Let f ( x ) = e^-x – 5. Does

converge or diverge?

It looks like this integral probably converges, since f ( x ) looks a lot like e^-x, and we know

converges. To show this for sure, we know that

e^-x – 5 < e^-x.

Since f ( x ) < e^-x on [1,∞) and

converges,

converges also.

There's one more property we should toss in.

Let c be a constant. Then we know with definite integrals

This extends to improper integrals where the function is badly behaved. It also extends to improper integrals where the limits are badly behaved. The integrals

and

either both converge or both diverge. Multiplying a function by a constant doesn't affect whether the integral of that function converges or diverges.

Let . Does

converge or diverge?

Since  and

diverges, so does

Does

converge or diverge?

We know that

-1 < sin x < 1.

This means

The part we care about is the part that says

Since

diverges, the integral

diverges also.

Does

converge or diverge?

If we make the denominator of a fraction bigger, we make the fraction smaller. This means

Since

diverges, so does

This trick of making the denominator bigger to get a smaller fraction, or making the denominator smaller to get a bigger fraction, is useful for finding functions with which to compare our integrand.

Remember that if we want to show an integral converges, we have to find an integral with a bigger integrand and the same limits of integration that also converges. If we want to show an integral diverges, we have to find an integral with a smaller integrand and the same limits of integration that diverges.

If we can't find a convergent integral with a bigger integrand, or a divergent integral with a smaller integrand, then there's nothing to be done.