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When does a function NOT have an antiderivative?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How can you prove that a function has no closed form integral?Does L'Hôpital's work the other way?Finding volumes - when to use double integrals and triple integrals?Why does a function have to be bounded to be integrable?Proof that order of integration does not matter for non-continuous functionswhy we say this function have closed form while the other doesn't?Antiderivative of unbounded function?antiderivative of jump-discontinuous functionWhy is it legal to take the antiderivative of both sides of an equation?why $int sqrt{(sin x)^2}, mathrm{d}x = int |sin x| ,mathrm{d}x$Why does the antiderivative of $frac{1}{x}$ have to be $0$ at precisely $x=1$? (when $C = 0$)












1












$begingroup$


I know this question may sound naïve but why can't we write $int e^{x^2} dx$ as $int e^{2x} dx$? The former does not have an antiderivative, while the latter has.



In light of this question, what are sufficient conditions for a function NOT to have an antiderivative. That is, do we need careful examination of a function to say it does not have an antiderivative or is there any way that once you see the function, you can right away say it does not have an antiderivative?






integration







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    Maybe because $x^2$ isn't the same as $2x$?
    $endgroup$
    – Lord Shark the Unknown
    3 hours ago










  • $begingroup$
    Can't we write $a^{b^c}$ as $a^{bc}$?
    $endgroup$
    – Rob
    3 hours ago






  • 1




    $begingroup$
    $$e^{x^2}=e^{xcdot x}neq e^xcdot e^x = e^{x+x}=e^{2x}$$
    $endgroup$
    – Don Thousand
    3 hours ago










  • $begingroup$
    $$(a^b)^c=a^{bcdot c}$$$$a^{(b^c)}neq a^{bcdot c}$$
    $endgroup$
    – Don Thousand
    3 hours ago










  • $begingroup$
    $(e^x)^2$ would be where you use the rule that you're thinking of.
    $endgroup$
    – Tartaglia's Stutter
    3 hours ago
















1












$begingroup$


I know this question may sound naïve but why can't we write $int e^{x^2} dx$ as $int e^{2x} dx$? The former does not have an antiderivative, while the latter has.



In light of this question, what are sufficient conditions for a function NOT to have an antiderivative. That is, do we need careful examination of a function to say it does not have an antiderivative or is there any way that once you see the function, you can right away say it does not have an antiderivative?






integration







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    Maybe because $x^2$ isn't the same as $2x$?
    $endgroup$
    – Lord Shark the Unknown
    3 hours ago










  • $begingroup$
    Can't we write $a^{b^c}$ as $a^{bc}$?
    $endgroup$
    – Rob
    3 hours ago






  • 1




    $begingroup$
    $$e^{x^2}=e^{xcdot x}neq e^xcdot e^x = e^{x+x}=e^{2x}$$
    $endgroup$
    – Don Thousand
    3 hours ago










  • $begingroup$
    $$(a^b)^c=a^{bcdot c}$$$$a^{(b^c)}neq a^{bcdot c}$$
    $endgroup$
    – Don Thousand
    3 hours ago










  • $begingroup$
    $(e^x)^2$ would be where you use the rule that you're thinking of.
    $endgroup$
    – Tartaglia's Stutter
    3 hours ago














1












1








1





$begingroup$


I know this question may sound naïve but why can't we write $int e^{x^2} dx$ as $int e^{2x} dx$? The former does not have an antiderivative, while the latter has.



In light of this question, what are sufficient conditions for a function NOT to have an antiderivative. That is, do we need careful examination of a function to say it does not have an antiderivative or is there any way that once you see the function, you can right away say it does not have an antiderivative?






integration







share|cite|improve this question









$endgroup$




I know this question may sound naïve but why can't we write $int e^{x^2} dx$ as $int e^{2x} dx$? The former does not have an antiderivative, while the latter has.



In light of this question, what are sufficient conditions for a function NOT to have an antiderivative. That is, do we need careful examination of a function to say it does not have an antiderivative or is there any way that once you see the function, you can right away say it does not have an antiderivative?






integration




integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









RobRob

333110




333110








  • 6




    $begingroup$
    Maybe because $x^2$ isn't the same as $2x$?
    $endgroup$
    – Lord Shark the Unknown
    3 hours ago










  • $begingroup$
    Can't we write $a^{b^c}$ as $a^{bc}$?
    $endgroup$
    – Rob
    3 hours ago






  • 1




    $begingroup$
    $$e^{x^2}=e^{xcdot x}neq e^xcdot e^x = e^{x+x}=e^{2x}$$
    $endgroup$
    – Don Thousand
    3 hours ago










  • $begingroup$
    $$(a^b)^c=a^{bcdot c}$$$$a^{(b^c)}neq a^{bcdot c}$$
    $endgroup$
    – Don Thousand
    3 hours ago










  • $begingroup$
    $(e^x)^2$ would be where you use the rule that you're thinking of.
    $endgroup$
    – Tartaglia's Stutter
    3 hours ago














  • 6




    $begingroup$
    Maybe because $x^2$ isn't the same as $2x$?
    $endgroup$
    – Lord Shark the Unknown
    3 hours ago










  • $begingroup$
    Can't we write $a^{b^c}$ as $a^{bc}$?
    $endgroup$
    – Rob
    3 hours ago






  • 1




    $begingroup$
    $$e^{x^2}=e^{xcdot x}neq e^xcdot e^x = e^{x+x}=e^{2x}$$
    $endgroup$
    – Don Thousand
    3 hours ago










  • $begingroup$
    $$(a^b)^c=a^{bcdot c}$$$$a^{(b^c)}neq a^{bcdot c}$$
    $endgroup$
    – Don Thousand
    3 hours ago










  • $begingroup$
    $(e^x)^2$ would be where you use the rule that you're thinking of.
    $endgroup$
    – Tartaglia's Stutter
    3 hours ago








6




6




$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
3 hours ago




$begingroup$
Maybe because $x^2$ isn't the same as $2x$?
$endgroup$
– Lord Shark the Unknown
3 hours ago












$begingroup$
Can't we write $a^{b^c}$ as $a^{bc}$?
$endgroup$
– Rob
3 hours ago




$begingroup$
Can't we write $a^{b^c}$ as $a^{bc}$?
$endgroup$
– Rob
3 hours ago




1




1




$begingroup$
$$e^{x^2}=e^{xcdot x}neq e^xcdot e^x = e^{x+x}=e^{2x}$$
$endgroup$
– Don Thousand
3 hours ago




$begingroup$
$$e^{x^2}=e^{xcdot x}neq e^xcdot e^x = e^{x+x}=e^{2x}$$
$endgroup$
– Don Thousand
3 hours ago












$begingroup$
$$(a^b)^c=a^{bcdot c}$$$$a^{(b^c)}neq a^{bcdot c}$$
$endgroup$
– Don Thousand
3 hours ago




$begingroup$
$$(a^b)^c=a^{bcdot c}$$$$a^{(b^c)}neq a^{bcdot c}$$
$endgroup$
– Don Thousand
3 hours ago












$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
3 hours ago




$begingroup$
$(e^x)^2$ would be where you use the rule that you're thinking of.
$endgroup$
– Tartaglia's Stutter
3 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

As you might have realised, exponentiation is not associative:



$$left(a^bright)^c ne a^left(b^cright)$$



So what should $a^{b^c}$ mean? The convention is that exponentiation is right associative:



$$a^{b^c} = a^left(b^cright)$$



Because the otherwise left-associative exponentiation is just less useful and redundant, as it can be represented by multiplication inside the power (again as you might have realised):



$$a^{bc} = left(a^bright)^c$$



Wikipedia on associativity of exponentiation.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The exponential expression $a^{b^c}$ is equal to $a^{(b^c)}$. It is not equal to $(a^b)^c=a^{bcdot c}$ as you seem to think it is. In general an exponential is evaluated from right to left with the highest term evaluated first. That is to say
    $$large{x_0^{x_1^{x_2^{dots^{x_n}}}}=x_0^{left(x_1^{left(x_2^{left(dots^{(x_n)}right)}right)}right)}}$$



    For the second part of your question see this duplicate.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I didn't think $a^{(b^c)} = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^{x^2}$ as $e^{2x}$, since without parentheses it can mean both things . Am I correct?
      $endgroup$
      – Rob
      3 hours ago










    • $begingroup$
      NO! Without parentheses the expression $a^{b^c}$ is always equal to $a^{left(b^cright)}$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
      $endgroup$
      – Peter Foreman
      3 hours ago












    • $begingroup$
      I see! Well, I learnt something new that bothered me for a long time. So $e ^ {x^2}$ is understood as $e^{(x^2)}$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
      $endgroup$
      – Rob
      3 hours ago



















    1












    $begingroup$

    To answer the titular question, there's a result in real analysis that shows that derivatives satisfy have the intermediate value property (just like continuous functions). It follows that a function that skips values cannot be the derivative of anything in the usual sense. This implies that functions with jump discontinuities (like the Heaviside step, for example) cannot be the derivative of anything.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Correct me if I am wrong, but that does not explain why $e^{x^2}$ has no antiderivative, right? I understand you are answering the titular question.
      $endgroup$
      – Rob
      2 hours ago










    • $begingroup$
      $e^{x^2}$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
      $endgroup$
      – MathIsFun
      2 hours ago












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    As you might have realised, exponentiation is not associative:



    $$left(a^bright)^c ne a^left(b^cright)$$



    So what should $a^{b^c}$ mean? The convention is that exponentiation is right associative:



    $$a^{b^c} = a^left(b^cright)$$



    Because the otherwise left-associative exponentiation is just less useful and redundant, as it can be represented by multiplication inside the power (again as you might have realised):



    $$a^{bc} = left(a^bright)^c$$



    Wikipedia on associativity of exponentiation.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      As you might have realised, exponentiation is not associative:



      $$left(a^bright)^c ne a^left(b^cright)$$



      So what should $a^{b^c}$ mean? The convention is that exponentiation is right associative:



      $$a^{b^c} = a^left(b^cright)$$



      Because the otherwise left-associative exponentiation is just less useful and redundant, as it can be represented by multiplication inside the power (again as you might have realised):



      $$a^{bc} = left(a^bright)^c$$



      Wikipedia on associativity of exponentiation.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        As you might have realised, exponentiation is not associative:



        $$left(a^bright)^c ne a^left(b^cright)$$



        So what should $a^{b^c}$ mean? The convention is that exponentiation is right associative:



        $$a^{b^c} = a^left(b^cright)$$



        Because the otherwise left-associative exponentiation is just less useful and redundant, as it can be represented by multiplication inside the power (again as you might have realised):



        $$a^{bc} = left(a^bright)^c$$



        Wikipedia on associativity of exponentiation.






        share|cite|improve this answer









        $endgroup$



        As you might have realised, exponentiation is not associative:



        $$left(a^bright)^c ne a^left(b^cright)$$



        So what should $a^{b^c}$ mean? The convention is that exponentiation is right associative:



        $$a^{b^c} = a^left(b^cright)$$



        Because the otherwise left-associative exponentiation is just less useful and redundant, as it can be represented by multiplication inside the power (again as you might have realised):



        $$a^{bc} = left(a^bright)^c$$



        Wikipedia on associativity of exponentiation.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        peterwhypeterwhy

        12.3k21229




        12.3k21229























            1












            $begingroup$

            The exponential expression $a^{b^c}$ is equal to $a^{(b^c)}$. It is not equal to $(a^b)^c=a^{bcdot c}$ as you seem to think it is. In general an exponential is evaluated from right to left with the highest term evaluated first. That is to say
            $$large{x_0^{x_1^{x_2^{dots^{x_n}}}}=x_0^{left(x_1^{left(x_2^{left(dots^{(x_n)}right)}right)}right)}}$$



            For the second part of your question see this duplicate.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I didn't think $a^{(b^c)} = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^{x^2}$ as $e^{2x}$, since without parentheses it can mean both things . Am I correct?
              $endgroup$
              – Rob
              3 hours ago










            • $begingroup$
              NO! Without parentheses the expression $a^{b^c}$ is always equal to $a^{left(b^cright)}$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
              $endgroup$
              – Peter Foreman
              3 hours ago












            • $begingroup$
              I see! Well, I learnt something new that bothered me for a long time. So $e ^ {x^2}$ is understood as $e^{(x^2)}$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
              $endgroup$
              – Rob
              3 hours ago
















            1












            $begingroup$

            The exponential expression $a^{b^c}$ is equal to $a^{(b^c)}$. It is not equal to $(a^b)^c=a^{bcdot c}$ as you seem to think it is. In general an exponential is evaluated from right to left with the highest term evaluated first. That is to say
            $$large{x_0^{x_1^{x_2^{dots^{x_n}}}}=x_0^{left(x_1^{left(x_2^{left(dots^{(x_n)}right)}right)}right)}}$$



            For the second part of your question see this duplicate.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I didn't think $a^{(b^c)} = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^{x^2}$ as $e^{2x}$, since without parentheses it can mean both things . Am I correct?
              $endgroup$
              – Rob
              3 hours ago










            • $begingroup$
              NO! Without parentheses the expression $a^{b^c}$ is always equal to $a^{left(b^cright)}$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
              $endgroup$
              – Peter Foreman
              3 hours ago












            • $begingroup$
              I see! Well, I learnt something new that bothered me for a long time. So $e ^ {x^2}$ is understood as $e^{(x^2)}$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
              $endgroup$
              – Rob
              3 hours ago














            1












            1








            1





            $begingroup$

            The exponential expression $a^{b^c}$ is equal to $a^{(b^c)}$. It is not equal to $(a^b)^c=a^{bcdot c}$ as you seem to think it is. In general an exponential is evaluated from right to left with the highest term evaluated first. That is to say
            $$large{x_0^{x_1^{x_2^{dots^{x_n}}}}=x_0^{left(x_1^{left(x_2^{left(dots^{(x_n)}right)}right)}right)}}$$



            For the second part of your question see this duplicate.






            share|cite|improve this answer









            $endgroup$



            The exponential expression $a^{b^c}$ is equal to $a^{(b^c)}$. It is not equal to $(a^b)^c=a^{bcdot c}$ as you seem to think it is. In general an exponential is evaluated from right to left with the highest term evaluated first. That is to say
            $$large{x_0^{x_1^{x_2^{dots^{x_n}}}}=x_0^{left(x_1^{left(x_2^{left(dots^{(x_n)}right)}right)}right)}}$$



            For the second part of your question see this duplicate.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            Peter ForemanPeter Foreman

            8,4771321




            8,4771321












            • $begingroup$
              I didn't think $a^{(b^c)} = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^{x^2}$ as $e^{2x}$, since without parentheses it can mean both things . Am I correct?
              $endgroup$
              – Rob
              3 hours ago










            • $begingroup$
              NO! Without parentheses the expression $a^{b^c}$ is always equal to $a^{left(b^cright)}$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
              $endgroup$
              – Peter Foreman
              3 hours ago












            • $begingroup$
              I see! Well, I learnt something new that bothered me for a long time. So $e ^ {x^2}$ is understood as $e^{(x^2)}$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
              $endgroup$
              – Rob
              3 hours ago


















            • $begingroup$
              I didn't think $a^{(b^c)} = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^{x^2}$ as $e^{2x}$, since without parentheses it can mean both things . Am I correct?
              $endgroup$
              – Rob
              3 hours ago










            • $begingroup$
              NO! Without parentheses the expression $a^{b^c}$ is always equal to $a^{left(b^cright)}$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
              $endgroup$
              – Peter Foreman
              3 hours ago












            • $begingroup$
              I see! Well, I learnt something new that bothered me for a long time. So $e ^ {x^2}$ is understood as $e^{(x^2)}$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
              $endgroup$
              – Rob
              3 hours ago
















            $begingroup$
            I didn't think $a^{(b^c)} = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^{x^2}$ as $e^{2x}$, since without parentheses it can mean both things . Am I correct?
            $endgroup$
            – Rob
            3 hours ago




            $begingroup$
            I didn't think $a^{(b^c)} = (a^b)^c$. Once we put parentheses it is all clear. But I think one would not be wrong for interpreting $e^{x^2}$ as $e^{2x}$, since without parentheses it can mean both things . Am I correct?
            $endgroup$
            – Rob
            3 hours ago












            $begingroup$
            NO! Without parentheses the expression $a^{b^c}$ is always equal to $a^{left(b^cright)}$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
            $endgroup$
            – Peter Foreman
            3 hours ago






            $begingroup$
            NO! Without parentheses the expression $a^{b^c}$ is always equal to $a^{left(b^cright)}$. See here - en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
            $endgroup$
            – Peter Foreman
            3 hours ago














            $begingroup$
            I see! Well, I learnt something new that bothered me for a long time. So $e ^ {x^2}$ is understood as $e^{(x^2)}$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
            $endgroup$
            – Rob
            3 hours ago




            $begingroup$
            I see! Well, I learnt something new that bothered me for a long time. So $e ^ {x^2}$ is understood as $e^{(x^2)}$. Also the link you sent, which I could not find at first, is very interesting but a little advanced for me.
            $endgroup$
            – Rob
            3 hours ago











            1












            $begingroup$

            To answer the titular question, there's a result in real analysis that shows that derivatives satisfy have the intermediate value property (just like continuous functions). It follows that a function that skips values cannot be the derivative of anything in the usual sense. This implies that functions with jump discontinuities (like the Heaviside step, for example) cannot be the derivative of anything.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Correct me if I am wrong, but that does not explain why $e^{x^2}$ has no antiderivative, right? I understand you are answering the titular question.
              $endgroup$
              – Rob
              2 hours ago










            • $begingroup$
              $e^{x^2}$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
              $endgroup$
              – MathIsFun
              2 hours ago
















            1












            $begingroup$

            To answer the titular question, there's a result in real analysis that shows that derivatives satisfy have the intermediate value property (just like continuous functions). It follows that a function that skips values cannot be the derivative of anything in the usual sense. This implies that functions with jump discontinuities (like the Heaviside step, for example) cannot be the derivative of anything.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Correct me if I am wrong, but that does not explain why $e^{x^2}$ has no antiderivative, right? I understand you are answering the titular question.
              $endgroup$
              – Rob
              2 hours ago










            • $begingroup$
              $e^{x^2}$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
              $endgroup$
              – MathIsFun
              2 hours ago














            1












            1








            1





            $begingroup$

            To answer the titular question, there's a result in real analysis that shows that derivatives satisfy have the intermediate value property (just like continuous functions). It follows that a function that skips values cannot be the derivative of anything in the usual sense. This implies that functions with jump discontinuities (like the Heaviside step, for example) cannot be the derivative of anything.






            share|cite|improve this answer









            $endgroup$



            To answer the titular question, there's a result in real analysis that shows that derivatives satisfy have the intermediate value property (just like continuous functions). It follows that a function that skips values cannot be the derivative of anything in the usual sense. This implies that functions with jump discontinuities (like the Heaviside step, for example) cannot be the derivative of anything.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            AllawonderAllawonder

            2,349617




            2,349617












            • $begingroup$
              Correct me if I am wrong, but that does not explain why $e^{x^2}$ has no antiderivative, right? I understand you are answering the titular question.
              $endgroup$
              – Rob
              2 hours ago










            • $begingroup$
              $e^{x^2}$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
              $endgroup$
              – MathIsFun
              2 hours ago


















            • $begingroup$
              Correct me if I am wrong, but that does not explain why $e^{x^2}$ has no antiderivative, right? I understand you are answering the titular question.
              $endgroup$
              – Rob
              2 hours ago










            • $begingroup$
              $e^{x^2}$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
              $endgroup$
              – MathIsFun
              2 hours ago
















            $begingroup$
            Correct me if I am wrong, but that does not explain why $e^{x^2}$ has no antiderivative, right? I understand you are answering the titular question.
            $endgroup$
            – Rob
            2 hours ago




            $begingroup$
            Correct me if I am wrong, but that does not explain why $e^{x^2}$ has no antiderivative, right? I understand you are answering the titular question.
            $endgroup$
            – Rob
            2 hours ago












            $begingroup$
            $e^{x^2}$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
            $endgroup$
            – MathIsFun
            2 hours ago




            $begingroup$
            $e^{x^2}$ does have an antiderivative, but it can't be expressed in terms of elementary functions. Having an antiderivative and having an elementary antiderivative are not the same thing.
            $endgroup$
            – MathIsFun
            2 hours ago


















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