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An elementary question on Riemann - Integration:

Under what conditions on $f$ is the following true:

$$\lim_{b\to a} \int_a^b f(x)dx=\int_a^af(x)dx=0$$

If $f$ is bounded in $[a,b]$, then this is simple to prove.

But what is the most general condition on $f$ for which this holds? Answers relating this to other types of integral (like Lebesgue, ...) are also welcome.

Taking the example of $ f(x) = {1\over x-a }$.

$ lim_{b\to a} lim_{t\to0} \int_{a+t}^b f(x) dx \to \infty$

but

$ lim_{t\to0}lim_{b\to a+t} \int_{a+t}^b f(x) dx \to 0 $

Enlighten me please.I'm pretty confused. The following questions are also related to this. It would be really helpful if somebody can resolve this:

Is there any notable difference between studying the Riemann integral over open intervals and studying it over closed intervals?

Why is the Riemann integral only defined on compact sets?

Is there any notable difference between studying the Riemann integral over open intervals and studying it over closed intervals?

Community
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Srinivas K
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    If the Riemann integrals in question are defined at all, ... – Hagen von Eitzen Jun 30 '14 at 10:00
  • I'm looking for a function like $ 1 \over x-a $ – Srinivas K Jun 30 '14 at 10:02
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    Such integrals are only defined as limits themselves, and then it can depend on which order you take the limits in, so there's no answer which is satisfying. – Adam Hughes Jun 30 '14 at 10:04
  • @Hagen von Eitzen for the function $1 \over x-a $ the function is not defined at $a$ but if we see intuitively the limit diverges to infinity and at the same time the second inequality always holds.That's why I'm confused. – Srinivas K Jun 30 '14 at 10:13
  • @AdamHughes Can you explain this for some examples like the one i've mentioned $1 \over x-a $ ? – Srinivas K Jun 30 '14 at 10:22
  • Integrating (in the sense of Riemann) ${1\over x-a}$ requires you to treat

    $$\int_a^b{dx\over x-a}=\lim_{t\to a}\int_t^b{dx\over x-a}$$

    Which gives you $\lim_{t\to a^+}\log\left|{b-t\over t-a}\right|$ which is divergent, so you'd be taking $\lim_{b\to a}(\infty)$ which is nonsense.

     – Adam Hughes Jun 30 '14 at 20:29

1 Answers1

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  1. using Riemann definition: the function $x\to \frac 1{x-a}$ is not defined on $a$. How is defined the integral in such case?
  2. using Lebesgue definition: if $f$ is integrable, then $$ \lim_{b\to a} \int_a^b f = \lim_{b\to a} \int f1_{a\le x\le b} $$and using the dominated convergence theorem, the limit is 0.
  3. if $f$ is not integrable and $\ge 0$, the terms of the sequence are all $\infty$, so will the limit.
mookid
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  • Thank you for (2) and (3) . For (1) That's precisely the question i'm asking.We know that the function is not defined at $a$. So,is there a way to define $\int_a^b f(x) dx $ as an improper integral and then try to take the limit. – Srinivas K Jun 30 '14 at 11:48
  • If you define the integral as the improper limit, anything can happen, basically. That's the whole problem with these things. – mookid Jun 30 '14 at 11:50