Besides the p-test, there are a few basic principles that go into the rest of this section.
If we have a convergent integral and we make its interval of integration smaller, the new integral will also converge.
Oddly enough, we can also make the interval of integration larger, as long as we don't include any points where the function is badly behaved. If
converges and f isn't badly behaved on [a, b], then
also converges.
This is because we changed the original improper integral by sticking on a finite area, and a definite integral
that isn't improper is always finite.
On the other hand, if
diverges then
will diverge also.
The moral of the story is that when we're looking at integrals of the form
,
it doesn't really matter what the lower limit of integration is so long as f (x) is well behaved on the interval of integration.
The integral
will converge if and only if
converges, if and only if
converges.
Sample Problem
We know that
converges. This means
also converges, because we've only added on a finite amount of area.
Sample Problem
We know that
converges. This means
also converges. The integral
is finite, so we've only increased the area by a finite amount.
If
0 ≤ f (x) < g(x)
for all x in [a, b] at which both functions are defined, and
converges, then
must converge as well. The area under g on [a, b] includes the area under f on [a, b], so if the area under g is finite the area under f must be finite as well. Similarly, if 0 ≤ f (x) < g(x) for all x in [a, b] at which both functions are defined, and
diverges, then
must diverge as well. The area under g on [a, b] includes the area under f on [a, b], so if the area under f is infinite the area under g must be infinite as well.
If 0 ≤ f (x) < g(x) for all x in [a, b] at which both functions are defined, and
converges, that doesn't help us figure out what the corresponding integral of g does. It doesn't help us to know that the area under g is larger than a finite area. Similarly, knowing
diverges doesn't help us figure out what the corresponding integral of f does. It doesn't help us to know that the area under f is smaller than an infinite area.
This property also extends to improper integrals with infinite limits.
If 0 ≤ f (x) < g(x) for all x in [a, ∞), then
If the integral of g is finite, so is the integral of f. If the integral of f is infinite, so is the integral of g.