Integral equation

[One-D]

what does an equation need so that the equation can be integrated?
not all of equation can be integarted right?
I was quite confuse if it can be integrated or not whether I found a difficult integration. Thanx

[cronxeh]

The function must be continuous on the interval you are integrating (x1 to x2) or continuous everywhere if you want an indefinite integral. The series must also converge

[mathwonk]

the answer is not correct, but the question is also not precise. what do you really want to know?

[dextercioby]

From the top of my head,the theory of intergral equations has been established about 100 yrs ago and we have enough tools to solve them.Give us an example & we'll see what to do.

Daniel.

[mruncleramos]

I think I know what he means, and I don't think he's talking about Integral Equations. One word: Differential Galois Theory :smile:

[Data]

That's not one word!

[One-D]

The function must be continuous on the interval you are integrating (x1 to x2) or continuous everywhere if you want an indefinite integral. The series must also converge
I thought the same, it's what I assumed (I heard from my teacher) before but I'm not quite sure. By the way, what does "The series must also converge" mean? and I don't know about Differential Galois Theory. Thanx for your reply :smile:

[Zurtex]

Any function f(x) that maps all x to a single f(x) can be integrated. Defining the integral of f(x) other than simple the integral of f(x) and doing anything with it can provide very tricky in general though.

[Data]

That's not true.

f(x) = \biggl\{ \begin{array}{ccc}0 & & x \ \mbox{rational} \\ 1 & & x \ \mbox{irrational}\end{array}

is not integrable.

[HallsofIvy]

Any function f(x) that maps all x to a single f(x) can be integrated.

I have no idea what it means for f(x) to map "all x to a single f(x)" means!

Certainly, every continuous function is integrable but that is not necessary.

If I remember correctly "every bounded function whose set of discontinuities is a set of measure 0" is Riemann integrable. Oddly enough, you need Lebesque measure theory to define "set of measure 0"!

[Zurtex]

I have no idea what it means for f(x) to map "all x to a single f(x)" means! To be honest, neither do I it was 4:30AM when I wrote it :rolleyes: Perhaps I should stop relying on theorems I come up with after 3:30AM and a bit of too much intoxicating drink :wink:

That's not true.

f(x) = \biggl\{ \begin{array}{ccc}0 & & x \ \mbox{rational} \\ 1 & & x \ \mbox{irrational}\end{array}

is not integrable.
Why can't that be integrated?

[Data]

Take a look at its upper and lower sums.

[Zurtex]

Take a look at its upper and lower sums.
If you're talking to me that means nothing to me.

[Data]

Well, if we're talking about Riemann integrability, then it's not integrable because, letting P = \{ I_1, I_2, ..., I_n\} be a partitioning of the interval I = [a, b], m_i = \min_{x \in I_i} f(x), M_i = \max_{x \in I_i} f(x) (Edit: really I should have used greatest lower bound and least upper bound here but it doesn't make a difference for this function), s to be the mesh fineness of P, and L_i to be the length of the interval I_i, we find

\lim_{s \rightarrow 0} \sum_{i=1}^n M_iL_i \neq \lim_{s \rightarrow 0} \sum_{i=1}^n m_iL_i

Edit 2: I should probably also make the important note that f(x) is Lebesgue-integrable.

[One-D]

thanx for your replies

[cronxeh]

One-D it would help to know for which class you want to know this.

Obviously if its for Real Analysis - my answer is not complete - and its more of a Galois theory you looking for

If it's for Calculus 2 then thats the correct answer - a function must be continuous in order to be integrated

If its for Multivariable Calculus then there are limits and continuity that you must check, for all variables

If its for Discrete Math its something along the lines of what Data said

[One-D]

now I m learning calculus 2, I just want to know, cause I got that questions on my mind. Do you know where I can find web so I can learn about calculus more, especially cal 2. thanx Cronxeh