Antiderivative of an odd function












5












$begingroup$



Is the antiderivative of an odd function even?




The answer given by the book is yes.

However, I found a counterexample defined in $mathbb{R}setminus {0}$:
$$f(x)=begin{cases}ln |x|+1& x<0\ln |x|&x>0end{cases}$$
Its derivative is $frac 1x$, which is an odd function.




Question: is my counterexample right?







real-analysis calculus integration







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:56










  • $begingroup$
    You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
    $endgroup$
    – apnorton
    Dec 17 '18 at 7:19












  • $begingroup$
    @apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
    $endgroup$
    – Kemono Chen
    Dec 17 '18 at 7:23






  • 3




    $begingroup$
    $x^2 + C$ is even for any $C$. For the opposite case of whether the antiderivative of an even function is odd, this point would be valid.
    $endgroup$
    – badjohn
    Dec 17 '18 at 11:28






  • 1




    $begingroup$
    @1123581321 Constants are even.
    $endgroup$
    – Dylan
    Dec 17 '18 at 19:19


















5












$begingroup$



Is the antiderivative of an odd function even?




The answer given by the book is yes.

However, I found a counterexample defined in $mathbb{R}setminus {0}$:
$$f(x)=begin{cases}ln |x|+1& x<0\ln |x|&x>0end{cases}$$
Its derivative is $frac 1x$, which is an odd function.




Question: is my counterexample right?







real-analysis calculus integration







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:56










  • $begingroup$
    You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
    $endgroup$
    – apnorton
    Dec 17 '18 at 7:19












  • $begingroup$
    @apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
    $endgroup$
    – Kemono Chen
    Dec 17 '18 at 7:23






  • 3




    $begingroup$
    $x^2 + C$ is even for any $C$. For the opposite case of whether the antiderivative of an even function is odd, this point would be valid.
    $endgroup$
    – badjohn
    Dec 17 '18 at 11:28






  • 1




    $begingroup$
    @1123581321 Constants are even.
    $endgroup$
    – Dylan
    Dec 17 '18 at 19:19
















5












5








5





$begingroup$



Is the antiderivative of an odd function even?




The answer given by the book is yes.

However, I found a counterexample defined in $mathbb{R}setminus {0}$:
$$f(x)=begin{cases}ln |x|+1& x<0\ln |x|&x>0end{cases}$$
Its derivative is $frac 1x$, which is an odd function.




Question: is my counterexample right?







real-analysis calculus integration







share|cite|improve this question









$endgroup$





Is the antiderivative of an odd function even?




The answer given by the book is yes.

However, I found a counterexample defined in $mathbb{R}setminus {0}$:
$$f(x)=begin{cases}ln |x|+1& x<0\ln |x|&x>0end{cases}$$
Its derivative is $frac 1x$, which is an odd function.




Question: is my counterexample right?







real-analysis calculus integration




real-analysis calculus integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 6:18









Kemono ChenKemono Chen

2,8831739




2,8831739








  • 2




    $begingroup$
    I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:56










  • $begingroup$
    You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
    $endgroup$
    – apnorton
    Dec 17 '18 at 7:19












  • $begingroup$
    @apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
    $endgroup$
    – Kemono Chen
    Dec 17 '18 at 7:23






  • 3




    $begingroup$
    $x^2 + C$ is even for any $C$. For the opposite case of whether the antiderivative of an even function is odd, this point would be valid.
    $endgroup$
    – badjohn
    Dec 17 '18 at 11:28






  • 1




    $begingroup$
    @1123581321 Constants are even.
    $endgroup$
    – Dylan
    Dec 17 '18 at 19:19
















  • 2




    $begingroup$
    I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:56










  • $begingroup$
    You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
    $endgroup$
    – apnorton
    Dec 17 '18 at 7:19












  • $begingroup$
    @apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
    $endgroup$
    – Kemono Chen
    Dec 17 '18 at 7:23






  • 3




    $begingroup$
    $x^2 + C$ is even for any $C$. For the opposite case of whether the antiderivative of an even function is odd, this point would be valid.
    $endgroup$
    – badjohn
    Dec 17 '18 at 11:28






  • 1




    $begingroup$
    @1123581321 Constants are even.
    $endgroup$
    – Dylan
    Dec 17 '18 at 19:19










2




2




$begingroup$
I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
$endgroup$
– Dylan
Dec 17 '18 at 6:56




$begingroup$
I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval
$endgroup$
– Dylan
Dec 17 '18 at 6:56












$begingroup$
You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
$endgroup$
– apnorton
Dec 17 '18 at 7:19






$begingroup$
You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$.
$endgroup$
– apnorton
Dec 17 '18 at 7:19














$begingroup$
@apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
$endgroup$
– Kemono Chen
Dec 17 '18 at 7:23




$begingroup$
@apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
$endgroup$
– Kemono Chen
Dec 17 '18 at 7:23




3




3




$begingroup$
$x^2 + C$ is even for any $C$. For the opposite case of whether the antiderivative of an even function is odd, this point would be valid.
$endgroup$
– badjohn
Dec 17 '18 at 11:28




$begingroup$
$x^2 + C$ is even for any $C$. For the opposite case of whether the antiderivative of an even function is odd, this point would be valid.
$endgroup$
– badjohn
Dec 17 '18 at 11:28




1




1




$begingroup$
@1123581321 Constants are even.
$endgroup$
– Dylan
Dec 17 '18 at 19:19






$begingroup$
@1123581321 Constants are even.
$endgroup$
– Dylan
Dec 17 '18 at 19:19












1 Answer
1






active

oldest

votes


















5












$begingroup$

I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that



$$F(x)=int_0^x f(t) dt+c.$$



If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=F(x).$



Try it !






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I only get that when c is zero...
    $endgroup$
    – lalala
    Dec 17 '18 at 9:28










  • $begingroup$
    @lalala If $f(t)=sin(t)$ then $int_0^x f(t) ,dt = 1-cos(t)$. Choose a $c$ such as $-1$ or if you prefer $1$ and do as Fred suggests to find $F(-x)$. What do you get?
    $endgroup$
    – Henry
    Dec 17 '18 at 13:13











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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that



$$F(x)=int_0^x f(t) dt+c.$$



If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=F(x).$



Try it !






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I only get that when c is zero...
    $endgroup$
    – lalala
    Dec 17 '18 at 9:28










  • $begingroup$
    @lalala If $f(t)=sin(t)$ then $int_0^x f(t) ,dt = 1-cos(t)$. Choose a $c$ such as $-1$ or if you prefer $1$ and do as Fred suggests to find $F(-x)$. What do you get?
    $endgroup$
    – Henry
    Dec 17 '18 at 13:13
















5












$begingroup$

I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that



$$F(x)=int_0^x f(t) dt+c.$$



If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=F(x).$



Try it !






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I only get that when c is zero...
    $endgroup$
    – lalala
    Dec 17 '18 at 9:28










  • $begingroup$
    @lalala If $f(t)=sin(t)$ then $int_0^x f(t) ,dt = 1-cos(t)$. Choose a $c$ such as $-1$ or if you prefer $1$ and do as Fred suggests to find $F(-x)$. What do you get?
    $endgroup$
    – Henry
    Dec 17 '18 at 13:13














5












5








5





$begingroup$

I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that



$$F(x)=int_0^x f(t) dt+c.$$



If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=F(x).$



Try it !






share|cite|improve this answer











$endgroup$



I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that



$$F(x)=int_0^x f(t) dt+c.$$



If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=F(x).$



Try it !







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 10:21

























answered Dec 17 '18 at 7:05









FredFred

44.6k1846




44.6k1846








  • 1




    $begingroup$
    I only get that when c is zero...
    $endgroup$
    – lalala
    Dec 17 '18 at 9:28










  • $begingroup$
    @lalala If $f(t)=sin(t)$ then $int_0^x f(t) ,dt = 1-cos(t)$. Choose a $c$ such as $-1$ or if you prefer $1$ and do as Fred suggests to find $F(-x)$. What do you get?
    $endgroup$
    – Henry
    Dec 17 '18 at 13:13














  • 1




    $begingroup$
    I only get that when c is zero...
    $endgroup$
    – lalala
    Dec 17 '18 at 9:28










  • $begingroup$
    @lalala If $f(t)=sin(t)$ then $int_0^x f(t) ,dt = 1-cos(t)$. Choose a $c$ such as $-1$ or if you prefer $1$ and do as Fred suggests to find $F(-x)$. What do you get?
    $endgroup$
    – Henry
    Dec 17 '18 at 13:13








1




1




$begingroup$
I only get that when c is zero...
$endgroup$
– lalala
Dec 17 '18 at 9:28




$begingroup$
I only get that when c is zero...
$endgroup$
– lalala
Dec 17 '18 at 9:28












$begingroup$
@lalala If $f(t)=sin(t)$ then $int_0^x f(t) ,dt = 1-cos(t)$. Choose a $c$ such as $-1$ or if you prefer $1$ and do as Fred suggests to find $F(-x)$. What do you get?
$endgroup$
– Henry
Dec 17 '18 at 13:13




$begingroup$
@lalala If $f(t)=sin(t)$ then $int_0^x f(t) ,dt = 1-cos(t)$. Choose a $c$ such as $-1$ or if you prefer $1$ and do as Fred suggests to find $F(-x)$. What do you get?
$endgroup$
– Henry
Dec 17 '18 at 13:13


















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