# Comparison Test for Improper Integral

## Theorem

Let $I = \left({a \,.\,.\, b}\right)$ be an open real interval.

Let $\phi$ be a real function which is continuous on $I$ and also non-negative on $I$.

Let $f$ be a real function which is continuous on $I$.

Let $f$ satisfy:

$\forall x \in I: \left\vert{f \left({x}\right)}\right\vert \le \phi \left({x}\right)$

If the improper integral of $\phi$ over $I$ exists, then so does that of $f$.

## Proof

Without loss of generality, we consider the case $I = \left({0 \,.\,.\, +\infty}\right)$ such that $\displaystyle l = \int_0^{\mathop \to +\infty} \phi \left({x}\right) \rd x$ exists.

Let:

$\displaystyle a_n = \int_{n - 1}^n f \left({x}\right) \rd x$
$\displaystyle b_n = \int_{n - 1}^n \phi \left({x}\right) \rd x$

for $n = 1, 2, \ldots$

Then the series:

$\displaystyle \sum_{n \mathop = 1}^\infty b_n$

is a convergent series of positive terms whose sum is $l$.

Also: $\left\vert{a_n}\right\vert \le b_n$ for $n = 1, 2, \ldots$

Hence $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ a convergent series by the Comparison Test.

That is:

$\displaystyle m = \lim_{N \mathop \to \infty} \int_0^N f \left({x}\right) \rd x$

exists.

Given any $X > 0$, $N$ can be taken to be the smallest natural number which satisfies $N > X$.

Then:

 $\displaystyle \left\vert{\int_1^X f \left({x}\right) \rd x - \int_1^{N-1} f \left({x}\right) \rd x}\right\vert$ $\le$ $\displaystyle \int_{N-1}^X \left\vert{f \left({x}\right)}\right\vert \rd x$ $\displaystyle$ $\le$ $\displaystyle b_N$ $\displaystyle$ $\to$ $\displaystyle 0 \text{ as } N \to \infty$

It follows that: $\displaystyle \lim_{X \mathop \to \infty} \int_1^X f \left({x}\right) \rd x = m$ as required.

The other cases are treated similarly.

$\blacksquare$