Questions about definite integrals

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While reading my calc book, I have developed a few questions about the situations in which definite integrals can exist. I've thought about these questions, and I feel that if I am able to answer some of them, I can make some other problems much easier, such as testing for convergence of a definite integral.

1) If a definite integral diverges, can we conclude that it does not exist?
2) If a definite integral does not exist, can we conclude that it diverges?
3) If [itex] f(a) [/itex] is not defined, does [itex]\int^{b}_{a}f(x)dx [/itex] necessarily not exist?
3) If [itex] f(a) [/itex] is not defined, but but [itex]\lim_{x→a+}f(x) [/itex] is, does [itex]\int^{b}_{a}f(x)dx [/itex] necessarily not exist?
5) If [itex] f(a) [/itex] is defined, but [itex]\lim_{x→a+}f(x) [/itex] does not, does [itex]\int^{b}_{a}f(x)dx [/itex] necessarily not exist?

I assume all these questions can be answered using the definition of the definite integral, along with the definition of continuity, but how exactly?

Thanks!

Note: Assume that b>a for the above integrals

BiP
 

Answers and Replies

  • #2
Simon Bridge
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An integral is just a fancy way of doing a sum.
So.

1. yes and no - depends what you mean by "exist".
2. no.
Can you think of where diverging integrals may come from - say - in Nature?

3. no.
f(x) only needs only to exist in (a,b) not [a,b] to be integrable.

4. Also answered by 3.

5. Can you come up with an example to illustrate this? I think you are describing a situation where f is not a function or is not peicewise continuous within (a,b).

Generally, in order to be integrable, f has to be a piecewise-continuous function in the range of the integration. Look these up.
 
  • #3
AlephZero
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3. no.
f(x) only needs only to exist in (a,b) not [a,b] to be integrable.

Generally, in order to be integrable, f has to be a piecewise-continuous function in the range of the integration. Look these up.

Those statements depend on how you define "integration", and you can define it so nether of them is required. But I guess the OP is thinking about the http://en.wikipedia.org/wiki/Riemann_integral (even if the calc textbook doesn't mention the name).
 
  • #4
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An integral is just a fancy way of doing a sum.
So.

1. yes and no - depends what you mean by "exist".
2. no.
Can you think of where diverging integrals may come from - say - in Nature?

3. no.
f(x) only needs only to exist in (a,b) not [a,b] to be integrable.

4. Also answered by 3.

5. Can you come up with an example to illustrate this? I think you are describing a situation where f is not a function or is not peicewise continuous within (a,b).

Generally, in order to be integrable, f has to be a piecewise-continuous function in the range of the integration. Look these up.

I see, thank you for all your help. If time permits, could you please offer an example of functions for the questions 3, 4, and 5?

Also, I have heard of something called Lipschitz continuity, does that have any bearing on this problem?

Also, for a Riemann integral, does the function have to be continuous in (a,b) or in [a,b] ? wikipedia proves the FTC assuming continuity in [a,b] but I assume that is superfluous condition and that continuity in (a,b) suffices to determine the existence of a riemann integral?
Thanks!

BiP
 
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  • #5
Simon Bridge
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The simplest way to create an undefined value is to divide by zero.

If I define a function u(x): u=0 for x≤0 and u=1 for x>0 then f(x)=1/u(x-a) would be undefined for x≤a ... but I can still do the integral from a to b if b>a. (It's b-a).

I'm pretty sure you can do the rest.
 
  • #6
Simon Bridge
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AlephZero said:
Those statements depend on how you define "integration", and you can define it so nether of them is required. But I guess the OP is thinking about the Riemann Integral.
Yah. I kinda figured a definition something like:
$$\int_a^bf(x)dx = \lim_{N\rightarrow \infty} \frac{b-a}{N} \sum_{n=0}^N f(\frac{b-a}{N}n + a)$$... but if I just bring the sum in each end by one unit ... in the limit N → ∞ it's the same.

$$\int_a^bf(x)dx = \lim_{N\rightarrow \infty} \frac{b-a}{N} \sum_{n=1}^{N-1} f(\frac{b-a}{N}n + a)$$

... though there is danger in that kind of argument of shaving a bit of each end until there's nothing left.

I can also take the sum to within a small distance ##\epsilon << (b-a)/N## of the limits
... so I do the sum in range ##[a+\epsilon, b-\epsilon]##, instead of ##[a,b]##, and see what happens in the limit ##\epsilon \rightarrow 0##.
$$\int_a^bf(x)dx = \lim_{N\rightarrow \infty}_{\epsilon \rightarrow 0}\frac{b-a-2\epsilon}{N} \sum_{n=0}^N f(\frac{b-a-2\epsilon}{N}n + a + \epsilon)$$
I'm sure AlephZero can refine this argument.

Few beginning courses are even this explicit anyway.
 
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