How can I prove that integral is zero precisely?












1














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







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  • 1




    What is your background?
    – Will Jagy
    52 mins ago






  • 3




    This is a divergent improper integral. WA might give you the principal value, not the exact value.
    – xbh
    51 mins ago










  • @EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
    – Tyberius
    50 mins ago










  • @Tyberius thank you! That's gonna help.
    – Est Mayhem
    45 mins ago
















1














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







share|cite|improve this question







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
    52 mins ago






  • 3




    This is a divergent improper integral. WA might give you the principal value, not the exact value.
    – xbh
    51 mins ago










  • @EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
    – Tyberius
    50 mins ago










  • @Tyberius thank you! That's gonna help.
    – Est Mayhem
    45 mins ago














1












1








1


1





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







share|cite|improve this question







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






share|cite|improve this question







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.











share|cite|improve this question







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.









share|cite|improve this question




share|cite|improve this question






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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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asked 54 mins ago









Est Mayhem

91




91




New contributor




Est Mayhem is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    52 mins ago






  • 3




    This is a divergent improper integral. WA might give you the principal value, not the exact value.
    – xbh
    51 mins ago










  • @EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
    – Tyberius
    50 mins ago










  • @Tyberius thank you! That's gonna help.
    – Est Mayhem
    45 mins ago














  • 1




    What is your background?
    – Will Jagy
    52 mins ago






  • 3




    This is a divergent improper integral. WA might give you the principal value, not the exact value.
    – xbh
    51 mins ago










  • @EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
    – Tyberius
    50 mins ago










  • @Tyberius thank you! That's gonna help.
    – Est Mayhem
    45 mins ago








1




1




What is your background?
– Will Jagy
52 mins ago




What is your background?
– Will Jagy
52 mins ago




3




3




This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
51 mins ago




This is a divergent improper integral. WA might give you the principal value, not the exact value.
– xbh
51 mins ago












@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
50 mins ago




@EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples.
– Tyberius
50 mins ago












@Tyberius thank you! That's gonna help.
– Est Mayhem
45 mins ago




@Tyberius thank you! That's gonna help.
– Est Mayhem
45 mins ago










2 Answers
2






active

oldest

votes


















8














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.






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New contributor




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
    45 mins ago










  • I got it. Thanks for the reply!
    – Est Mayhem
    41 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
    39 mins ago



















0














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.$$begin{align*}intlimits_{-1}^1frac {mathrm dx}x & =limlimits_{varepsilonto0}left[intlimits_{-1}^{-varepsilon}frac {mathrm dx}x+intlimits_{varepsilon}^1frac {mathrm dx}xright]\ & =limlimits_{varepsilonto0}Bigr[log(-varepsilon)-log(-1)+log 1-log(varepsilon)Bigr]\ & =limlimits_{varepsilonto0}logleft(frac {-varepsilon}{-1}frac 1{varepsilon}right)\ & =log 1\ & =0end{align*}$$






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8














    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.






    share|cite|improve this answer








    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
      45 mins ago










    • I got it. Thanks for the reply!
      – Est Mayhem
      41 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
      39 mins ago
















    8














    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.






    share|cite|improve this answer








    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
      45 mins ago










    • I got it. Thanks for the reply!
      – Est Mayhem
      41 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
      39 mins ago














    8












    8








    8






    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.






    share|cite|improve this answer








    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.







    share|cite|improve this answer








    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.









    share|cite|improve this answer



    share|cite|improve this answer






    New contributor




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    answered 52 mins ago









    ItsJustLogicBro

    1561




    1561




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    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.
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    • What if this a Lebesgue integral?
      – Est Mayhem
      45 mins ago










    • I got it. Thanks for the reply!
      – Est Mayhem
      41 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
      39 mins ago


















    • What if this a Lebesgue integral?
      – Est Mayhem
      45 mins ago










    • I got it. Thanks for the reply!
      – Est Mayhem
      41 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
      39 mins ago
















    What if this a Lebesgue integral?
    – Est Mayhem
    45 mins ago




    What if this a Lebesgue integral?
    – Est Mayhem
    45 mins ago












    I got it. Thanks for the reply!
    – Est Mayhem
    41 mins ago




    I got it. Thanks for the reply!
    – Est Mayhem
    41 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
    39 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
    39 mins ago











    0














    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.$$begin{align*}intlimits_{-1}^1frac {mathrm dx}x & =limlimits_{varepsilonto0}left[intlimits_{-1}^{-varepsilon}frac {mathrm dx}x+intlimits_{varepsilon}^1frac {mathrm dx}xright]\ & =limlimits_{varepsilonto0}Bigr[log(-varepsilon)-log(-1)+log 1-log(varepsilon)Bigr]\ & =limlimits_{varepsilonto0}logleft(frac {-varepsilon}{-1}frac 1{varepsilon}right)\ & =log 1\ & =0end{align*}$$






    share|cite|improve this answer


























      0














      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.$$begin{align*}intlimits_{-1}^1frac {mathrm dx}x & =limlimits_{varepsilonto0}left[intlimits_{-1}^{-varepsilon}frac {mathrm dx}x+intlimits_{varepsilon}^1frac {mathrm dx}xright]\ & =limlimits_{varepsilonto0}Bigr[log(-varepsilon)-log(-1)+log 1-log(varepsilon)Bigr]\ & =limlimits_{varepsilonto0}logleft(frac {-varepsilon}{-1}frac 1{varepsilon}right)\ & =log 1\ & =0end{align*}$$






      share|cite|improve this answer
























        0












        0








        0






        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.$$begin{align*}intlimits_{-1}^1frac {mathrm dx}x & =limlimits_{varepsilonto0}left[intlimits_{-1}^{-varepsilon}frac {mathrm dx}x+intlimits_{varepsilon}^1frac {mathrm dx}xright]\ & =limlimits_{varepsilonto0}Bigr[log(-varepsilon)-log(-1)+log 1-log(varepsilon)Bigr]\ & =limlimits_{varepsilonto0}logleft(frac {-varepsilon}{-1}frac 1{varepsilon}right)\ & =log 1\ & =0end{align*}$$






        share|cite|improve this answer












        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.$$begin{align*}intlimits_{-1}^1frac {mathrm dx}x & =limlimits_{varepsilonto0}left[intlimits_{-1}^{-varepsilon}frac {mathrm dx}x+intlimits_{varepsilon}^1frac {mathrm dx}xright]\ & =limlimits_{varepsilonto0}Bigr[log(-varepsilon)-log(-1)+log 1-log(varepsilon)Bigr]\ & =limlimits_{varepsilonto0}logleft(frac {-varepsilon}{-1}frac 1{varepsilon}right)\ & =log 1\ & =0end{align*}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 42 mins ago









        Frank W.

        3,0461320




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