What is wrong with this false proof of $pi=0$?












4












$begingroup$


Consider the integral
$$I=int_{-1}^{1}frac{1}{x^2+1}mathrm{d}x$$
Now, from the standard integral results we know,
$$intfrac{1}{x^2+1}mathrm{d}x=arctan(x)+c$$
So,
$$int_{-1}^{1}frac{1}{x^2+1}mathrm{d}x=arctan(1)-arctan(-1)=frac{pi}{4}-(-frac{pi}{4})=frac{pi}{2}$$
Now, if you are bored and just doing random things you might evaluate this integral in a roundabout way by the substitution $u=dfrac{1}{x}$.



This gives the bound for the integrals as $u=dfrac{1}{-1}=-1$ to $u=dfrac{1}{1}=1$ and the differential becomes $mathrm{d}{x}=dfrac{-1}{u^2}mathrm{d}{u}$. So, the integral becomes
$$int_{-1}^{1}frac{1}{frac{1}{u^2}+1}frac{-1}{u^2}mathrm{d}u = -int_{-1}^{1}frac{1}{u^2+1}mathrm{d}u=-I$$.



Now, $$I=-I implies 2I=0implies pi=0$$.



So, there is your false proof.



I think that the problem in the proof comes when we do the substitution $x=dfrac{1}{u}$. But, I dunno what it is. Does it have to do something with the continuity of the substitution, or is it something else.






integration fake-proofs







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Think about the bounds. What happens when $xto0$?
    $endgroup$
    – Peter Foreman
    yesterday










  • $begingroup$
    @PeterForeman So, if the substituted variable becomes discontinuous at some point where the original variable was continuous so we cant proceed with the integration with substitution. Am I correct to infer that?
    $endgroup$
    – user350331
    yesterday






  • 1




    $begingroup$
    You can change limits, if variable is undefined somewhere. 1/x is discontinuous (to say more precisely) in range [-1,1]. For this you have to seperate into two integrals $int_{-1}^{0^-} + int_{0^+}^1$
    $endgroup$
    – Tojrah
    yesterday










  • $begingroup$
    BTW, with TeX and MathJax, you want the punctuation inside the double dollar sign, since that ends the line. Otherwise, you end up with the punctuation all by itself on the next line. (With a single dollar sign, you want the punctuation outside, since the punctuation isn't part of the math.)
    $endgroup$
    – Teepeemm
    yesterday
















4












$begingroup$


Consider the integral
$$I=int_{-1}^{1}frac{1}{x^2+1}mathrm{d}x$$
Now, from the standard integral results we know,
$$intfrac{1}{x^2+1}mathrm{d}x=arctan(x)+c$$
So,
$$int_{-1}^{1}frac{1}{x^2+1}mathrm{d}x=arctan(1)-arctan(-1)=frac{pi}{4}-(-frac{pi}{4})=frac{pi}{2}$$
Now, if you are bored and just doing random things you might evaluate this integral in a roundabout way by the substitution $u=dfrac{1}{x}$.



This gives the bound for the integrals as $u=dfrac{1}{-1}=-1$ to $u=dfrac{1}{1}=1$ and the differential becomes $mathrm{d}{x}=dfrac{-1}{u^2}mathrm{d}{u}$. So, the integral becomes
$$int_{-1}^{1}frac{1}{frac{1}{u^2}+1}frac{-1}{u^2}mathrm{d}u = -int_{-1}^{1}frac{1}{u^2+1}mathrm{d}u=-I$$.



Now, $$I=-I implies 2I=0implies pi=0$$.



So, there is your false proof.



I think that the problem in the proof comes when we do the substitution $x=dfrac{1}{u}$. But, I dunno what it is. Does it have to do something with the continuity of the substitution, or is it something else.






integration fake-proofs







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Think about the bounds. What happens when $xto0$?
    $endgroup$
    – Peter Foreman
    yesterday










  • $begingroup$
    @PeterForeman So, if the substituted variable becomes discontinuous at some point where the original variable was continuous so we cant proceed with the integration with substitution. Am I correct to infer that?
    $endgroup$
    – user350331
    yesterday






  • 1




    $begingroup$
    You can change limits, if variable is undefined somewhere. 1/x is discontinuous (to say more precisely) in range [-1,1]. For this you have to seperate into two integrals $int_{-1}^{0^-} + int_{0^+}^1$
    $endgroup$
    – Tojrah
    yesterday










  • $begingroup$
    BTW, with TeX and MathJax, you want the punctuation inside the double dollar sign, since that ends the line. Otherwise, you end up with the punctuation all by itself on the next line. (With a single dollar sign, you want the punctuation outside, since the punctuation isn't part of the math.)
    $endgroup$
    – Teepeemm
    yesterday














4












4








4





$begingroup$


Consider the integral
$$I=int_{-1}^{1}frac{1}{x^2+1}mathrm{d}x$$
Now, from the standard integral results we know,
$$intfrac{1}{x^2+1}mathrm{d}x=arctan(x)+c$$
So,
$$int_{-1}^{1}frac{1}{x^2+1}mathrm{d}x=arctan(1)-arctan(-1)=frac{pi}{4}-(-frac{pi}{4})=frac{pi}{2}$$
Now, if you are bored and just doing random things you might evaluate this integral in a roundabout way by the substitution $u=dfrac{1}{x}$.



This gives the bound for the integrals as $u=dfrac{1}{-1}=-1$ to $u=dfrac{1}{1}=1$ and the differential becomes $mathrm{d}{x}=dfrac{-1}{u^2}mathrm{d}{u}$. So, the integral becomes
$$int_{-1}^{1}frac{1}{frac{1}{u^2}+1}frac{-1}{u^2}mathrm{d}u = -int_{-1}^{1}frac{1}{u^2+1}mathrm{d}u=-I$$.



Now, $$I=-I implies 2I=0implies pi=0$$.



So, there is your false proof.



I think that the problem in the proof comes when we do the substitution $x=dfrac{1}{u}$. But, I dunno what it is. Does it have to do something with the continuity of the substitution, or is it something else.






integration fake-proofs







share|cite|improve this question











$endgroup$




Consider the integral
$$I=int_{-1}^{1}frac{1}{x^2+1}mathrm{d}x$$
Now, from the standard integral results we know,
$$intfrac{1}{x^2+1}mathrm{d}x=arctan(x)+c$$
So,
$$int_{-1}^{1}frac{1}{x^2+1}mathrm{d}x=arctan(1)-arctan(-1)=frac{pi}{4}-(-frac{pi}{4})=frac{pi}{2}$$
Now, if you are bored and just doing random things you might evaluate this integral in a roundabout way by the substitution $u=dfrac{1}{x}$.



This gives the bound for the integrals as $u=dfrac{1}{-1}=-1$ to $u=dfrac{1}{1}=1$ and the differential becomes $mathrm{d}{x}=dfrac{-1}{u^2}mathrm{d}{u}$. So, the integral becomes
$$int_{-1}^{1}frac{1}{frac{1}{u^2}+1}frac{-1}{u^2}mathrm{d}u = -int_{-1}^{1}frac{1}{u^2+1}mathrm{d}u=-I$$.



Now, $$I=-I implies 2I=0implies pi=0$$.



So, there is your false proof.



I think that the problem in the proof comes when we do the substitution $x=dfrac{1}{u}$. But, I dunno what it is. Does it have to do something with the continuity of the substitution, or is it something else.






integration fake-proofs




integration fake-proofs






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









YuiTo Cheng

2,43841037




2,43841037










asked yesterday









user350331user350331

1,32321232




1,32321232








  • 1




    $begingroup$
    Think about the bounds. What happens when $xto0$?
    $endgroup$
    – Peter Foreman
    yesterday










  • $begingroup$
    @PeterForeman So, if the substituted variable becomes discontinuous at some point where the original variable was continuous so we cant proceed with the integration with substitution. Am I correct to infer that?
    $endgroup$
    – user350331
    yesterday






  • 1




    $begingroup$
    You can change limits, if variable is undefined somewhere. 1/x is discontinuous (to say more precisely) in range [-1,1]. For this you have to seperate into two integrals $int_{-1}^{0^-} + int_{0^+}^1$
    $endgroup$
    – Tojrah
    yesterday










  • $begingroup$
    BTW, with TeX and MathJax, you want the punctuation inside the double dollar sign, since that ends the line. Otherwise, you end up with the punctuation all by itself on the next line. (With a single dollar sign, you want the punctuation outside, since the punctuation isn't part of the math.)
    $endgroup$
    – Teepeemm
    yesterday














  • 1




    $begingroup$
    Think about the bounds. What happens when $xto0$?
    $endgroup$
    – Peter Foreman
    yesterday










  • $begingroup$
    @PeterForeman So, if the substituted variable becomes discontinuous at some point where the original variable was continuous so we cant proceed with the integration with substitution. Am I correct to infer that?
    $endgroup$
    – user350331
    yesterday






  • 1




    $begingroup$
    You can change limits, if variable is undefined somewhere. 1/x is discontinuous (to say more precisely) in range [-1,1]. For this you have to seperate into two integrals $int_{-1}^{0^-} + int_{0^+}^1$
    $endgroup$
    – Tojrah
    yesterday










  • $begingroup$
    BTW, with TeX and MathJax, you want the punctuation inside the double dollar sign, since that ends the line. Otherwise, you end up with the punctuation all by itself on the next line. (With a single dollar sign, you want the punctuation outside, since the punctuation isn't part of the math.)
    $endgroup$
    – Teepeemm
    yesterday








1




1




$begingroup$
Think about the bounds. What happens when $xto0$?
$endgroup$
– Peter Foreman
yesterday




$begingroup$
Think about the bounds. What happens when $xto0$?
$endgroup$
– Peter Foreman
yesterday












$begingroup$
@PeterForeman So, if the substituted variable becomes discontinuous at some point where the original variable was continuous so we cant proceed with the integration with substitution. Am I correct to infer that?
$endgroup$
– user350331
yesterday




$begingroup$
@PeterForeman So, if the substituted variable becomes discontinuous at some point where the original variable was continuous so we cant proceed with the integration with substitution. Am I correct to infer that?
$endgroup$
– user350331
yesterday




1




1




$begingroup$
You can change limits, if variable is undefined somewhere. 1/x is discontinuous (to say more precisely) in range [-1,1]. For this you have to seperate into two integrals $int_{-1}^{0^-} + int_{0^+}^1$
$endgroup$
– Tojrah
yesterday




$begingroup$
You can change limits, if variable is undefined somewhere. 1/x is discontinuous (to say more precisely) in range [-1,1]. For this you have to seperate into two integrals $int_{-1}^{0^-} + int_{0^+}^1$
$endgroup$
– Tojrah
yesterday












$begingroup$
BTW, with TeX and MathJax, you want the punctuation inside the double dollar sign, since that ends the line. Otherwise, you end up with the punctuation all by itself on the next line. (With a single dollar sign, you want the punctuation outside, since the punctuation isn't part of the math.)
$endgroup$
– Teepeemm
yesterday




$begingroup$
BTW, with TeX and MathJax, you want the punctuation inside the double dollar sign, since that ends the line. Otherwise, you end up with the punctuation all by itself on the next line. (With a single dollar sign, you want the punctuation outside, since the punctuation isn't part of the math.)
$endgroup$
– Teepeemm
yesterday










2 Answers
2






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4












$begingroup$

What you should actually get is that
$$I=int_0^1 frac1{x^2+1}dx+int_{-1}^0 frac1{x^2+1}dx$$
Then the substitution will give
$$I=-int_infty^1 frac1{u^2+1}du-int_{-1}^{-infty} frac1{u^2+1}du$$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    The flaw is in changing the integration interval.



    $$-1le xle1iff frac1xle-1lor frac1xge 1$$






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

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      2 Answers
      2






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      active

      oldest

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      4












      $begingroup$

      What you should actually get is that
      $$I=int_0^1 frac1{x^2+1}dx+int_{-1}^0 frac1{x^2+1}dx$$
      Then the substitution will give
      $$I=-int_infty^1 frac1{u^2+1}du-int_{-1}^{-infty} frac1{u^2+1}du$$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        What you should actually get is that
        $$I=int_0^1 frac1{x^2+1}dx+int_{-1}^0 frac1{x^2+1}dx$$
        Then the substitution will give
        $$I=-int_infty^1 frac1{u^2+1}du-int_{-1}^{-infty} frac1{u^2+1}du$$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          What you should actually get is that
          $$I=int_0^1 frac1{x^2+1}dx+int_{-1}^0 frac1{x^2+1}dx$$
          Then the substitution will give
          $$I=-int_infty^1 frac1{u^2+1}du-int_{-1}^{-infty} frac1{u^2+1}du$$






          share|cite|improve this answer









          $endgroup$



          What you should actually get is that
          $$I=int_0^1 frac1{x^2+1}dx+int_{-1}^0 frac1{x^2+1}dx$$
          Then the substitution will give
          $$I=-int_infty^1 frac1{u^2+1}du-int_{-1}^{-infty} frac1{u^2+1}du$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Peter ForemanPeter Foreman

          7,8731320




          7,8731320























              4












              $begingroup$

              The flaw is in changing the integration interval.



              $$-1le xle1iff frac1xle-1lor frac1xge 1$$






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                The flaw is in changing the integration interval.



                $$-1le xle1iff frac1xle-1lor frac1xge 1$$






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  The flaw is in changing the integration interval.



                  $$-1le xle1iff frac1xle-1lor frac1xge 1$$






                  share|cite|improve this answer









                  $endgroup$



                  The flaw is in changing the integration interval.



                  $$-1le xle1iff frac1xle-1lor frac1xge 1$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Yves DaoustYves Daoust

                  133k676232




                  133k676232






























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