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. \(\int\limits_{a}^{b} f(x)\, dx\) exists.

• If we define \(F(x)\) by

  \(F(x) = \int\limits_{a}^xf(x)\, dx\)

  for any \(x, a\leq x \leq b\), 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

\(F(x) = \int\limits_{a}^x f(x)\, dx, \quad a\leq x\leq b\),

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 \(\varphi(t) = f(t) - yt\) 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 \(\varphi\) is a continuous function on \([a,b]\). Then at \(c\), \(\varphi\) 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 \(\left[\dfrac{\pi}{2}-\epsilon, \dfrac{\pi}{2}\right]\), with \(\epsilon\) as small as possible, then for no point \(c\) in \(\left(\dfrac{\pi}{2}-\epsilon, \dfrac{\pi}{2}\right)\), we would have had

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

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

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

if we happen to choose \(\epsilon < \frac{1}{4}\).

Hence, if at all \(f\) had an antiderivative, say \(F\) on \(\left[\dfrac{\pi}{2}-\epsilon, \frac{\pi}{2}\right]\), 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

\(F(x) = \int\limits_{a}^xf(x)\, dx, \quad a\leq x \leq b\).

It was in the step

\( \frac{F(x_0+h)-F(x_0)}{h} = \frac{\int\limits_{x_0}^{x_0+h} f(x)\, dx}{h}.\)

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 \(\mathbb{R}\)?