Unlearn and relearn - I: Integration

    I pity myself sometimes for not giving enough attention to minute details of a statement and also not worrying about lifting various conditions off a theorem. Recently I encountered a problem on stackexchange which asked if the function

$f(x)=2x-\lfloor \sin (x)\rfloor$

had an antiderivative. I had an adrenalin rush and I immediately posted a comment:

Some properties of Riemann-Stieltje's integral may help us in answering this problem. Following are they

• If $f(x)$ is a continuous function on $[a,b]$ except possibly at finitely many points, then $f(x)$ is integrable on $[a,b]$, i.e. $\underset{a}{\overset{b}{\int }}f(x)dx$ exists.

• If we define $F(x)$ by

  $F(x)=\underset{a}{\overset{x}{\int }}f(x)dx$

  for any $x,a\le x\le b$, then $F(x)$ is differentable on $[a,b]$ and that $F^{\prime }(x)=f(x)$.

    I did not give an answer in `Yes' or `No' although I had in my mind that the answer was a 'Yes'. But, one of the comments from another user (with good reputation) read - `No, as derivatives satisfy intermediate value theorem'. Although this comment was posted before that of mine, I hadn't considered it seriously. But, I soon got to thinking and started worrying about my deteriorating knowledge on integrals. What might be the reason for it - I asked myself. Probably, one reason that emereged out strongly was that I never had studied integrals with certain intermediary goals. Further, very little I asked myself questions such as what would have happenned if some condition was lifted off in a given statement. Was this because I was overjoyous upon understanding something or because I didn't want to take further risk, I do not know. I remember till today, I preferred Mathematics to any other subject because of the joy I got on understanding the Riemanninan view of integration.

     I soon realized that I had made a mistake in the second statmenet. I had missed one condition on $f$ - that continuity of $f$. I remember, I never had asked myself the question - `does there exist a function $f(x)$ discontinuous at some point in $[a,b]$, some closed interval such that if we define $F(x)$ by

$F(x)=\underset{a}{\overset{x}{\int }}f(x)dx,a\le x\le b$,

then, $F(x)$ is not differentaible at some point in $[a,b]$?' Had I asked this then, I might have had landed upon what was called Darboux's theorem on the intermediate property of derivatives, which I was not aware of till this day. The statement of Darboux's is as follows:

If $f$ is differentiable on an interval $[a,b]$ and that $y$ lies between $f^{\prime }(a)$ and $f^{\prime }(b)$, that is $y\in (f^{\prime }(a),f^{\prime }(b))$, assuming that $f^{\prime }(a)<f^{\prime }(b)$, then there exists $c\in (a,b)$ such that

$f^{\prime }(c)=y$.

Even if $f^{\prime }$ isn't continuous, the intermediate property holds good for $f^{\prime }$. The proof of this is very simple, as one can see in the wiki article. Simply define a new function $\phi (t)=f(t)-yt$ on $[a,b]$ and show that this function cannot attain maximum at $a$ or $b$ and hence must attain a maximum in $(a,b)$, say at $c$ as $\phi$ is a continuous function on $[a,b]$. Then at $c$, $\phi$ must have a zero derivative.

This was a very simple theorem. Now, I remodified the question on stackexchange. Does $f(x)$ as defined in the beginning have an antiderivative on $\mathbb{R}$? If we consider an interval of the kind $[{\displaystyle \frac{\pi }{2}}-ϵ,{\displaystyle \frac{\pi }{2}}]$, with $ϵ$ as small as possible, then for no point $c$ in $({\displaystyle \frac{\pi }{2}}-ϵ,{\displaystyle \frac{\pi }{2}})$, we would have had

$f(c)=2c-\lfloor \sin (c)\rfloor =\pi -{\displaystyle \frac{1}{2}}.$

This is because, for any $c\in ({\displaystyle \frac{\pi }{2}}-ϵ,{\displaystyle \frac{\pi }{2}})$, we have $f(c)=2c>2(\frac{\pi }{2}-ϵ)=\pi -2ϵ$. If we choose $ϵ<\frac{1}{4}$, then $f(c)>\pi -2ϵ>\pi -\frac{1}{2}$. But observe that $\pi -\frac{1}{2}$ lies between $f\left({\displaystyle \frac{\pi }{2}}\right)$ and $f({\displaystyle \frac{\pi }{2}}-ϵ)$. Thus, for every $c\in (\frac{\pi }{2}-ϵ,\frac{\pi }{2})$, we would have had

$f(c)\ne \pi -\frac{1}{2}$,

if we happen to choose $ϵ<\frac{1}{4}$.

Hence, if at all $f$ had an antiderivative, say $F$ on $[{\displaystyle \frac{\pi }{2}}-ϵ,\frac{\pi }{2}]$, then $F$ should have satisfied Darboux's theorem, that is for some $c$ in that interval, we would have had

$F^{\prime }(c)=f(c)=\pi -\frac{1}{2}$,

which cannot happen.

I now recall very clearly where the definition of continuity was used in the proof of the following statement:

If $f$ is continuous on $[a,b]$, then the function $F$ is differentiable on $[a,b]$, where

$F(x)=\underset{a}{\overset{x}{\int }}f(x)dx,a\le x\le b$.

It was in the step

$\frac{F(x_0+h)-F(x_0)}{h}=\frac{\underset{x_0}{\overset{x_0+h}{\int }}f(x)dx}{h}.$

Now I started pondering on the question - 'Can there exist a function $f$ on $\mathbb{R}$ which is discontinuous at some point, but has antiderivative on $\mathbb{R}$?' There are a lot of functions $f$ such that $f$ is differentiable on $\mathbb{R}$ with derivative discontinuous at some point. I am now pondering on the question - 'Is there a function $f$ on $\mathbb{R}$ which has anti derivative on each closed interval but has no antiderivative on $\mathbb{R}$?