1. Sep 14, 2012

### Bipolarity

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 $f(a)$ is not defined, does $\int^{b}_{a}f(x)dx$ necessarily not exist?
3) If $f(a)$ is not defined, but but $\lim_{x→a+}f(x)$ is, does $\int^{b}_{a}f(x)dx$ necessarily not exist?
5) If $f(a)$ is defined, but $\lim_{x→a+}f(x)$ does not, does $\int^{b}_{a}f(x)dx$ 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

2. Sep 14, 2012

### Simon Bridge

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.

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. Sep 14, 2012

### AlephZero

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. Sep 14, 2012

### Bipolarity

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

Last edited: Sep 14, 2012
5. Sep 14, 2012

### Simon Bridge

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. Sep 14, 2012

### Simon Bridge

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.

Last edited: Sep 14, 2012