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






active

oldest

votes


















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









    $endgroup$














      Your Answer








      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
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3188675%2fwhat-is-wrong-with-this-false-proof-of-pi-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









      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






























                      draft saved

                      draft discarded




















































                      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.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3188675%2fwhat-is-wrong-with-this-false-proof-of-pi-0%23new-answer', 'question_page');
                      }
                      );

                      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







                      -

                      Popular posts from this blog

                      數位音樂下載

                      Image
                      數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
                      Read more

                      When can things happen in Etherscan, such as the picture below?

                      Image
                      2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi 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  |  ...
                      Read more

                      格利澤436b

                      Image
                      格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
                      Read more