Comparison Test for Improper Integrals
Here now that we've seen how to actually calculate improper integrals we should to address one more topic about them. Frequently we aren't concerned along with the actual value of these integrals. In place of it we might just only be interested in if the integral is convergent or divergent. As well, there will be some integrals which we simply won't be capable to integrate and yet we would still such as to know if they converge or diverge.
To deal along with this we have got a test for convergence or divergence which we can use to assist us answer the question of convergence for a not proper integral.
We will provide this test only for a sub-case of the infinite interval integral, though versions of the test exist for the other sub-cases of the infinite interval integrals also integrals with discontinuous integrands.
If f (x) ≥ g (x) > 0 on the interval [a, ∞] then,
1. If ∫∞a f(x) converges then so does ∫∞a g(x) dx.
2. If ∫∞a g(x) dx diverges then so does ∫∞a f (x) dx.
Note: If you think in terms of area the Comparison Test makes a lot of sense. Determine if f (x) is larger than g(x) then the area within f (x) must as well be larger than the area under g(x). Thus, if the area within the larger function is finite after that the area under the smaller function has to be finite. Similarly, if the area under the smaller function is infinite after that the area within the larger function must as well be infinite. Be cautious not to misuse this test. If the smaller function converges there is no basis to believe that the larger will as well converge (after all infinity is larger as compared to a finite number...) and determine if the larger function diverges there is no reason to believe that the smaller function will also diverge.