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  1. #1 Integration 
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    Im having problems with integration. Im not sure how to integrate the following statement: (4x) / (2x^2+1) Does anyone know how to do this?


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  3. #2 Re: Integration 
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    Quote Originally Posted by chris083210
    Im having problems with integration. Im not sure how to integrate the following statement: (4x) / (2x^2+1) Does anyone know how to do this?
    well looking at the statement the thought of integrating using the substitution method comes screaming at me.

    however. if you can integrate 1/u du then be sure to let me know, because i sure as hell can't.
    that and your function is not continuous for all x. i think thats another reason why you can't obtain an indefinate integral, however i'm not overwhelmingly sure of this.


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    Ok, hold on to your hats because here comes an integration lecture

    Wallaby, your function does not need to be continuous for you to integrate it. For example the function f(x) = 1 if x is rational and 0 otherwise is integrable (i.e. its integral exists) with ∫ f dx = 0. Now as "easy" functions go, that one is quite discontinuous. This leads you into the murky waters of integration theory and eventually measure theory.

    Secondly, ∫ 1/u du = log(u) + C where log here the the logarithm to the natural base (engineers may prefer ln)

    Finally, the integral of (4x) / (2x<sup>2</sup>+1) is just log(2x<sup>2</sup> + 1) + C. If you feel daring, start with the substitution x = 1/Sqrt(2) tan(t)
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  5. #4  
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    Quote Originally Posted by river_rat
    Ok, hold on to your hats because here comes an integration lecture

    Wallaby, your function does not need to be continuous for you to integrate it. For example the function f(x) = 1 if x is rational and 0 otherwise is integrable (i.e. its integral exists) with ∫ f dx = 0. Now as "easy" functions go, that one is quite discontinuous. This leads you into the murky waters of integration theory and eventually measure theory.
    that i didn't know.

    Quote Originally Posted by river_rat
    Secondly, ∫ 1/u du = log(u) + C where log here the the logarithm to the natural base (engineers may prefer ln)
    that i should have known
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  6. #5  
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    One of the things that make the integral powerful is the fact that you can integrate functions which are not continuous and get continuous functions out at the end - which is really handy for studying differential equations for example. Integrals make functions nicer (i.e. more and more smooth) while derivatives do the exact opposite which makes the derivative a difficult operator to use in general.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    unfortunately at this stage integral calculus is just involves doing the questions they tell me to do, practising the techniques of integration basically.

    hard to appreciate it or know when it comes in handy, appart from the area under a graph buisness. (which is where i erroneously came to the conclusion that discontinuities were a bad thing for integration)
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  8. #7  
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    Well let me take this moment to tie in some of the stuff guitarist has been showing you with integration

    Do you remember what a linear operator is? Well the integral (along with the derivative) happen to be linear operators acting on the space of all smooth functions. Now that vector space is quite messy so lets take a simpler subspace, namely the vector space of all polynomials (i think guitarist mentioned that earlier as well). Now the question here is what does the derivative and integral operators look like in this example? (i.e. if you what does the matrix that represents these two operators look like?)

    The hint here is to consider how each operator acts on the basis elements - and start with small finite dimensional cases before attacking the bigger problem.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  9. #8  
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    Quote Originally Posted by river_rat
    Well let me take this moment to tie in some of the stuff guitarist has been showing you with integration

    Do you remember what a linear operator is? Well the integral (along with the derivative) happen to be linear operators acting on the space of all smooth functions. Now that vector space is quite messy so lets take a simpler subspace, namely the vector space of all polynomials (i think guitarist mentioned that earlier as well). Now the question here is what does the derivative and integral operators look like in this example? (i.e. if you what does the matrix that represents these two operators look like?)

    The hint here is to consider how each operator acts on the basis elements - and start with small finite dimensional cases before attacking the bigger problem.
    polynomials forming a vector space was mentioned, mainly to dispell the notion of pointed arrows. so far i'd have to say that the matrix by which we represent the integral or derrivative operator would have non-zero values along the principle diagonal, i've only considered square matrices at this point. IF that were true i don't think the operators would be orthogonal or unitary, thinking back on integration and differentiating i'm sure of this.

    but i think its a bit late at night/early in the morning to be thinking straight.
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  10. #9  
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    Quote Originally Posted by river_rat
    your function does not need to be continuous for you to integrate it. For example the function f(x) = 1 if x is rational and 0 otherwise is integrable (i.e. its integral exists) with ∫ f dx = 0.
    Um. I recognize this integral as being of the Lebesgue flavour. I never really understood this, let me paraphrase from the first crappy book here I found.

    Suppose us to be in some function space. Then the IP is defined as (f,g) ≡ ∫f*(x)g(x)dx, where the integral is definite. Fine.

    It follows that (f,f) = ∫|f|<sup>2</sup>dx ≥ 0. Yep, no probs

    We have a definite integral, let's say with bounds in [a,b], But, says this book, (f,f) = 0 need not imply that f(x) = 0 for all x in [a,b]. So far so good. "the function f(x) can be non-zero at any finite number of points, and the integral will not notice this". I take this to mean that, even though the integrand is not zero for all x in [a,b], this makes no significant contribution to the integral.

    We then get into a discussion of the Lebesgue integral, f(x) = 0 "almost everywhere", which I sort of understand, but have (so far) failed to develop an intuition for.

    Can you spare a dime, buddy?

    P.S. By later EDIT: By your set-up, this has something to do with the irrationals being dense in R, right?
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  11. #10  
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    Its easier to handle null sets if you keep the discussion to the real line as i can give a little neat characterization of those sets there that may help.

    A set A is termed null (or has measure zero, assuming you have a complete measure) if given any number epsilon > 0 you can find a countable covering of that set consisting of open intervals I<sub>n</sub> = (a<sub>n</sub>, b<sub>n</sub>) such that the sum ∑ (b<sub>n</sub> - a</sub>n</sub>) is less then epsilon.

    So a null set has too much space between its elements for it to have a nonzero length. Now its easy to see that any singleton is a null set, and as a measure is countably additive that every countable set is also null (i.e. that the rationals are a null set). Now the reason the Dirichlet function is integrable is because the rationals are countable and the Lebesgue integral cannot "see" what happens on a countable set. I should stress though that not every null set is countable, the cantor set is also null for example.

    The denseness of the rationals or irrationals is not the important part here, its more the fact that the irrationals are very tightly packed together while the rationals are not.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  12. #11 The answer 
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    The integral of (4x)/(2x^2+1) is log(2x^2+1) base e + Constant.

    If the derivate of the denominator of a fraction is in the numerator, as is the case with this function, the integral is the log of the denominator.

    If the derivative is not in the numerator the denominator has to be a linear function for you to integrate easily.

    ...hopefully
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  13. #12  
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    Hey river_rat! Thanks for that. It wasn't quite what I was after, but it's always good to get a different perspective on things.

    I was after developing an intuition about the Lebesgue integral similar to that we all have about Riemann's. See if you can agree with this.

    Let X be a set with measure, with x, y in X. Let S<sub>i</sub> denote the collection of subsets in a partition of X. For some function f on X, let f have a step discontinuity at y in X, but otherwise be smooth. I mentally draw the graph (as one does).

    I now consider the interval f(x) - δf(x) ≤ f(x) ≤ f(x) + δf(x) on the f(x) axis. Then for all parts of my graph, I will have, by "projecting" this interval onto it's pre-image, some interval (set) S<sub>x</sub> with some measure. I will call this the set of measure m(S<sub>x</sub>).

    Clearly, for the smooth parts of my graph, all m(S<sub>i</sub>) ≠ 0. I now consider the discontinuity at y on X. Evidently, here,
    m(S<sub>y</sub>) = 0, and I will say this is a set with "measure zero".

    Now the Lebesgue integral is defined to be the sum of all the f(m(S<sub>i</sub>)), and clearly the subsets where m(S<sub>j</sub>) = 0 make no contribution to this sum. So that, for any union of subsets S<sub>i</sub> of X where the Lebesgue integral exists, one may say that the integral is defined almost everywhere on X

    One may also say that, whereas the Riemann sum assumes a definite value f(x) for all x in X, that is f(x<sub>i</sub>) ≠ f(x<sub>j</sub>) ==> x<sub>i</sub> ≠ x<sub>j</sub> (not conversely, obviously), Lebesgue merely asserts that for all x in m(S<sub>x</sub>), all f(x) in [f(x) - δf(x), f(x) + δf(x)] take on the same value, and that for the Lebesgue sum to be an integral, we may make the [f(x) - δf(x),f(x) + δf(x)] as small was we like.

    I'm Ok with this, I think. But then you handed me the Dirichlet function, and tell me that it is only by virtue of their countability that the rationals in [0,1] make no contribution to the Lebesgue integral. This is the bit I don't get, and why I suggested it has to do with the denseness of the irrationals in R (by which I simply meant that, for any pair of rational numbers, however close, there will always be uncountably many irrationals between them).

    Ugh. I hate analysis.
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  14. #13  
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    I'm Ok with this, I think. But then you handed me the Dirichlet function, and tell me that it is only by virtue of their countability that the rationals in [0,1] make no contribution to the Lebesgue integral. This is the bit I don't get, and why I suggested it has to do with the denseness of the irrationals in R (by which I simply meant that, for any pair of rational numbers, however close, there will always be uncountably many irrationals between them).
    Another way of handling this function is as the limit of simple functions (which we can do as the rationals are countable, the countability of the rationals is central to this specific function). So consider {a<sub>n</sub>} as an enumeration of the rationals and the sequence of functions {f<sub>n</sub>} with f<sub>n+1</sub>(x) = f<sub>n</sub>(x) + Χ<sub>{a<sub>n</sub>}</sub>(x) where Χ<sub>{a<sub>n</sub>}</sub> is the characteristic function for the singleton {a<sub>n</sub>}. Now our Dirichlet function is just the limits of these (Riemann) integrable functions and ∫f<sub>n</sub> = 0 ∀ n. Now the Lebesgue integral is the limit of these integrals in this case and thus ∫f = 0

    My problem with trying to explain this using denseness is that both sets here are dense (in the normal sense) in (0, 1) and you still end up using a counting argument to distinguish between them.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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