The Unapologetic Mathematician

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”.

April 22, 2008 - Posted by | Analysis, Calculus