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Please, compare and contrast different spaces, it depends on the space! I cannot formulate this mathematically but I think that $\mathbb R$ extended with infinities behave totally differently.

$\int_{0}^{\infty}\frac{1}{kx^{2}+1} dx=\int_{0}^{\infty} dx -\int_{0}^{\infty}\frac{kx^{2}}{kx^{2}+1}$ where the term $\int_{0}^{\infty} dx$ diverges.


Let's consider $k=1$ then $\int_{0}^{\infty}\frac{1}{x^{2}+1} dx= Arctan(x)_{0}^{\infty} =\frac{\pi}{2}$ so I must have some mistake in

$$\int_{0}^{\infty}\frac{1}{kx^{2}+1} dx=\int_{0}^{\infty}\frac{(kx^{2}+1)-kx^{2}}{kx^{2}+1} dx=\int_{0}^{\infty} dx -\int_{0}^{\infty}\frac{kx^{2}}{kx^{2}+1}$$

but I cannot see it, where am I doing a mistake?


Let' s take simpler example:

$$\int_{0}^{\infty} \frac{1+x+x^{2}+x^{3}}{x^{k}} dx= \int_{0}^{\infty} x^{-k} dx+\int_{0}^{\infty} x^{-k+1} dx+ \int_{0}^{\infty} x^{-k+2} dx+ \int_{0}^{\infty} x^{-k+3}dx$$

So this is also wrong? I cannot break it up like this? What does it mean to be $indefinite$? If I have extended real space $\mathbb R\cup\{\infty\}\cup\{-\infty\}$, what about now? Now look $\infty$ is just a normal number now, it is not indefinite so does it converge?

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$\int_0^\infty 0dx=\int_0^\infty (x-x)dx=\int_0^\infty xdx-\int_0^\infty xdx$ – Evgeny Savinov Feb 11 '12 at 23:54
up vote 1 down vote accepted

Suppose $A=B+C$. If the integrals of $A,B,C$ all exist, then $\int A=\int B+\int C$. The trouble is, it is possible that one of the integrals exists and the other two don't. For example, $${1\over3(x-3)}={1\over3x}+{1\over x(x-3)}$$ is an algebraic identity, but if you write $$\int_{-1}^1{dx\over3(x-3)}=\int_{-1}^1{dx\over3x}+\int_{-1}^1{dx\over x(x-3)}$$ you're in trouble because the two integrals on the right don't exist. (Actually, in this case the difficulty already arises before you even integrate; the left side of the identity makes sense for $x=0$, while the right side doesn't.)

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I like this answer because it is a generic and simple answer -- if I have understood now this (after consultation), I need to be particularly careful with any singular point (where zero is a divisor or where the border is mock-infinite while it actually nears some big M) -- so besides the zero -divisors (not sure whether I can scale this to algebraic zero-divisors $ab=0$ where $a\not=0 \vee b\not=0$ -- probably not, do not be misguided by this term now then here). Thanks. – hhh Feb 15 '12 at 19:08

$\infty-\infty$ is undefined (the right hand side of your equation).

If $\int_0^\infty f(x)\, dx $ converges and if $\int_0^\infty g(x)\, dx $ converges, then you can write $\int_0^\infty[ f(x)+g(x)]\, dx=\int_0^\infty f(x)\, dx +\int_0^\infty g(x)\, dx $.

But you can't split a convergent improper integral into two pieces as you did unless both (or just one) pieces converge, for essentially the reason I gave at the outset. (Note that both the integrals on the right hand side of your equation are infinite. But you can't do anything with them, the right hand side is not defined, and the step you made going from the left hand side to the right hand side is not justified.)

That $\infty-\infty$ is undefined should become clear if you consider the limits of the expressions: $(n)-(n+1)$, or $(n+17)-(n-42)$, or $(n^2)-(n)$.

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...can you somehow break up the point or elaborate on it? (I know that I can solve this with basic u -substitution but I cannot understand this undefined thing yet, thinking...) – hhh Feb 11 '12 at 23:39
Look at my updated question -- I cannot understand this at all. Please, compare to extended real space. I think it depends on your space. I am making now a case that I can do what I am doing in an extented space but I cannot still see why I cannot do it in non-extended space...there must be some theory about this...not just $indefinite$ -word. – hhh Feb 12 '12 at 0:03
To expand on David's point: you know $\lim_{n\to\infty}4=4$ and you know $4=(n+4)-n$ but if you try to evaluate $\lim_{n\to\infty}4$ by writing $\lim_{n\to\infty}4=\lim_{n\to\infty}(n+4)-\lim_{n\to\infty}n$ then you've traded in something that converges for two things that don't. That's essentially what you've done with your integrals. – Gerry Myerson Feb 12 '12 at 0:05
Now look $\infty$ is just a normal number now... To expand on @David's answer on this specific point: $\infty$ is not a normal number, whatever normal means. It is true that some arithmetic properties of the real line can be extended, for example, one can safely use $\infty+3=\infty-11=\infty+\infty=\infty$ and $4-\infty=-7-\infty=-\infty$. But other properties cannot be extended, in any way whatsoever. For example, $\infty-\infty$ has no meaning, as the basic counterexamples provided by others show. And the example in your title is based on this case. – Did Feb 12 '12 at 11:21
By the way, your second example is kosher because all the functions involved are nonnegative, and addition (but not subtraction) is well defined on $[0,+\infty)\cup\{+\infty\}$. This means that all the parts on the RHS are finite iff the LHS is, and then the (finite) sum of the parts on the RHS equals the LHS. The other case is when the LHS is infinite, then at least one part on the RHS is infinite as well. Note finally that, at a more advanced level, one can manipulate finite negative values as well, as long as $\infty-\infty$ does not enter the picture. – Did Feb 12 '12 at 11:30

You're missing the fact that for $k >0$

$$\int\limits_0^\infty {\frac{{dx}}{{k{x^2} + 1}} < } \int\limits_0^\infty {\frac{{dx}}{{{x^2} + 1}} = \frac{\pi }{2}} $$

so the integral converges.

And that putting

$$\sqrt k x = u$$

$$\int\limits_0^\infty {\frac{{dx}}{{k{x^2} + 1}}} = \frac{1}{{\sqrt k }}\int\limits_0^\infty {\frac{{du}}{{{u^2} + 1}}} = \frac{1}{{\sqrt k }}\frac{\pi }{2}$$

There is not "mistake" so to call it, but rather that you're not thinking about

$$\int\limits_0^\infty {f\left( x \right)dx} = \mathop {\lim }\limits_{u \to \infty } \int\limits_0^u {f\left( x \right)dx} $$

and that when the function is not continuous at $x=c\in(a,b)$ then the integral should be interpreted as:

$$\int\limits_a^b {f\left( x \right)dx} = \mathop {\lim }\limits_{m \to c} \int\limits_a^m {f\left( x \right)dx} + \int\limits_m^d {f\left( x \right)dx} $$

In the case of the polynomials you have a discontinuity at $x=0$ so when you take limits you have

$$\int\limits_0^\infty {{x^{1 - k}}dx} + \int\limits_0^\infty {{x^{2 - k}}dx} + \int\limits_0^\infty {{x^{3 - k}}dx} = \left( {\mathop {\lim }\limits_{u \to \infty } \frac{{{u^{2 - k}}}}{{2 - k}} + \frac{{{u^{3 - k}}}}{{3 - k}} + \frac{{{u^{4 - k}}}}{{4 - k}}} \right) - \left( {\mathop {\lim }\limits_{u \to 0} \frac{{{u^{2 - k}}}}{{2 - k}} + \frac{{{u^{3 - k}}}}{{3 - k}} + \frac{{{u^{4 - k}}}}{{4 - k}}} \right)$$

And you can radily check that the limit at infinity is zero for all three expressions but for zero things go to infinity. If $k<0$ then the problem would be at infinity.

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