Is there a way to tell whether an integral exists or not without graphing the function?












1












$begingroup$


For example, trying to integrate the function 1/(t-4)^2 from x=1 to x=5. If you just go through the motions and apply u-substitution and part two of the fundamental theorem, you can arrive at an answer for this just fine. But it was only after graphing it that I saw that there was an asymptote within the limits of the integral, and so obviously the integral would be undefined.



Is there any way to tell whether this will be the case at first glance?



Is the only way to just get good at recognizing when functions have an asymptote? Isn't this sometimes very difficult to predict without fulling graphing the function?



Any help is appreciated.



Edit: one thing that should've been obvious to me is that: the function will have an asymptote if there's an opportunity for division by zero. So now my question is: is there anything else that can cause asymptotic behavior, and are there any situations in which integrals are undefined other than when the graph has an asymptote?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    For example, trying to integrate the function 1/(t-4)^2 from x=1 to x=5. If you just go through the motions and apply u-substitution and part two of the fundamental theorem, you can arrive at an answer for this just fine. But it was only after graphing it that I saw that there was an asymptote within the limits of the integral, and so obviously the integral would be undefined.



    Is there any way to tell whether this will be the case at first glance?



    Is the only way to just get good at recognizing when functions have an asymptote? Isn't this sometimes very difficult to predict without fulling graphing the function?



    Any help is appreciated.



    Edit: one thing that should've been obvious to me is that: the function will have an asymptote if there's an opportunity for division by zero. So now my question is: is there anything else that can cause asymptotic behavior, and are there any situations in which integrals are undefined other than when the graph has an asymptote?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      For example, trying to integrate the function 1/(t-4)^2 from x=1 to x=5. If you just go through the motions and apply u-substitution and part two of the fundamental theorem, you can arrive at an answer for this just fine. But it was only after graphing it that I saw that there was an asymptote within the limits of the integral, and so obviously the integral would be undefined.



      Is there any way to tell whether this will be the case at first glance?



      Is the only way to just get good at recognizing when functions have an asymptote? Isn't this sometimes very difficult to predict without fulling graphing the function?



      Any help is appreciated.



      Edit: one thing that should've been obvious to me is that: the function will have an asymptote if there's an opportunity for division by zero. So now my question is: is there anything else that can cause asymptotic behavior, and are there any situations in which integrals are undefined other than when the graph has an asymptote?










      share|cite|improve this question











      $endgroup$




      For example, trying to integrate the function 1/(t-4)^2 from x=1 to x=5. If you just go through the motions and apply u-substitution and part two of the fundamental theorem, you can arrive at an answer for this just fine. But it was only after graphing it that I saw that there was an asymptote within the limits of the integral, and so obviously the integral would be undefined.



      Is there any way to tell whether this will be the case at first glance?



      Is the only way to just get good at recognizing when functions have an asymptote? Isn't this sometimes very difficult to predict without fulling graphing the function?



      Any help is appreciated.



      Edit: one thing that should've been obvious to me is that: the function will have an asymptote if there's an opportunity for division by zero. So now my question is: is there anything else that can cause asymptotic behavior, and are there any situations in which integrals are undefined other than when the graph has an asymptote?







      calculus integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 20:48







      James Ronald

















      asked Dec 9 '18 at 20:27









      James RonaldJames Ronald

      1207




      1207






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Well, it is not true that the answer is not defined; consider $int_0^1 x^{-1/2},dx$.



          Generally speaking, do not rely on purely graphical methods to determine anything! Yes, the presence of an asymptote should make you pause, and carefully check something, but pictures are lies as my analysis professor used to say.



          As far as recognizing when a function will have a singularity, really you are just hunting for the possibility of dividing by zero; with enough time and patience you will find that you do not need to graph anything at all to recognize that.



          EDIT: To add more information based on conversations and updates to the OP, let me discuss two phenomena that can also cause an integral to fail to have a value.



          First, we turn our attention to the indicator function of the rationals:
          $$
          chi_{mathbb{Q}}(x) = left{
          begin{array}{ll}
          1 & quad x inmathbb{Q} \
          0 & quad x notinmathbb{Q}
          end{array}
          right.
          $$

          This function, if you somehow were able to graph it, would look like two lines -- the line $y=1$ and the line $y=0$. But it isn't! The rationals are what we call "dense" in the reals, and so in reality there is just a very fine "cloud" of points, everywhere, that is hovering at $1$, and another cloud of points down at $0$. If you try and apply your Darboux or Riemann integral definition to this function on any interval $[a,b]$, you will find that the limits/sups/infs/whatevers don't agree, and thus the integral doesn't exist.



          Second, we consider improper integrals over oscillatory functions. Consider the integral $int_0^infty sin(2pi x),dx$. The value (if it were to exist) of this integral would be $lim_{brightarrowinfty}1-cos(2pi b)/2pi$. But this limit doesn't exist, because $cos(2pi x)$ just "wiggles" off into infinity. Even though there doesn't appear to be a singularity in this function, there is; it has been obfuscated and pushed out to the upper limit of integration!






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It depends on your intergral though. If we're using an improper Riemann Integral then the integral exists, but if we're using Darboux integral then it does not, for Darboux requires boundedness for the upper/lower sums to make sense. In general I agree with your sentiment though.
            $endgroup$
            – Melody
            Dec 9 '18 at 20:33










          • $begingroup$
            @Melody Does anyone truly care about the Darboux integral though? Honest question -- I've never really used it for anything beyond a simplifying assumption leading up to the Riemann integral.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 20:35










          • $begingroup$
            Isn't it not sometimes easier to show that a function is Darboux integral, and to compute it's integral that way? Sometimes you lose simplicity when you generalize.
            $endgroup$
            – Melody
            Dec 9 '18 at 20:42










          • $begingroup$
            @Melody Perhaps, I'm just saying that I've never done it, except perhaps when I was taking my low-powered calculus course way back when. My real analysis course focused heavily on Riemann integration; we didn't fall back on Darboux at all.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 20:44










          • $begingroup$
            Well actually I was looking for a way to determine whether or not without graphing it, as graphing it isn't always an option. Is there something about the function itself that can tell me whether it's integral will be undefined/nonexistent anywhere?
            $endgroup$
            – James Ronald
            Dec 9 '18 at 20:44











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          1












          $begingroup$

          Well, it is not true that the answer is not defined; consider $int_0^1 x^{-1/2},dx$.



          Generally speaking, do not rely on purely graphical methods to determine anything! Yes, the presence of an asymptote should make you pause, and carefully check something, but pictures are lies as my analysis professor used to say.



          As far as recognizing when a function will have a singularity, really you are just hunting for the possibility of dividing by zero; with enough time and patience you will find that you do not need to graph anything at all to recognize that.



          EDIT: To add more information based on conversations and updates to the OP, let me discuss two phenomena that can also cause an integral to fail to have a value.



          First, we turn our attention to the indicator function of the rationals:
          $$
          chi_{mathbb{Q}}(x) = left{
          begin{array}{ll}
          1 & quad x inmathbb{Q} \
          0 & quad x notinmathbb{Q}
          end{array}
          right.
          $$

          This function, if you somehow were able to graph it, would look like two lines -- the line $y=1$ and the line $y=0$. But it isn't! The rationals are what we call "dense" in the reals, and so in reality there is just a very fine "cloud" of points, everywhere, that is hovering at $1$, and another cloud of points down at $0$. If you try and apply your Darboux or Riemann integral definition to this function on any interval $[a,b]$, you will find that the limits/sups/infs/whatevers don't agree, and thus the integral doesn't exist.



          Second, we consider improper integrals over oscillatory functions. Consider the integral $int_0^infty sin(2pi x),dx$. The value (if it were to exist) of this integral would be $lim_{brightarrowinfty}1-cos(2pi b)/2pi$. But this limit doesn't exist, because $cos(2pi x)$ just "wiggles" off into infinity. Even though there doesn't appear to be a singularity in this function, there is; it has been obfuscated and pushed out to the upper limit of integration!






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It depends on your intergral though. If we're using an improper Riemann Integral then the integral exists, but if we're using Darboux integral then it does not, for Darboux requires boundedness for the upper/lower sums to make sense. In general I agree with your sentiment though.
            $endgroup$
            – Melody
            Dec 9 '18 at 20:33










          • $begingroup$
            @Melody Does anyone truly care about the Darboux integral though? Honest question -- I've never really used it for anything beyond a simplifying assumption leading up to the Riemann integral.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 20:35










          • $begingroup$
            Isn't it not sometimes easier to show that a function is Darboux integral, and to compute it's integral that way? Sometimes you lose simplicity when you generalize.
            $endgroup$
            – Melody
            Dec 9 '18 at 20:42










          • $begingroup$
            @Melody Perhaps, I'm just saying that I've never done it, except perhaps when I was taking my low-powered calculus course way back when. My real analysis course focused heavily on Riemann integration; we didn't fall back on Darboux at all.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 20:44










          • $begingroup$
            Well actually I was looking for a way to determine whether or not without graphing it, as graphing it isn't always an option. Is there something about the function itself that can tell me whether it's integral will be undefined/nonexistent anywhere?
            $endgroup$
            – James Ronald
            Dec 9 '18 at 20:44
















          1












          $begingroup$

          Well, it is not true that the answer is not defined; consider $int_0^1 x^{-1/2},dx$.



          Generally speaking, do not rely on purely graphical methods to determine anything! Yes, the presence of an asymptote should make you pause, and carefully check something, but pictures are lies as my analysis professor used to say.



          As far as recognizing when a function will have a singularity, really you are just hunting for the possibility of dividing by zero; with enough time and patience you will find that you do not need to graph anything at all to recognize that.



          EDIT: To add more information based on conversations and updates to the OP, let me discuss two phenomena that can also cause an integral to fail to have a value.



          First, we turn our attention to the indicator function of the rationals:
          $$
          chi_{mathbb{Q}}(x) = left{
          begin{array}{ll}
          1 & quad x inmathbb{Q} \
          0 & quad x notinmathbb{Q}
          end{array}
          right.
          $$

          This function, if you somehow were able to graph it, would look like two lines -- the line $y=1$ and the line $y=0$. But it isn't! The rationals are what we call "dense" in the reals, and so in reality there is just a very fine "cloud" of points, everywhere, that is hovering at $1$, and another cloud of points down at $0$. If you try and apply your Darboux or Riemann integral definition to this function on any interval $[a,b]$, you will find that the limits/sups/infs/whatevers don't agree, and thus the integral doesn't exist.



          Second, we consider improper integrals over oscillatory functions. Consider the integral $int_0^infty sin(2pi x),dx$. The value (if it were to exist) of this integral would be $lim_{brightarrowinfty}1-cos(2pi b)/2pi$. But this limit doesn't exist, because $cos(2pi x)$ just "wiggles" off into infinity. Even though there doesn't appear to be a singularity in this function, there is; it has been obfuscated and pushed out to the upper limit of integration!






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It depends on your intergral though. If we're using an improper Riemann Integral then the integral exists, but if we're using Darboux integral then it does not, for Darboux requires boundedness for the upper/lower sums to make sense. In general I agree with your sentiment though.
            $endgroup$
            – Melody
            Dec 9 '18 at 20:33










          • $begingroup$
            @Melody Does anyone truly care about the Darboux integral though? Honest question -- I've never really used it for anything beyond a simplifying assumption leading up to the Riemann integral.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 20:35










          • $begingroup$
            Isn't it not sometimes easier to show that a function is Darboux integral, and to compute it's integral that way? Sometimes you lose simplicity when you generalize.
            $endgroup$
            – Melody
            Dec 9 '18 at 20:42










          • $begingroup$
            @Melody Perhaps, I'm just saying that I've never done it, except perhaps when I was taking my low-powered calculus course way back when. My real analysis course focused heavily on Riemann integration; we didn't fall back on Darboux at all.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 20:44










          • $begingroup$
            Well actually I was looking for a way to determine whether or not without graphing it, as graphing it isn't always an option. Is there something about the function itself that can tell me whether it's integral will be undefined/nonexistent anywhere?
            $endgroup$
            – James Ronald
            Dec 9 '18 at 20:44














          1












          1








          1





          $begingroup$

          Well, it is not true that the answer is not defined; consider $int_0^1 x^{-1/2},dx$.



          Generally speaking, do not rely on purely graphical methods to determine anything! Yes, the presence of an asymptote should make you pause, and carefully check something, but pictures are lies as my analysis professor used to say.



          As far as recognizing when a function will have a singularity, really you are just hunting for the possibility of dividing by zero; with enough time and patience you will find that you do not need to graph anything at all to recognize that.



          EDIT: To add more information based on conversations and updates to the OP, let me discuss two phenomena that can also cause an integral to fail to have a value.



          First, we turn our attention to the indicator function of the rationals:
          $$
          chi_{mathbb{Q}}(x) = left{
          begin{array}{ll}
          1 & quad x inmathbb{Q} \
          0 & quad x notinmathbb{Q}
          end{array}
          right.
          $$

          This function, if you somehow were able to graph it, would look like two lines -- the line $y=1$ and the line $y=0$. But it isn't! The rationals are what we call "dense" in the reals, and so in reality there is just a very fine "cloud" of points, everywhere, that is hovering at $1$, and another cloud of points down at $0$. If you try and apply your Darboux or Riemann integral definition to this function on any interval $[a,b]$, you will find that the limits/sups/infs/whatevers don't agree, and thus the integral doesn't exist.



          Second, we consider improper integrals over oscillatory functions. Consider the integral $int_0^infty sin(2pi x),dx$. The value (if it were to exist) of this integral would be $lim_{brightarrowinfty}1-cos(2pi b)/2pi$. But this limit doesn't exist, because $cos(2pi x)$ just "wiggles" off into infinity. Even though there doesn't appear to be a singularity in this function, there is; it has been obfuscated and pushed out to the upper limit of integration!






          share|cite|improve this answer











          $endgroup$



          Well, it is not true that the answer is not defined; consider $int_0^1 x^{-1/2},dx$.



          Generally speaking, do not rely on purely graphical methods to determine anything! Yes, the presence of an asymptote should make you pause, and carefully check something, but pictures are lies as my analysis professor used to say.



          As far as recognizing when a function will have a singularity, really you are just hunting for the possibility of dividing by zero; with enough time and patience you will find that you do not need to graph anything at all to recognize that.



          EDIT: To add more information based on conversations and updates to the OP, let me discuss two phenomena that can also cause an integral to fail to have a value.



          First, we turn our attention to the indicator function of the rationals:
          $$
          chi_{mathbb{Q}}(x) = left{
          begin{array}{ll}
          1 & quad x inmathbb{Q} \
          0 & quad x notinmathbb{Q}
          end{array}
          right.
          $$

          This function, if you somehow were able to graph it, would look like two lines -- the line $y=1$ and the line $y=0$. But it isn't! The rationals are what we call "dense" in the reals, and so in reality there is just a very fine "cloud" of points, everywhere, that is hovering at $1$, and another cloud of points down at $0$. If you try and apply your Darboux or Riemann integral definition to this function on any interval $[a,b]$, you will find that the limits/sups/infs/whatevers don't agree, and thus the integral doesn't exist.



          Second, we consider improper integrals over oscillatory functions. Consider the integral $int_0^infty sin(2pi x),dx$. The value (if it were to exist) of this integral would be $lim_{brightarrowinfty}1-cos(2pi b)/2pi$. But this limit doesn't exist, because $cos(2pi x)$ just "wiggles" off into infinity. Even though there doesn't appear to be a singularity in this function, there is; it has been obfuscated and pushed out to the upper limit of integration!







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 21:12

























          answered Dec 9 '18 at 20:31









          GenericMathematicianGenericMathematician

          863




          863












          • $begingroup$
            It depends on your intergral though. If we're using an improper Riemann Integral then the integral exists, but if we're using Darboux integral then it does not, for Darboux requires boundedness for the upper/lower sums to make sense. In general I agree with your sentiment though.
            $endgroup$
            – Melody
            Dec 9 '18 at 20:33










          • $begingroup$
            @Melody Does anyone truly care about the Darboux integral though? Honest question -- I've never really used it for anything beyond a simplifying assumption leading up to the Riemann integral.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 20:35










          • $begingroup$
            Isn't it not sometimes easier to show that a function is Darboux integral, and to compute it's integral that way? Sometimes you lose simplicity when you generalize.
            $endgroup$
            – Melody
            Dec 9 '18 at 20:42










          • $begingroup$
            @Melody Perhaps, I'm just saying that I've never done it, except perhaps when I was taking my low-powered calculus course way back when. My real analysis course focused heavily on Riemann integration; we didn't fall back on Darboux at all.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 20:44










          • $begingroup$
            Well actually I was looking for a way to determine whether or not without graphing it, as graphing it isn't always an option. Is there something about the function itself that can tell me whether it's integral will be undefined/nonexistent anywhere?
            $endgroup$
            – James Ronald
            Dec 9 '18 at 20:44


















          • $begingroup$
            It depends on your intergral though. If we're using an improper Riemann Integral then the integral exists, but if we're using Darboux integral then it does not, for Darboux requires boundedness for the upper/lower sums to make sense. In general I agree with your sentiment though.
            $endgroup$
            – Melody
            Dec 9 '18 at 20:33










          • $begingroup$
            @Melody Does anyone truly care about the Darboux integral though? Honest question -- I've never really used it for anything beyond a simplifying assumption leading up to the Riemann integral.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 20:35










          • $begingroup$
            Isn't it not sometimes easier to show that a function is Darboux integral, and to compute it's integral that way? Sometimes you lose simplicity when you generalize.
            $endgroup$
            – Melody
            Dec 9 '18 at 20:42










          • $begingroup$
            @Melody Perhaps, I'm just saying that I've never done it, except perhaps when I was taking my low-powered calculus course way back when. My real analysis course focused heavily on Riemann integration; we didn't fall back on Darboux at all.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 20:44










          • $begingroup$
            Well actually I was looking for a way to determine whether or not without graphing it, as graphing it isn't always an option. Is there something about the function itself that can tell me whether it's integral will be undefined/nonexistent anywhere?
            $endgroup$
            – James Ronald
            Dec 9 '18 at 20:44
















          $begingroup$
          It depends on your intergral though. If we're using an improper Riemann Integral then the integral exists, but if we're using Darboux integral then it does not, for Darboux requires boundedness for the upper/lower sums to make sense. In general I agree with your sentiment though.
          $endgroup$
          – Melody
          Dec 9 '18 at 20:33




          $begingroup$
          It depends on your intergral though. If we're using an improper Riemann Integral then the integral exists, but if we're using Darboux integral then it does not, for Darboux requires boundedness for the upper/lower sums to make sense. In general I agree with your sentiment though.
          $endgroup$
          – Melody
          Dec 9 '18 at 20:33












          $begingroup$
          @Melody Does anyone truly care about the Darboux integral though? Honest question -- I've never really used it for anything beyond a simplifying assumption leading up to the Riemann integral.
          $endgroup$
          – GenericMathematician
          Dec 9 '18 at 20:35




          $begingroup$
          @Melody Does anyone truly care about the Darboux integral though? Honest question -- I've never really used it for anything beyond a simplifying assumption leading up to the Riemann integral.
          $endgroup$
          – GenericMathematician
          Dec 9 '18 at 20:35












          $begingroup$
          Isn't it not sometimes easier to show that a function is Darboux integral, and to compute it's integral that way? Sometimes you lose simplicity when you generalize.
          $endgroup$
          – Melody
          Dec 9 '18 at 20:42




          $begingroup$
          Isn't it not sometimes easier to show that a function is Darboux integral, and to compute it's integral that way? Sometimes you lose simplicity when you generalize.
          $endgroup$
          – Melody
          Dec 9 '18 at 20:42












          $begingroup$
          @Melody Perhaps, I'm just saying that I've never done it, except perhaps when I was taking my low-powered calculus course way back when. My real analysis course focused heavily on Riemann integration; we didn't fall back on Darboux at all.
          $endgroup$
          – GenericMathematician
          Dec 9 '18 at 20:44




          $begingroup$
          @Melody Perhaps, I'm just saying that I've never done it, except perhaps when I was taking my low-powered calculus course way back when. My real analysis course focused heavily on Riemann integration; we didn't fall back on Darboux at all.
          $endgroup$
          – GenericMathematician
          Dec 9 '18 at 20:44












          $begingroup$
          Well actually I was looking for a way to determine whether or not without graphing it, as graphing it isn't always an option. Is there something about the function itself that can tell me whether it's integral will be undefined/nonexistent anywhere?
          $endgroup$
          – James Ronald
          Dec 9 '18 at 20:44




          $begingroup$
          Well actually I was looking for a way to determine whether or not without graphing it, as graphing it isn't always an option. Is there something about the function itself that can tell me whether it's integral will be undefined/nonexistent anywhere?
          $endgroup$
          – James Ronald
          Dec 9 '18 at 20:44


















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