# The Unapologetic Mathematician

## Convergence Tests for Improper Integrals

We have a few tests that will come in handy for determining if an improper integral converges. In all of these we’ll have an integrator $\alpha$ on the ray $\left[a,\infty\right)$, and a function $f$ which is integrable with respect to $\alpha$ on the interval $\left[a,b\right]$ for all $b>a$.

First, say $\alpha$ is nondecreasing and $f$ is nonnegative. Then the integral $\int_a^\infty f\,d\alpha$ converges if and only if there is a constant $M>0$ so that

$\displaystyle I(b)=\int\limits_a^bfd\alpha\leq M$

for every $b\geq a$. This follows because the function $I(b)$ is then nondecreasing, and a nondecreasing function bounded above must have a finite limit at infinity. Indeed, the set of values of $I$ must be bounded above, and so there is a least upper bound $\sup\limits_{x\geq a}I(x)$. It’s straightforward to show that the limit $\lim\limits_{x\rightarrow\infty}I(x)$ is this least upper bound.

Now if $\alpha$ is nondecreasing and $f(x)\leq g(x)$ are two nonnegative functions, then if the improper integral of $g$ converges then so does that of $f$, and we have the inequality

$\displaystyle\int\limits_a^\infty fd\alpha\leq\int\limits_a^\infty gd\alpha$

since for every $b\geq a$ we have

$\displaystyle\int\limits_a^bfd\alpha\leq\int\limits_a^bgd\alpha\leq\int\limits_a^\infty gd\alpha$

On the other hand, if the improper integral of $f$ diverges, then that of $g$ must diverge.

If $\alpha$ is nondecreasing and we have two nonnegative functions $f$ and $g$ so that

$\displaystyle\lim\limits_{x\rightarrow\infty}\frac{f(x)}{g(x)}=1$

then their improper integrals either both converge or both diverge. This limit implies there must be some $R$ beyond which we have $\frac{1}{2}\leq\frac{f(x)}{g(x)}\leq2$. Equivalently, for $x\geq R$ we have $\frac{1}{2}g(x)\leq f(x)\leq2g(x)$, and the result follows by two applications of the previous theorem.

Notice that this last theorem also follows if the limit of the ratio converges to any nonzero number. Also notice how the convergence of the integral only depends on the behavior of our functions in some neighborhood of $\infty$. We use their behavior in the ray $\left[R,\infty\right)$ when we started by looking for convergence over the ray $\left[a,\infty\right)$.

April 21, 2008 - Posted by | Analysis, Calculus