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Using a trivial example to illustrate the question -

$$\int_0^\infty 2 dx$$

$=2x \mid_0^\infty = 2(\infty) - 2(0)$

Can we actually say $2(\infty)$? It doesn't seem valid me.

csss
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  • what you can say is the integral is unbounded. i.e. $\lim_{x\to \infty}\int^{x}_{0}2dx\rightarrow\infty$ If we had $\int \mathrm{e}^{-x}dx$ then this converges for example. – Chinny84 Mar 03 '14 at 10:02
  • This is just a trivial example...in more complex situations I would have to go to the step of saying, eg. $x^3e^-(\frac{4x^5}{2})^2\mid_0^\infty$...and then I would like to show the "evaulation" step where we sub in $\infty$ and $0$ so it is clear what is happening...but I just have a problem with subbing in $\infty$... – csss Mar 03 '14 at 10:17

2 Answers2

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An integral, one (or both) of limits of which is infinite, is an improper integral. For such type of integral you can't take Newton-Leibniz formula as is, you'll have to use the definition of improper integral:

$$I=\int_0^\infty 2dx\equiv\lim_{t\to\infty}\int_0^t2dx.$$

Now $\int_0^t2dx=2t$, and, following the definition,

$$I=\lim_{t\to\infty} 2t=\infty,$$

so the integral diverges.

Ruslan
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  • Although we do still have the appealing fact that if $f(x)$ is continuous on $[0, +\infty]$ and differentiable on $(0, +\infty)$, then $\int_0^{+\infty} f'(x) , dx = f(+\infty) - f(0)$, even if it is improper at $0$ too. (including the integral being undefined should the right hand side be undefined) –  Mar 03 '14 at 17:26
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The expression $2 \cdot (+\infty) - 2 \cdot 0$ does indeed make sense. It is rather useful to do arithmetic/analysis with $\pm \infty$ alongside the other real numbers; e.g $\lim_{x \to +\infty}$ has all of the same properties that $\lim_{x \to a}$ does, and $\lim_{x \to a} f(x) = +\infty$ behaves much more like a convergent limit than other sorts of divergence (e.g. that of $\lim_{x \to 0} \sin(1/x)$), and $[-\infty, +\infty]$ is a compact interval.

The name of the structure that includes $\pm \infty$ at the endpoints of the real line is the "extended real numbers".