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Studying hyperbolic partial differential equations, we arrive, in a certain calculation, to the following doubt: every integrable function has a primitive?

If $ u_0 $ is integrable, then $ \exists v_0 $ such that $ \int_b^a v'_0 dt = \int_b^a u_0 dt $?

I think no, but maybe there is some theorem that determines when this holds. And maybe this theorem aplies to our case.

Many thanks for any help!

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    $\begingroup$ For fixed $a,b$, there is a constant $C$ such that $\int_b^a C dt = \int_b^a u_0 dt$. So perhaps you mean something more difficult. Such as: there exists $v_0$ such that your equation holds for all $a,b$? $\endgroup$ – GEdgar Apr 20 at 20:39
  • $\begingroup$ @Gedgar, yes, this is what i am trying, for all a and b. Thanks. $\endgroup$ – Na'omi Apr 21 at 19:15
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Something similar is correct. We cannot arrange that $v_0$ is differentiable everywhere. But we can arrange that is it differentiable almost everywhere.

If $u_0$ is integrable on the interval $[0,1]$, then $$ v_0(x) = \int_a^x u_0(t)\;dt,\qquad 0 \le x \le 1 $$ satisfies $v_0'(x) = u_0(x)$ for almost all $x \in [0,1]$. And therefore $$ \int_a^b v_0'(x)\;dx = \int_a^b u_0(x)\;dx,\qquad 0 \le a \le b \le 1 $$

This should be found in textbooks which cover Lebesgue integration.

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  • $\begingroup$ Many thanks for the help! $\endgroup$ – Na'omi Apr 23 at 13:30

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