Inconsistency in Mathematica of whether integral diverges or not?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
3
down vote

favorite












Consider the following integral



Integrate[(1 - x)^a x^b, x, 0, 1]



ConditionalExpression[(Gamma[1 + a] Gamma[1 + b])/Gamma[2 + a + b], Re[b]>-1&&Re[a]>-1]




The above general case states that the integral only converges when Re[b]>-1 and Re[a]>-1. However, if we plug the following explicit constants into the integral, we get:



Integrate[(1 - x)^(-(1/2) - 2 I) x^(-2 + 7 I), x, 0, 1]



(Gamma[-1 + 7 I] Gamma[1/2 - 2 I])/Gamma[-(1/2) + 5 I]




In which case mathematica seems to return a perfectly finite result that is the same as the general result above, except the parameters have been chosen such that the general case would not converge.



Now I am a bit confused and would like to ask:




Is the explicit result obtained above a bug in Mathematica and the integral actually should not converge at all?
Or maybe the result and convergence is fine, and the constraints in the general case are wrong?




Which one is it?






calculus-and-analysis







share|improve this question



























    up vote
    3
    down vote

    favorite












    Consider the following integral



    Integrate[(1 - x)^a x^b, x, 0, 1]



    ConditionalExpression[(Gamma[1 + a] Gamma[1 + b])/Gamma[2 + a + b], Re[b]>-1&&Re[a]>-1]




    The above general case states that the integral only converges when Re[b]>-1 and Re[a]>-1. However, if we plug the following explicit constants into the integral, we get:



    Integrate[(1 - x)^(-(1/2) - 2 I) x^(-2 + 7 I), x, 0, 1]



    (Gamma[-1 + 7 I] Gamma[1/2 - 2 I])/Gamma[-(1/2) + 5 I]




    In which case mathematica seems to return a perfectly finite result that is the same as the general result above, except the parameters have been chosen such that the general case would not converge.



    Now I am a bit confused and would like to ask:




    Is the explicit result obtained above a bug in Mathematica and the integral actually should not converge at all?
    Or maybe the result and convergence is fine, and the constraints in the general case are wrong?




    Which one is it?






    calculus-and-analysis







    share|improve this question

























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      Consider the following integral



      Integrate[(1 - x)^a x^b, x, 0, 1]



      ConditionalExpression[(Gamma[1 + a] Gamma[1 + b])/Gamma[2 + a + b], Re[b]>-1&&Re[a]>-1]




      The above general case states that the integral only converges when Re[b]>-1 and Re[a]>-1. However, if we plug the following explicit constants into the integral, we get:



      Integrate[(1 - x)^(-(1/2) - 2 I) x^(-2 + 7 I), x, 0, 1]



      (Gamma[-1 + 7 I] Gamma[1/2 - 2 I])/Gamma[-(1/2) + 5 I]




      In which case mathematica seems to return a perfectly finite result that is the same as the general result above, except the parameters have been chosen such that the general case would not converge.



      Now I am a bit confused and would like to ask:




      Is the explicit result obtained above a bug in Mathematica and the integral actually should not converge at all?
      Or maybe the result and convergence is fine, and the constraints in the general case are wrong?




      Which one is it?






      calculus-and-analysis







      share|improve this question















      Consider the following integral



      Integrate[(1 - x)^a x^b, x, 0, 1]



      ConditionalExpression[(Gamma[1 + a] Gamma[1 + b])/Gamma[2 + a + b], Re[b]>-1&&Re[a]>-1]




      The above general case states that the integral only converges when Re[b]>-1 and Re[a]>-1. However, if we plug the following explicit constants into the integral, we get:



      Integrate[(1 - x)^(-(1/2) - 2 I) x^(-2 + 7 I), x, 0, 1]



      (Gamma[-1 + 7 I] Gamma[1/2 - 2 I])/Gamma[-(1/2) + 5 I]




      In which case mathematica seems to return a perfectly finite result that is the same as the general result above, except the parameters have been chosen such that the general case would not converge.



      Now I am a bit confused and would like to ask:




      Is the explicit result obtained above a bug in Mathematica and the integral actually should not converge at all?
      Or maybe the result and convergence is fine, and the constraints in the general case are wrong?




      Which one is it?






      calculus-and-analysis




      calculus-and-analysis






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 3 hours ago









      Henrik Schumacher

      39.5k254118




      39.5k254118










      asked 3 hours ago









      Kagaratsch

      4,34031246




      4,34031246




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          Use:



          Integrate[(1 - x)^(-(1/2) - 2 I) x^(-2 + 7 I), x, m, 1]



          ConditionalExpression[(Gamma[-1 + 7 I] Gamma[1/2 - 2 I])/
          Gamma[-(1/2) + 5 I] + (1/50 + (7 I)/50) m^(-1 + 7 I)
          Hypergeometric2F1[-1 + 7 I, 1/2 + 2 I, 7 I, m],
          0 < Re[m] < 1 && Im[m] == 0]




          and note that the first term is the result from your second integral, while the second, Hypergeometric2F1 term diverges for $m to 0$ due to the factor $m^(-1+7i)$. So, the second example looks like a bug in Integrate.



          Addendum



          Consider the Series expansion about the origin:



          Series[(1 - x)^a x^b, x, 0, 1] //TeXForm



          $x^b left(1-a x+Oleft(x^2right)right)$




          So, the integral will only converge at the origin if $Re(b)>-1$. Similarly, consider the Series expansion about $x=1$:



          Series[(1 - x)^a x^b, x, 1, 1] //TeXForm



          $(1-x)^a left(1+b (x-1)+Oleft((x-1)^2right)right)$




          This integral will converge at $x=1$ only if $Re(a)>-1$.






          share|improve this answer


















          • 1




            To clarify: it's not the hypergeometric function that causes the divergence in the second term, it's the factor m^(-1 + 7 I).
            – J. M. is somewhat okay.♦
            3 hours ago










          • @J.M.issomewhatokay. Agreed, my original wording was confusing.
            – Carl Woll
            3 hours ago










          • The result is certainly confusing, but is it absolutely wrong in all cases? With a suitable (non-standard) branch cut, I think you might be able to choose a spiral path (in the complex plane) to reach x=0 without the integrand diverging.
            – mikado
            3 hours ago










          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "387"
          ;
          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',
          convertImagesToLinks: false,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f182667%2finconsistency-in-mathematica-of-whether-integral-diverges-or-not%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          4
          down vote



          accepted










          Use:



          Integrate[(1 - x)^(-(1/2) - 2 I) x^(-2 + 7 I), x, m, 1]



          ConditionalExpression[(Gamma[-1 + 7 I] Gamma[1/2 - 2 I])/
          Gamma[-(1/2) + 5 I] + (1/50 + (7 I)/50) m^(-1 + 7 I)
          Hypergeometric2F1[-1 + 7 I, 1/2 + 2 I, 7 I, m],
          0 < Re[m] < 1 && Im[m] == 0]




          and note that the first term is the result from your second integral, while the second, Hypergeometric2F1 term diverges for $m to 0$ due to the factor $m^(-1+7i)$. So, the second example looks like a bug in Integrate.



          Addendum



          Consider the Series expansion about the origin:



          Series[(1 - x)^a x^b, x, 0, 1] //TeXForm



          $x^b left(1-a x+Oleft(x^2right)right)$




          So, the integral will only converge at the origin if $Re(b)>-1$. Similarly, consider the Series expansion about $x=1$:



          Series[(1 - x)^a x^b, x, 1, 1] //TeXForm



          $(1-x)^a left(1+b (x-1)+Oleft((x-1)^2right)right)$




          This integral will converge at $x=1$ only if $Re(a)>-1$.






          share|improve this answer


















          • 1




            To clarify: it's not the hypergeometric function that causes the divergence in the second term, it's the factor m^(-1 + 7 I).
            – J. M. is somewhat okay.♦
            3 hours ago










          • @J.M.issomewhatokay. Agreed, my original wording was confusing.
            – Carl Woll
            3 hours ago










          • The result is certainly confusing, but is it absolutely wrong in all cases? With a suitable (non-standard) branch cut, I think you might be able to choose a spiral path (in the complex plane) to reach x=0 without the integrand diverging.
            – mikado
            3 hours ago














          up vote
          4
          down vote



          accepted










          Use:



          Integrate[(1 - x)^(-(1/2) - 2 I) x^(-2 + 7 I), x, m, 1]



          ConditionalExpression[(Gamma[-1 + 7 I] Gamma[1/2 - 2 I])/
          Gamma[-(1/2) + 5 I] + (1/50 + (7 I)/50) m^(-1 + 7 I)
          Hypergeometric2F1[-1 + 7 I, 1/2 + 2 I, 7 I, m],
          0 < Re[m] < 1 && Im[m] == 0]




          and note that the first term is the result from your second integral, while the second, Hypergeometric2F1 term diverges for $m to 0$ due to the factor $m^(-1+7i)$. So, the second example looks like a bug in Integrate.



          Addendum



          Consider the Series expansion about the origin:



          Series[(1 - x)^a x^b, x, 0, 1] //TeXForm



          $x^b left(1-a x+Oleft(x^2right)right)$




          So, the integral will only converge at the origin if $Re(b)>-1$. Similarly, consider the Series expansion about $x=1$:



          Series[(1 - x)^a x^b, x, 1, 1] //TeXForm



          $(1-x)^a left(1+b (x-1)+Oleft((x-1)^2right)right)$




          This integral will converge at $x=1$ only if $Re(a)>-1$.






          share|improve this answer


















          • 1




            To clarify: it's not the hypergeometric function that causes the divergence in the second term, it's the factor m^(-1 + 7 I).
            – J. M. is somewhat okay.♦
            3 hours ago










          • @J.M.issomewhatokay. Agreed, my original wording was confusing.
            – Carl Woll
            3 hours ago










          • The result is certainly confusing, but is it absolutely wrong in all cases? With a suitable (non-standard) branch cut, I think you might be able to choose a spiral path (in the complex plane) to reach x=0 without the integrand diverging.
            – mikado
            3 hours ago












          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Use:



          Integrate[(1 - x)^(-(1/2) - 2 I) x^(-2 + 7 I), x, m, 1]



          ConditionalExpression[(Gamma[-1 + 7 I] Gamma[1/2 - 2 I])/
          Gamma[-(1/2) + 5 I] + (1/50 + (7 I)/50) m^(-1 + 7 I)
          Hypergeometric2F1[-1 + 7 I, 1/2 + 2 I, 7 I, m],
          0 < Re[m] < 1 && Im[m] == 0]




          and note that the first term is the result from your second integral, while the second, Hypergeometric2F1 term diverges for $m to 0$ due to the factor $m^(-1+7i)$. So, the second example looks like a bug in Integrate.



          Addendum



          Consider the Series expansion about the origin:



          Series[(1 - x)^a x^b, x, 0, 1] //TeXForm



          $x^b left(1-a x+Oleft(x^2right)right)$




          So, the integral will only converge at the origin if $Re(b)>-1$. Similarly, consider the Series expansion about $x=1$:



          Series[(1 - x)^a x^b, x, 1, 1] //TeXForm



          $(1-x)^a left(1+b (x-1)+Oleft((x-1)^2right)right)$




          This integral will converge at $x=1$ only if $Re(a)>-1$.






          share|improve this answer














          Use:



          Integrate[(1 - x)^(-(1/2) - 2 I) x^(-2 + 7 I), x, m, 1]



          ConditionalExpression[(Gamma[-1 + 7 I] Gamma[1/2 - 2 I])/
          Gamma[-(1/2) + 5 I] + (1/50 + (7 I)/50) m^(-1 + 7 I)
          Hypergeometric2F1[-1 + 7 I, 1/2 + 2 I, 7 I, m],
          0 < Re[m] < 1 && Im[m] == 0]




          and note that the first term is the result from your second integral, while the second, Hypergeometric2F1 term diverges for $m to 0$ due to the factor $m^(-1+7i)$. So, the second example looks like a bug in Integrate.



          Addendum



          Consider the Series expansion about the origin:



          Series[(1 - x)^a x^b, x, 0, 1] //TeXForm



          $x^b left(1-a x+Oleft(x^2right)right)$




          So, the integral will only converge at the origin if $Re(b)>-1$. Similarly, consider the Series expansion about $x=1$:



          Series[(1 - x)^a x^b, x, 1, 1] //TeXForm



          $(1-x)^a left(1+b (x-1)+Oleft((x-1)^2right)right)$




          This integral will converge at $x=1$ only if $Re(a)>-1$.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 1 hour ago

























          answered 3 hours ago









          Carl Woll

          58.7k276150




          58.7k276150







          • 1




            To clarify: it's not the hypergeometric function that causes the divergence in the second term, it's the factor m^(-1 + 7 I).
            – J. M. is somewhat okay.♦
            3 hours ago










          • @J.M.issomewhatokay. Agreed, my original wording was confusing.
            – Carl Woll
            3 hours ago










          • The result is certainly confusing, but is it absolutely wrong in all cases? With a suitable (non-standard) branch cut, I think you might be able to choose a spiral path (in the complex plane) to reach x=0 without the integrand diverging.
            – mikado
            3 hours ago












          • 1




            To clarify: it's not the hypergeometric function that causes the divergence in the second term, it's the factor m^(-1 + 7 I).
            – J. M. is somewhat okay.♦
            3 hours ago










          • @J.M.issomewhatokay. Agreed, my original wording was confusing.
            – Carl Woll
            3 hours ago










          • The result is certainly confusing, but is it absolutely wrong in all cases? With a suitable (non-standard) branch cut, I think you might be able to choose a spiral path (in the complex plane) to reach x=0 without the integrand diverging.
            – mikado
            3 hours ago







          1




          1




          To clarify: it's not the hypergeometric function that causes the divergence in the second term, it's the factor m^(-1 + 7 I).
          – J. M. is somewhat okay.♦
          3 hours ago




          To clarify: it's not the hypergeometric function that causes the divergence in the second term, it's the factor m^(-1 + 7 I).
          – J. M. is somewhat okay.♦
          3 hours ago












          @J.M.issomewhatokay. Agreed, my original wording was confusing.
          – Carl Woll
          3 hours ago




          @J.M.issomewhatokay. Agreed, my original wording was confusing.
          – Carl Woll
          3 hours ago












          The result is certainly confusing, but is it absolutely wrong in all cases? With a suitable (non-standard) branch cut, I think you might be able to choose a spiral path (in the complex plane) to reach x=0 without the integrand diverging.
          – mikado
          3 hours ago




          The result is certainly confusing, but is it absolutely wrong in all cases? With a suitable (non-standard) branch cut, I think you might be able to choose a spiral path (in the complex plane) to reach x=0 without the integrand diverging.
          – mikado
          3 hours ago

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f182667%2finconsistency-in-mathematica-of-whether-integral-diverges-or-not%23new-answer', 'question_page');

          );

          Post as a guest
































































          Comments

          Post a Comment

          Popular posts from this blog

          What does second last employer means? [closed]

          Image
          Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
          Read more

          Installing NextGIS Connect into QGIS 3?

          Image
          Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS 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  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
          Read more

          One-line joke

          A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
          Read more