a easy explanation to comparison and limit comparison for improper integrals
In: Mathematics
It’d help to be a little more specific for math questions because math is taught differently around the world. I’ll use one version, though the idea is the same for other kinds of improper integrals. LaTeX note: `infty` = infinity, `int_a^b` = integral from a to b, `lim` = limit, `ge` = greater than (>=), etc. I presume you are familiar with notation for intervals using [], (), and with basic limit definitions (if not, comment).
(Direct) Comparison Test: Let continuous `f(x) ge g(x) ge 0` for `x in [a, infty)`, then:
* If `F = int_a^infty f(x) dx` converges, then `G = int_a^infty g(x) dx ge 0` converges with `F ge G`.
* If `G` diverges, then `F` diverges.
This one is a direct comparison of the area of the graph (you can find pictures online), and the conditions mean that the functions don’t change sign. Then at each point `x`, since `f(x)` is at least `g(x)` (and both are non-negative), the area corresponding to the improper integral `F` must be at least that of `G`, which implies the convergence/divergence results.
Limit Comparison Test: Let continuous `f(x) ge 0, g(x) > 0` for `x in [a, infty)` and integrable on the same interval (keeping the notation of `F, G` as in above) and suppose that
`lim_{x to infty} f(x) / g(x) = L` exists. Then:
* `0 < L < infty`: then `F` converges if and only if `G` converges.
* `L = 0`: If `G` converges, then `F` converges (contrapositive equivalent: if `F` diverges, then `G` diverges).
* `L = infty`: If `G` diverges, then `F` diverges (contrapositive equivalent: if `F` converges, then `G` converges).
The first case is the main part of the result. The idea is to use the formal definition of a limit to create a new function that is a multiple of `g(x)` that we can use to bound `f(x)` by so that we can use the direct comparison test. By definition of limit, for every `epsilon > 0`, there exists some (sufficiently large but finite) `x_0` such that `| f(x)/g(x) – L| < epsilon` whenever `x ge x_0`. This gives `(L – epsilon) g(x) < f(x) < (L + epsilon) g(x)` for `x ge x_0`, for which we can apply the (direct) comparison test on the integral starting at `x_0` by taking `epsilon < L`. The convergence/divergence will not be affected by adding the integral between `a` and `x_0` because that interval is finite and the assumption of continuity.
The 0 and infinity cases are similar, except that for each, there is only one inequality and so we only get one direction of implication. It’s not hard to find counterexamples to the reverse directions. For instance, `f(x) = 1/x^2` and `g(x) = 1/x` gives `L = 0` but `F` converges while `G` diverges;
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