Intuition of improper integrals of type II

  • Thread starter Bipolarity
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  • #1
Bipolarity
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Let's say you have a function that is continuous in (a,b] but discontinuous at x=a and you integrate it from a to b.

For example, [itex] \int^{1}_{0} \frac{1}{\sqrt{x}}dx [/itex]

I understand that the integral exists, and it can be easily computed by using the limit as x approaches 0 from the positive side.

What bothers me is the intuition of the problem. If we divide the area up into infinite rectangles, it seems that some of the rectangles approaching the lower limit are of infinite length, so that their area must also be infinite? I can't fathom how you can add up these rectangles that approach lengths of infinity.

BiP
 

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  • #2
Stephen Tashi
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it seems that some of the rectangles approaching the lower limit are of infinite length

A clearer view is that some of the rectangles approaching the lower limit "have heights that are approaching infinite length", not that they "are" of infinite length.

What's your intuition about [itex] \lim_{x \rightarrow a} f(x) g(x) [/itex] when [itex] \lim_{x \rightarrow a} f(x) = \infty [/itex] ? Do you think a [itex] g(x) [/itex] can get small "fast enough" to make the product [itex] f(x)g(x) [/itex] stay finite?
 
  • #3
Bipolarity
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A clearer view is that some of the rectangles approaching the lower limit "have heights that are approaching infinite length", not that they "are" of infinite length.

What's your intuition about [itex] \lim_{x \rightarrow a} f(x) g(x) [/itex] when [itex] \lim_{x \rightarrow a} f(x) = \infty [/itex] ? Do you think a [itex] g(x) [/itex] can get small "fast enough" to make the product [itex] f(x)g(x) [/itex] stay finite?

Why not?

BiP
 
  • #4
dumbQuestion
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Just a side note, but in the example you gave above, can we really say the function is discontinuous at x=0? In this case x=0 is simply not part of the domain. It's my understanding that we can only really think of continuity of a function in terms of regions it's defined upon. Perhaps this has something to do with the answer?
 
  • #5
Stephen Tashi
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Why not?

BiP

Why not what? If you think [itex] g(x) [/itex] can decrease quickly enough so that [itex] \lim_{x \rightarrow a} f(x) g(x) [/itex] is finite, then what problem do you have beleiveing that the bases of rectangles can decrease "fast enough" so that the areas of the rectangles remains finite even though the height approaches infinity?

As to the definition of continuity, the usual definition is "A real valued function [itex] f(x) [/itex] whose domain is a subset of the real numbers is continuous at [itex] x = a [/itex] if and only if [itex] \lim_{x \rightarrow a} f(x) = f(a) [/itex]". Hence the definition requires that [itex] f(a) [/itex] exists. Hence if [itex] f(x) [/itex] does not exist at [itex] x = a [/itex] then [itex] f(x) [/itex] is not continuous at [itex] x = a [/itex].
 

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