 21 mins ago
Frank W.
3,0461320
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 21 mins ago
Frank W.
3,0461320
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 21 mins ago
Frank W.
3,0461320
answered 21 mins ago
Frank W.
3,0461320
answered 21 mins ago
Frank W.
3,0461320
answered 21 mins ago
Frank W.
3,0461320
3,0461320
add a comment |
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.
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What is your background?
– Will Jagy
32 mins ago
3
This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
31 mins ago
@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
30 mins ago
@Tyberius thank you! That's gonna help.
– Est Mayhem
25 mins ago