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When does a function NOT have an antiderivative?

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

asked 3 hours ago

RobRob

333110

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  
  
  

  
  $begingroup$
  $$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
  $endgroup$
  – 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
  

  
  
  
  

 | 
show 8 more comments

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

asked 3 hours ago

RobRob

333110

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  
  
  

  
  $begingroup$
  $$e^x^2=e^xcdot xneq e^xcdot e^x = e^x+x=e^2x$$
  $endgroup$
  – 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
  

  
  
  
  

 | 
show 8 more comments

$begingroup$

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

asked 3 hours ago

RobRob

333110

$endgroup$

share|cite|improve this question

asked 3 hours ago

RobRob

333110

share|cite|improve this question

asked 3 hours ago

RobRob

333110

asked 3 hours ago

RobRob

333110

asked 3 hours ago

RobRob

333110

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

answered 2 hours ago

peterwhypeterwhy

12.3k21229

$endgroup$

add a comment | 

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
$$largex_0^x_1^x_2^dots^x_n=x_0^left(x_1^left(x_2^left(dots^(x_n)right)right)right)$$

For the second part of your question see this duplicate.

share|cite|improve this answer

answered 3 hours ago

Peter ForemanPeter Foreman

8,4671321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  

  
  
  
  

add a comment | 

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

answered 3 hours ago

AllawonderAllawonder

2,349617

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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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

answered 2 hours ago

peterwhypeterwhy

12.3k21229

$endgroup$

add a comment | 

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

answered 2 hours ago

peterwhypeterwhy

12.3k21229

$endgroup$

add a comment | 

$begingroup$

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

answered 2 hours ago

peterwhypeterwhy

12.3k21229

$endgroup$

share|cite|improve this answer

answered 2 hours ago

peterwhypeterwhy

12.3k21229

answered 2 hours ago

peterwhypeterwhy

12.3k21229

answered 2 hours ago

peterwhypeterwhy

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
$$largex_0^x_1^x_2^dots^x_n=x_0^left(x_1^left(x_2^left(dots^(x_n)right)right)right)$$

For the second part of your question see this duplicate.

share|cite|improve this answer

answered 3 hours ago

Peter ForemanPeter Foreman

8,4671321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  

  
  
  
  

add a comment | 

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
$$largex_0^x_1^x_2^dots^x_n=x_0^left(x_1^left(x_2^left(dots^(x_n)right)right)right)$$

For the second part of your question see this duplicate.

share|cite|improve this answer

answered 3 hours ago

Peter ForemanPeter Foreman

8,4671321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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
$$largex_0^x_1^x_2^dots^x_n=x_0^left(x_1^left(x_2^left(dots^(x_n)right)right)right)$$

For the second part of your question see this duplicate.

share|cite|improve this answer

answered 3 hours ago

Peter ForemanPeter Foreman

8,4671321

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Peter ForemanPeter Foreman

8,4671321

answered 3 hours ago

Peter ForemanPeter Foreman

8,4671321

answered 3 hours ago

Peter ForemanPeter Foreman

8,4671321

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

answered 3 hours ago

AllawonderAllawonder

2,349617

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 3 hours ago

AllawonderAllawonder

2,349617

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 3 hours ago

AllawonderAllawonder

2,349617

$endgroup$

share|cite|improve this answer

answered 3 hours ago

AllawonderAllawonder

2,349617

answered 3 hours ago

AllawonderAllawonder

2,349617

answered 3 hours ago

AllawonderAllawonder

2,349617