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Probably easy but I'm not very sure. If f(x) has an antiderivative F(x) then F(x) has also an antiderivative. True or False?

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Hint: an antiderivative is differentiable, therefore continuous.

Robert Israel
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All Riemann integrable functions have antiderivatives. Given $f(x)$,$$F(x) =\int_0^x f(t) dt$$ is an antiderivative. (Convince yourself by taking the derivative by FTOC.)

Then the second antiderivative would be $$F_2(x)=\int_0^x \int_0^\tau f(t) \,dt \,d\tau $$

Teoc
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    Please give a proof or citation of the first sentence. – vadim123 Feb 08 '15 at 20:18
  • I already have in the answer. – Teoc Feb 08 '15 at 20:20
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    Not true. For example, a function with a jump discontinuity doesn't have an antiderivative. FTOC applies only to continuous functions. – Robert Israel Feb 08 '15 at 20:21
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    See also this, which shows that not all integrable functions have antiderivatives. – vadim123 Feb 08 '15 at 20:23
  • Also, @MathNoob, "where $\tau = x$" is unnecessary. You should not have $\int_0^x f(x)dx$ where $x$ appears both in the limits of integration and as the variable against which you are integrating, hence the use of a dummy variable, just like is necessary in the single integral case (replacing with $t$). In the double integral you need to replace it twice (once with $t$, and again with something else, often $\tau$). – JMoravitz Feb 08 '15 at 20:23
  • Is f(x)=-2x*e^(-x^2) a counterexample? –  Feb 08 '15 at 20:24
  • While it can be integrated elementrary with a simple $u$ substitution into $e^{-x^2}$, that cannot be integrated elementray. However it does have antiderivative $\int_0^x e^{-t^2}, dt$ – Teoc Feb 08 '15 at 20:26
  • Take the example: $f(x) = 1$ if $x\geq 0$ and $f(x) = 0$ if $x<0$. Then $f$ is Riemann integrable with $F(x) = \int_{0}^x f(x')dx' = x$ if $x\geq 0$ and $F(x) = \int_{0}^x f(x')dx' = 0$ if $x<0$. However $F'(x) \not = f(x)$ for all $x$ (it does not exist for $x=0$) which is required for it to be called an antiderivative. – Winther Feb 08 '15 at 20:28
  • @Winther The antiderivative is merely $xf(x)$ – Teoc Feb 08 '15 at 20:29
  • Th point is that the definition of an antiderivative is a $F$ that satisfy $F'(x) = f(x)$ for all $x$. For the example above it is not the case. – Winther Feb 08 '15 at 20:30
  • But$ \frac{d}{dx}F(x)= \frac{d}{dx}xf(x)=x' f(x)+f'(x) x$. Now $f'(x)=\delta(x)$, which is $0$ everywhere except at $x=0$. So we have that $F'(x)$ does indeed equal $f(x)$ – Teoc Feb 08 '15 at 20:33
  • What is the value for $f'(0)$? $\delta(0)$ is not a real number. – Winther Feb 08 '15 at 20:33
  • Let us continue this discussion in chat. – Teoc Feb 08 '15 at 20:35