I am reading a notes in which following is written:

Let $f$ be continuous periodic function defined on $ \mathbb R$ with period 2l. Furthermore assume that $f’$ is piecewise continuous then $\int_ l^x f’(t) dt= f(x) $

I think the author is applying Fundamental Theorem of Calculus but $f’$ is just piecewise continuous and I know that Fundamental theorem of calculus is not valid for piecewise continuous function.Could you please help me verifying the above claim?

  • $\begingroup$ If it really says "$\int_ l^x f’(t) dt= f(x) dx$" you may want to find another set of notes to read - that makes no sense. A revised version $\int_ l^x f’(t) dt= f(x) $ makes sense but isn't true. (Exactly what do the notes say here?) But anyway, if $f'$ is piecewise continuous then it's certainly true that $f(x)-f(y)=\int_y^x f'(t)\,dt$; who told you this was not so? $\endgroup$ – David C. Ullrich Mar 2 '18 at 16:03
  • $\begingroup$ @DavidC.Ullrich: Sorry for the typo. Could you please say how does the claim you mentioned in the last holds ? I know the proof of this only for continuous function. $\endgroup$ – Math Lover Mar 2 '18 at 16:27
  • $\begingroup$ Sorry, I should have said that if $f$ is continuous and $f'$ is piecewise continuous then $f(x)-f(y)=\int_y^x f'(t)\,dt$. Here we're given that $f$ is continuous. (In any case, what you say it says, as corrected, now makes sense but it's still false - that intergral is $f(x)-f(l)$, not $f(x)$.) $\endgroup$ – David C. Ullrich Mar 2 '18 at 16:40
  • $\begingroup$ @DavidC.Ullrich: I only know the “ordinary” fundamental theorem of Calculus which assumes continuity of f and shows that the function F defined by $F(x)= \int_ a^x f(t) dt $ is differentiable and $F’(x)= f(x)$. Could you please give the reference for the result you mentioned? $\endgroup$ – Math Lover Mar 2 '18 at 16:54
  • $\begingroup$ I just posted a proof in an Answer... $\endgroup$ – David C. Ullrich Mar 2 '18 at 16:57

I tend to suspect that what's actually written in the notes is not the same as what's written above. For example, the integral is actually $f(x)-f(l)$, not $f(x)$. Also it's assumed that $f$ is periodic, but this is totally irrelevant to the conclusion.

If $f$ is continuous, $f$ is differentiable except at finitely many points, and $f'$ is piecewise continuous then it's certainly true that $$f(y)-f(x)=\int_x^yf'(t)\,dt.$$

Suppose for example that $f'$ is continuous on $[x,a)$, continuous on $(a,y]$, and has one-sided limits at $a$. (Note that the existence of one-sided limits is part of the definition of "piecewise continuous".) For small $\delta>0$ we have $$(\int_x^{a-\delta}+\int_{a+\delta}^y)f'(t)\,dt=f(y)-f(a+\delta)+f(a-\delta)-f(x).$$Now let $\delta\to0$: The fact that $f'$ has one-sided limits at $a$ shows that $$(\int_x^{a-\delta}+\int_{a+\delta}^y)f'(t)\,dt\to\int_x^y f'(t)\,dt,$$while the continuity of $f$ shows that $$-f(a+\delta)+f(a-\delta)\to0.$$

Similarly if $f'$ has more than one (but only finitely many) points of discontinuity.

  • $\begingroup$ Thank you. Could you just explain the part “ The fact that f’ has ...” ? $\endgroup$ – Math Lover Mar 2 '18 at 17:54
  • $\begingroup$ @MathLover That's the same as saying that $\int_{a-\delta}^{a+\delta}f'(t)\,dt\to0$. Since $f$ has one-sided limits it is bounded near $a$; now if $|f'|\le M$ then $\left|\int_{a-\delta}^{a+\delta}f'(t)\,dt\right|\le 2\delta M$. $\endgroup$ – David C. Ullrich Mar 2 '18 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.