6

Let, $g(x) =\int_{a}^{x} f(t) dt$ be an integral function.

What can we say about $g'(c)$ when,

a) $f$ is removable discontinuous at $c \in[a,x]$?

b) $f$ is infinite discontinuous at $c\in [a,x]$?

c) $f$ is jump discontinuous at $c\in [a,x]$?

Here is what I think.

a) Given, $$f(x) = \begin{cases} {x^3} & \text{if $x\in \mathbb{R} - \{2\} $} \\ 10 & \text{if $x=2$} \end{cases}$$ Here, $x=2=c$ is a point of removable discontinuity. The right-hand derivative of $g(x)$ at $c$ is given as, $$g'(c) =\lim_{h\to 0^{+} } \frac{g(c+h)-g(c)}{h}= \lim_{h\to 0^{+} }\frac{\int_{c}^{c+h}f(t) dt}{h}$$ The left-hand derivative of $g(x)$ at $c$ is given as, $$g'(c) = \lim_{h\to 0^{+} } \frac{g(c-h)-g(c)}{-h}= \lim_{h\to 0^{+} }\frac{\int_{c-h}^{c}f(t) dt}{-h}$$

We see that the left and right derivatives converge at 8. This means that $g'(2)=8$. We know that, "for continuous integrands", $g'(x)=f(x)$. Here, $f(2)\neq 8$. This tells me that the derivative does not exist at $x=2$.

Am I thinking in the right direction?

Paramanand Singh
  • 87,309
R004
  • 983
  • 1
    You need to give us more on this: Your thoughts, some work, some context ... – zhw. Jun 23 '17 at 06:27
  • Will do. Let me work on it. – R004 Jun 23 '17 at 06:30
  • Didn't you just prove in your case that $g'(2)=8$? Then why are you saying that $g'(x)=f(x)$? Does your case meet the assumptions of the theorem that you are using? – Li Chun Min Jun 23 '17 at 07:10
  • 1
    For continuous functions in general, $g'(x)=f(x)$. Here, I have proved that it's not the case for discontinuous functions. I must go and edit that statement. – R004 Jun 23 '17 at 07:12
  • If the discontinuity of $f$ at $c$ is removable $g'(c) =\lim_{x\to c} f(x) $. In fact more is true. If $f(x) \to L$ as $x\to c^{+} $ then right hand derivative of $g$ at $c$ is equal to $L$ and similar remarks apply to the left hand derivative of $g$. – Paramanand Singh Jun 23 '17 at 09:13

1 Answers1

5

To expand on my comment which covers only cases a) and c) I show that the Fundamental Theorem of Calculus can be applied in case a) also. Let $f$ have a removable discontinuity at $c$ and let $h(x) =f(x) $ for all $x\neq c$ and $h(c) =\lim_{x\to c} f(x) $ so that $h$ is continuous at $c$. Now changing integrand at a finite number of points does not change the integral so $$g(x) =\int_{a} ^{x} f(t) \, dt=\int_{a} ^{x} h(t) \, dt$$ and by Fundamental Theorem of Calculus we have $g'(c) =h(c) =\lim_{x\to c} f(x) $.

For case b) the integral does not make sense as the integrand is unbounded. For that you will need to deal with improper Riemann integral and then if the integral exists the derivative $g'(c) $ will also be infinite. You can see for example the case when $$g(x) =\arcsin x, f(x) =1/\sqrt{1-x^{2}},a=0,c=1$$

For case c) the derivative $g'(c) $ does not exist (based on my comment, the left and right derivatives are unequal).

There is another case d) which you have not considered where $f$ has oscillatory discontinuity at $c$. For that case you have the favorite example $$f(x) =\cos(1/x),f(0)=0,a=c=0, g'(0)=0$$ It is possible in this case that $g'(c)$ may not exist, but I don't have an example with me right now.

Paramanand Singh
  • 87,309
  • I believe I made a mistake while computing $\int_{c}^{c+h} f(t) dt$ & $\int_{c-h}^{c} f(t) dt$. In either case, if my thinking is not wrong, the Riemann sums( taken for different points within a rectangle ) do not converge, thereby suggesting that $f$ is not integrable in those intervals. This makes the derivative undefined. – R004 Jun 23 '17 at 10:03
  • Even in the case of jump discontinuity, the sums do not converge in the intervals containing the point of discontinuity. – R004 Jun 23 '17 at 10:04
  • 1
    @R004: Your approach is essentially using fundamental theorem of calculus. And it's fine for removable discontinuity. For jump discontinuity the left and right derivatives will exist but not be equal and therefore the derivative ceases to exist. The existence of left and right derivatives again comes from fundamental theorem of calculus. – Paramanand Singh Jun 23 '17 at 10:08
  • In your approach, you have extended the definition of the function by making $h(c) =\lim_{x\to c} f(x) $. The way I have defined my removal discontinuity is by stating that $h(c) \neq\lim_{x\to c} f(x) $. I was more interested in working on that type of discontinuity. – R004 Jun 23 '17 at 10:08
  • @R004: note that discontinuities at single points do not matter at all for convergence or divergence of Riemann sums, so that reasoning is incorrect. – Paramanand Singh Jun 23 '17 at 10:09
  • @R004: Your and mine working for case a) is same but expressed in different words. – Paramanand Singh Jun 23 '17 at 10:10
  • Let us continue this discussion in chat. – Paramanand Singh Jun 23 '17 at 10:12