Absolute Convergence

Let’s apply one of the tests from last time. Let $\alpha$ be a nondecreasing integrator on the ray $\left[a,\infty\right)$ , and $f$ be any function integrable with respect to $\alpha$ through the whole ray. Then if the improper integral $\int_a^\infty|f|d\alpha$ converges, then so does $\int_a^\infty f\,d\alpha$ .

To see this, notice that $-|f(x)|\leq f(x)\leq|f(x)|$ , and so $0\leq|f(x)|+f(x)\leq2|f(x)|$ . Then since $\int_a^\infty2|f|\,d\alpha$ converges we see that $\int_a^\infty|f|+f\,d\alpha$ converges. Subtracting off the integral of $|f|$ we get our result. (Technically to do this, we need to extend the linearity properties of Riemann-Stieltjes integrals to improper integrals, but this is straightforward).

When the integral of $|f|$ converges like this, we say that the integral of $f$ is “absolutely convergent”. The above theorem shows us that absolute convergence implies convergence, but it doesn’t necessarily hold the other way around. If the integral of $f$ converges, but that of $|f|$ doesn’t, we say that the former is “conditionally convergent”.

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April 22, 2008 - Posted by John Armstrong | Analysis, Calculus

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[…] integral, we have analogues of the direct comparison and limit comparison tests, and of the idea of absolute convergence. Each of these is exactly the same as before, replacing limits as approaches with limits as […]

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