# Show that $\int_{\mathbb{R}} \phi (x) dx= 0$

Let $\phi \in C_{c}^{\infty}\mathbb{(R)}$ . I need to show that $\int_{\mathbb{R}} \phi (x) dx = 0$ iff there exists a function $\psi \in C_{c}^{\infty}\mathbb{(R)}$ such that $\phi(x) = \psi ' (x)$.

I have absolutely no clue how to begin. Any help with this is appreciated!

• What is $C_c^\infty$ the Riemann Sphere? – caverac Apr 23 '17 at 16:41
• Is it right that $\;\int_{\Bbb R}=\int_{-\infty}^\infty\;$ , an improper integral? – DonAntonio Apr 23 '17 at 16:45
• I was thinking it was the extended complex plane. In that case if $\phi(x) = \phi'(x)$ then the integral only depends on the end points, which are the same on this space – caverac Apr 23 '17 at 16:46
• @Dark_Knight Even if this is Lebesgue Integral we have a problem here as improper Lebesgue integrals are defined by means of improper Riemann integrals, and this last doesn't exist in this case. – DonAntonio Apr 23 '17 at 17:00
• If nothing is mentioned then I suppose they follow the standard use for $C_c^\infty$, then we can define $\psi(x) = \int _{-\infty}^x \phi(u) du$ and since $\int _{-\infty}^\infty \phi(u) du=0$ we see that $\psi(x) \in C_c^\infty$ – clark Apr 23 '17 at 17:29

Hint:

Say $\operatorname{supp}\psi \subseteq [-M, M]$. Then, it follows $$\int_{\mathbb R} \phi = \int_{\mathbb R} \psi' = \int_{-M}^M \psi' = \psi(M) - \psi(-M).$$

On other hand, assume $\operatorname{supp} \phi \subseteq [-M, M]$. Then, $$\psi(x) = \int_{-M}^x \phi$$ has compact support if and only if $$\psi(M) = \int_{\mathbb R} \phi = 0.$$