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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 ?

prolea
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  • When $k=n$ you have $0$ in the denominator. – zhw. Aug 24 '20 at 17:39
  • sorry yeah it was in the first sum, it's the right node. I corrected – prolea Aug 24 '20 at 20:00
  • What is the Cauchy integral in this context? I'm not familiar with it. – zhw. Aug 24 '20 at 21:10
  • @zhw: It is Cauchy's definition as a limit of approximating sums of the form $\sum_{k=1}^nf(x_{k-1})(x_k - x_{k-1})$ as the partition norm tends to $0$ -- always evaluating the integrand at an endpoint of a subinterval. See for example here and here. – RRL Aug 24 '20 at 22:03
  • I think the point of Bressoud's exercise is that a Cauchy sum can converge to an improper Riemann integral when the function is unbounded and the Riemann integral does not exist. This always works if the integrand is monotone as it is here, but not in general. – RRL Aug 24 '20 at 22:08

1 Answers1

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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 $

$$\left| \sum_{k=1}^{n} \frac{x_k - x_{k-1}}{\sqrt{-x_{k-1}}} -2 \right| \leqslant \left| \sum_{k=1}^{n} \frac{x_k - x_{k-1}}{\sqrt{-x_{k-1}}} -(2-\sqrt{c}) \right| + \sqrt{c}\leqslant \sqrt{c} + \epsilon/2 < \epsilon$$

RRL
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