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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.

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