Antiderivative of an odd function











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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?











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  • 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
    – Dylan
    48 mins ago










  • 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 $f(x) = x^3$.
    – apnorton
    24 mins ago












  • @apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
    – Kemono Chen
    21 mins ago















up vote
2
down vote

favorite













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?











share|cite|improve this question






















  • 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
    – Dylan
    48 mins ago










  • 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 $f(x) = x^3$.
    – apnorton
    24 mins ago












  • @apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
    – Kemono Chen
    21 mins ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite












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?











share|cite|improve this question














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






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









Kemono Chen

2,137435




2,137435












  • 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
    – Dylan
    48 mins ago










  • 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 $f(x) = x^3$.
    – apnorton
    24 mins ago












  • @apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
    – Kemono Chen
    21 mins ago


















  • 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
    – Dylan
    48 mins ago










  • 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 $f(x) = x^3$.
    – apnorton
    24 mins ago












  • @apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
    – Kemono Chen
    21 mins ago
















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
– Dylan
48 mins ago




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
– Dylan
48 mins ago












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 $f(x) = x^3$.
– apnorton
24 mins ago






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 $f(x) = x^3$.
– apnorton
24 mins ago














@apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
– Kemono Chen
21 mins ago




@apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative.
– Kemono Chen
21 mins ago










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










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




























    up vote
    2
    down vote













    For a function $f$ to be even, we must have $f(-x) = f(x)$.



    In your function $f$, consider $f(1)$ and $f(-1)$:



    $$f(1) = ln|1| = ln(1) = 0$$
    $$f(-1) = ln|-1|+1 = ln(1)+1 = 0+1 = 1$$



    $f$ is not even, since $f(1) neq f(-1)$.



    Or odd, for that matter, since $f(1) neq -f(-1)$.






    share|cite|improve this answer





















    • What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
      – Kemono Chen
      1 hour ago










    • This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
      – apnorton
      26 mins ago











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    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

























      up vote
      2
      down vote



      accepted










      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























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        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












        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










        answered 39 mins ago









        Fred

        43.7k1644




        43.7k1644






















            up vote
            2
            down vote













            For a function $f$ to be even, we must have $f(-x) = f(x)$.



            In your function $f$, consider $f(1)$ and $f(-1)$:



            $$f(1) = ln|1| = ln(1) = 0$$
            $$f(-1) = ln|-1|+1 = ln(1)+1 = 0+1 = 1$$



            $f$ is not even, since $f(1) neq f(-1)$.



            Or odd, for that matter, since $f(1) neq -f(-1)$.






            share|cite|improve this answer





















            • What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
              – Kemono Chen
              1 hour ago










            • This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
              – apnorton
              26 mins ago















            up vote
            2
            down vote













            For a function $f$ to be even, we must have $f(-x) = f(x)$.



            In your function $f$, consider $f(1)$ and $f(-1)$:



            $$f(1) = ln|1| = ln(1) = 0$$
            $$f(-1) = ln|-1|+1 = ln(1)+1 = 0+1 = 1$$



            $f$ is not even, since $f(1) neq f(-1)$.



            Or odd, for that matter, since $f(1) neq -f(-1)$.






            share|cite|improve this answer





















            • What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
              – Kemono Chen
              1 hour ago










            • This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
              – apnorton
              26 mins ago













            up vote
            2
            down vote










            up vote
            2
            down vote









            For a function $f$ to be even, we must have $f(-x) = f(x)$.



            In your function $f$, consider $f(1)$ and $f(-1)$:



            $$f(1) = ln|1| = ln(1) = 0$$
            $$f(-1) = ln|-1|+1 = ln(1)+1 = 0+1 = 1$$



            $f$ is not even, since $f(1) neq f(-1)$.



            Or odd, for that matter, since $f(1) neq -f(-1)$.






            share|cite|improve this answer












            For a function $f$ to be even, we must have $f(-x) = f(x)$.



            In your function $f$, consider $f(1)$ and $f(-1)$:



            $$f(1) = ln|1| = ln(1) = 0$$
            $$f(-1) = ln|-1|+1 = ln(1)+1 = 0+1 = 1$$



            $f$ is not even, since $f(1) neq f(-1)$.



            Or odd, for that matter, since $f(1) neq -f(-1)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Eevee Trainer

            3,340225




            3,340225












            • What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
              – Kemono Chen
              1 hour ago










            • This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
              – apnorton
              26 mins ago


















            • What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
              – Kemono Chen
              1 hour ago










            • This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
              – apnorton
              26 mins ago
















            What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
            – Kemono Chen
            1 hour ago




            What I unsure is the answer given by the book. $int_0^x f(x)dx=int_0^{-x}f(x)dx$. But the integral does not converge for this case. So is this "proof" given by the book wrong?
            – Kemono Chen
            1 hour ago












            This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
            – apnorton
            26 mins ago




            This answer isn't relevant to the question, as the question is claiming to show an odd function ($f(x) = 1/x$) with an anti-derivative that isn't even ($F(x) = text{piecewise function}$).
            – apnorton
            26 mins ago


















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