Take the following exercise taken from the book :
http://93.174.95.29/main/DD7D07B152D8622B441B7E63F9D4461E
https://libgen.lc/ads.php?md5=091431F03323DE6C347E43D0475FC387
(see page 14).
Consider the function defined on $[-1,0]$ by $f(x) = 1/\sqrt(-x)$ on $[-1,0)$ and, $f(0) =0$. Since this function is not bounded on $[-1,0]$, the Riemann integral does not exist. Show that, nevertheless, the Cauchy integral of this function over this interval does exist.
For the Riemann part, it's ok. For the Cauchy part, I would like to show that for any sequence $(x_k)_{1\le k\le n}$ such that $-1=x_0<x_1<\dots<x_n=0$, we have $$\sum_{k=1}^n\frac{x_k-x_{k-1}}{\sqrt{-x_{k-1}}}<\infty$$ when $k\to \infty$.
I cheat and write that $$\sum_{k=1}^n\frac{x_k-x_{k-1}}{\sqrt{-x_{k-1}}}<\int_{-1}^0\frac{1}{\sqrt{x}}=2$$ and I conclude.
What would be the right way to do it ?
The Riemann and Cauchy integrals both exist over $[-1,-c]$ where $c > 0$, since $f$ is bounded and continuous on that interval.
For any $\epsilon > 0$ there exists $\delta > 0$ such that for a partition $P: -1 = x_0 < x_1 < \ldots < x_{n-1} = -c$ with $\|P\| < \delta $, we have
$$2(1- \sqrt{c})- \epsilon = \int_{-1}^{-c}\frac{dx}{\sqrt{-x}}- \epsilon \leqslant \sum_{k=1}^{n-1} \frac{x_k - x_{k-1}}{\sqrt{-x_{k-1}}}\leqslant \int_{-1}^{-c}\frac{dx}{\sqrt{-x}}+ \epsilon = 2(1- \sqrt{c})+ \epsilon$$
With $x_n = 0$ we have
$$\sum_{k=1}^{n} \frac{x_k - x_{k-1}}{\sqrt{-x_{k-1}}}= \frac{0- (-c)}{\sqrt{c}}+ \sum_{k=1}^{n-1} \frac{x_k - x_{k-1}}{\sqrt{-x_{k-1}}}= \sqrt{c}+ \sum_{k=1}^{n-1} \frac{x_k - x_{k-1}}{\sqrt{-x_{k-1}}}$$
Thus,
$$-\epsilon/2 \leqslant \sum_{k=1}^{n} \frac{x_k - x_{k-1}}{\sqrt{-x_{k-1}}} -(2 - \sqrt{c}) \leqslant \epsilon/2,$$
and when the norm of the full partition $P': -1 = x_0 < x_2 < \ldots < x_{n-1} < x_n = 0$ is sufficiently small we have both $\|P\| < \delta$ and $\sqrt{c} < \epsilon /2 $
$$\left| \sum_{k=1}^{n} \frac{x_k - x_{k-1}}{\sqrt{-x_{k-1}}} -2 \right| \leqslant \left| \sum_{k=1}^{n} \frac{x_k - x_{k-1}}{\sqrt{-x_{k-1}}} -(2-\sqrt{c}) \right| + \sqrt{c}\leqslant \sqrt{c} + \epsilon/2 < \epsilon$$