Riemann Integrability and the Fundamental Theorem of Calculus

287 Views Asked by At

James Stewart's text Calculus with Early Transcendentals 7ed. pg 388 states:

FTC: If $f$ is continuous on $[a,b]$, then the function defined by $g(x) = \int_a^xf(t)dt,\: x\in[a,b]$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and $g'=f$.

Charles Pugh relaxes the conditions in his book Real Mathematical Analysis on pg. 171 by stating:

FTC: If $f:[a,b]\rightarrow\mathbb{R}$ is Riemann integrable then its indefinite integral $F(x) = \int_a^xf(t)dt$ is a continuous function of $x$ and $F'(x)=f(x)$ at all points $x$ at which $f$ is continuous.

On pg. 163 Pugh defines Riemann Integrability as follows:

A function $f:[a,b]\rightarrow\mathbb{R}$ is Riemann integrable if and only if it is bounded and its set of discontinuity points is a zero set(measure $0$).

Now consider the function $f(x) = x^{-1/3}$ and define the function $g(x) = \int_a^xf(t)dt = \int_a^xt^{-1/3}dt$.

$f(x)$ is not bounded at $x=0$ where we have a vertical asymptote.$f(x) = x^{-1/3}$

However, suppose $a<0$ and $x>0$ or $a>0$ and $x<0$, then the integral function $g(x)$ will, in a sense, cross this vertical asymptote. The improper integral limits from left and right at zero converge and so it seems we may write $g(x) = \frac{3}{2}\left(x^{2/3}-a^{2/3}\right)$ which is a continuous function for all $x\in\mathbb{R}$ that has a cusp at zero.enter image description here

Questions:

  1. If the improper integral converges at $x_0$ where $f(x_0)=\pm\infty$ can we relax the boundedness requirement in Pugh's definition?

  2. Would the resulting integral function always either have a finite discontinuity or cusp at $x_0$?

  3. What resources would you suggest I read to investigate related phenomena?