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By the denition of Riemann integral, we have $$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}f(a + \frac{b-a}{n}k)\frac{b-a}{n} = \int_{a}^{b}f(x)dx$$

In a particular problem, for $k<n$, $f$ is defined, and for $k=n$, $f$ is not defined, so my question is that

In this situation, is the following relationship valid? $$\lim_{n\rightarrow \infty}\sum_{k=1}^{n-1}f(a + \frac{b-a}{n}k)\frac{b-a}{n} = \int_{a}^{b}f(x)dx$$

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  • $\begingroup$ $\sum_{k=1}^{n}f(a + \frac{b-a}{n}k)\frac{b-a}{n} =\sum_{k=1}^{n-1}f(a + \frac{b-a}{n}k)\frac{b-a}{n}+f(b)\frac{b-a}{n} $. $\endgroup$
    – Riemann
    May 13, 2018 at 7:08
  • $\begingroup$ @Riemann $f(b)$ is not defined. Indeed, in my problem, $b=1$ and $f(x)=\frac{1}{1-x}$, thus $f(b) = \frac{1}{1-b} = \frac{1}{1-1} = \frac{1}{0}$. $\endgroup$ May 13, 2018 at 7:23

1 Answer 1

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If $f$ is unbounded on $[0,1]$ then it is not Riemann integrable, as shown here, and we can't be sure that Riemann sums converge to anything.

If the improper integral is convergent,

$$\lim_{c \to 1}\int_0^c f(x) \, dx = \underbrace{\int_0^1 f(x) \, dx }_{\text{improper integral}},$$

then it is possible for the Riemann sums to converge if the intermediate points (tags) are carefully selected. This would be the case for $f(x) = 1/\sqrt{1-x},$ where

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\sqrt{1 - (k-1)/n}} = \lim_{c \to 1}\int_0^c \frac{dx}{\sqrt{1-x}} = \lim_{c \to 1}(2 - 2\sqrt{1-c}) = 2$$

Note that it is not entirely trivial to prove that the limit of the sum on the LHS is $2$. It is proved here for the equivalent improper integral $\displaystyle\int_0^1 \frac{dx}{\sqrt{x}}$.

However, in your example $f(x) = 1/(1-x)$, not even the improper integral is convergent, since

$$\lim_{c \to 1}\int_0^c \frac{dx}{1-x} = \lim_{c \to 1}[\log(1-0)-\log(1-c)] = +\infty$$

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