Comparison Test for Improper Integral

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Theorem

Let $I = \openint a b$ 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:

$\forall x \in I: \size {\map f x} \le \map \phi x$


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


Proof

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

Let:

$\ds a_n = \int_{n - 1}^n \map f x \rd x$
$\ds b_n = \int_{n - 1}^n \map \phi x \rd x$

for $n = 1, 2, \ldots$

Then the series:

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

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

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

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

That is:

$\ds m = \lim_{N \mathop \to \infty} \int_0^N \map f x \rd x$

exists.

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

Then:

\(\ds \size {\int_1^X \map f x \rd x - \int_1^{N - 1} \map f x \rd x}\) \(\le\) \(\ds \int_{N - 1}^X \size {\map f x} \rd x\)
\(\ds \) \(\le\) \(\ds b_N\)
\(\ds \) \(\to\) \(\ds 0 \text{ as } N \to \infty\)

It follows that:

$\ds \lim_{X \mathop \to \infty} \int_1^X \map f x \rd x = m$

as required.


The other cases are treated similarly.

$\blacksquare$


Sources