4.23 Improper Integrals, comparison and ratio tests

As we said in Lecture 13, sometimes it is difficult to determine if a given improper integral converges or not. So, one idea to determine the convergence is to compare it with an integral that we already know its behavior.

As before, to compare integrals, we assume that the functions are non-negative.

Convention. We often write $\displaystyle \int_a^\infty f(x)\,dx < \infty$ to indicate that an improper integral of a non-negative function converges.

We begin by stating our first comparison test.

Theorem (Comparison Test)

Let $f, g$ be two locally integrable functions on $[a, \infty)$. Assume that $0\leq f(x)\leq g(x)\quad\forall a\leq x<b$

Then

${\displaystyle\int_{a}^{b}f(x)\,dx<\infty}$ if ${\displaystyle\int_{a}^{b}g(x)\,dx<\infty}$
${\displaystyle\int_{a}^{b}g(x)\,dx=\infty}$ if ${\displaystyle\int_{a}^{b}f(x)\,dx=\infty}$

Main idea: These are the only possibilities and \displaystyle\int_{a}^{b}f(x)dx = \sup_{a \le x < b}F(x)...

Proof

Consider $\displaystyle F(x)=\int^{x}_{a}f(t)dt, G(x)=\int^x_ag(t)dt$, then by theorem $F(x),G(x)$ are increasing and $\displaystyle\int^{b}_{a}f(x)dx=\sup\limits_{a\leq x<b}F(x),\int^{b}_{a}g(x)dx=\sup \limits_{a\leq x<b}G(x)$

Now, since $f(x)\leq g(x)$, we know that $\displaystyle F(x)=\int^x_af(t)dt\leq \int^x_ag(t)dt=G(x)$

Then $\sup\limits_{a\leq x<b}F(x)<\sup\limits_{a\leq x<b}G(x)$, if $\sup\limits_{a\leq x<b}G(x)$ is finite, then $\sup\limits_{a\leq x<b}F(x)$ is also finite.

And if $\sup\limits_{a\leq x<b}F(x)=\infty$, then $\sup\limits_{a\leq x<b}G(x)=\infty$

Example

Determine the convergence of the following improper integral $\displaystyle I=\int^{1}_{0}\frac{2+\sin(\pi x)}{x^{p}}dx$

Since $f(x)=\frac{2+\sin(\pi x)}{x^{p}}\geq0$ in $(0,1]$, then $\frac{2+\sin(\pi x)}{x^{p}}\leq \frac{3}{x^p}$

Since $\displaystyle \int^{1}_{0}\frac{3}{x^{p}}=3\int^{1}_{0}\frac{1}{x^{p}}$, if $p<1$, then $\displaystyle\int_{0}^{1}\frac{3}{x^{p}}<\infty$ since $\displaystyle\int_{c}^{1}\frac{1}{x^{p}}\,dx=\left.\frac{x^{-p+1}}{-p+1}\right|_{c}^{1}=\left .\frac{1}{(-p+1)x^{p-1}}\right|_{c}^{1}$, then ${\displaystyle\int_0^1\frac{2+\sin(\pi x)}{x^{p}}dx}<\infty$

And since $\frac{1}{x^{p}}\leq\frac{2+\sin(\pi x)}{x^{p}}$, then if $p>1$, we know ${\displaystyle\int_0^1\frac{1}{x^{p}}=\infty}$, then ${\displaystyle\int_0^1\frac{2+\sin(\pi x)}{x^{p}}dx}=\infty$

Thus $\displaystyle \int_{0}^{1}\frac{2 + \sin(\pi x)}{x^{p}}\, dx = \begin{cases} < \infty & \text{if } p < 1 \\ = \infty & \text{if } p \geq 1 \end{cases}$

If $p=1$, it is also divergent easily.

Remark

Since we are studying the convergence of the improper integrals, we may relax the condition $0\leq f(x)\leq g(x)\quad\forall a\leq x<b$

Since ${\displaystyle\int_{a}^{b}f(x)\,dx={\displaystyle\int_{a}^{a_1}f(x)\,dx+{\displaystyle\int_{a_1}^{b}f(x)\,dx}}}$, it follows that ${\displaystyle\int_{a}^{b}f(x)\,dx}$ converges if and only if the improper integral ${\displaystyle\int_{a_1}^{b}f(x)\,dx}$ also converges.

Thus, the condition may be replaced by the condition $0\leq f(x)\leq g(x)\quad\forall a_{1}\leq x<b$ for some $a_{1}\in[a,b)$.

Now we introduce our second comparison test.

Theorem (Ratio Test)

Let $f, g$ be two locally integrable functions on $[a,b)$. Assume that $f(x) \geq 0$ and $g(x) > 0$ on some subinterval $[a_{1},b)$ of $[a,b)$ and $\lim_{x\to b^{-}}\frac{f(x)}{g(x)}=M$

If $0<M<\infty$, then ${{\displaystyle\int_{a}^{b}f(x)\,dx}}$ and ${\displaystyle\int_{a}^{b}g(x)\,dx}$ converge or diverge together.
If $M=\infty$ and ${\displaystyle\int_{a}^{b}g(x)\,dx=\infty}$, then ${\displaystyle\int_{a}^{b}f(x)\,dx=\infty}$.
If $M=0$ and ${\displaystyle\int_{a}^{b}g(x)\,dx<\infty}$, then ${\displaystyle\int_{a}^{b}f(x)\,dx<\infty}$.

Main Idea: We can bound function by definition of limits

Proof

We know that $\lim_{x\to b^{-}}\frac{f(x)}{g(x)}=M$, this means that $\forall \varepsilon>0,\exists \delta>0:\left|\frac{f(x) }{g(x)}-M\right|<\varepsilon$ if $|b-x|<\delta$

Then $-\varepsilon<\frac{f(x)}{g(x)}-M<\varepsilon$, then $M-\varepsilon<\frac{f(x)}{g(x)}<M+\varepsilon$

Suppose $M\neq 0,\infty$ and $\varepsilon \leq \frac{M}{2}$, then $\frac{M}{2}<\frac{f(x)}{g(x)}<\frac{3M}{2}$, then since $g(x)>0$ in $[a,b)$, then $\frac{M}{2}g(x)<f(x)<\frac{3M}{2}g(x)$, $\forall x$ in $[a,b)$

Assume first that ${{\displaystyle\int_{a}^{b}f(x)\,dx}}$ is convergent, then $0<\frac{M}{2}g\left(x\right)<f\left(x\right),\forall x\in[a,b)$

By comparison test, ${{{\displaystyle\int_{a}^{b}\frac{M}{2}g(x)\,dx}}}$ is convergent, then ${{\displaystyle\int_{a}^{b}g(x)\,dx}}$ is convergent

Assume ${{\displaystyle\int_{a}^{b}f(x)\,dx}}$ is divergent, then $0\leq f(x)\leq \frac{3}{2}Mg(x)$, again ${{\displaystyle\int_{a}^{b}g(x)\,dx}}$ is divergent

If $M=\infty$, thus $\frac{f(x)}{g(x)}>1$ for $x\in[a_{1},b)$, then $f(x)>g(x)$ for $x\in [a_1,b)$

By comparison test, if ${{\displaystyle\int_{a}^{b}g(x)\,dx}}$ is divergent, then ${{{\displaystyle\int_{a}^{b}f(x)\,dx}}}$ is divergent

If $M=0$, then $\frac{f(x)}{g(x)}<1$ for $x\in [a_1,b)$, then $f(x)<g(x)$ for $x\in [a_1,b)$

Again by the same theorem, if ${{\displaystyle\int_{a}^{b}g(x)\,dx}}$ is convergent, then ${{\displaystyle\int_{a}^{b}f(x)\,dx}}$ is convergent

Example

$f(x)=\frac{2+\sin(\pi x)}{x^{p}}$ in $(0,1]$

Let $g(x)=\frac{1}{x^p}$, then $\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{\frac{2+\sin(\pi x)}{x^{p}}}{\frac{1}{x^{p}}} =2+\sin(0)=2=M$

Then ${\displaystyle\int_{a}^{b}\frac{2+\sin(\pi x)}{x^{p}}\,dx=\begin{cases}<\infty & \text{if }p<1\\ =\infty & \text{if }p\geq1\end{cases}}$
Let $a,b\in \R,a\neq 0$. Analyze the convergence of $\displaystyle\int^{+\infty}_c\frac{1}{(ax+b)^p}dx$ for $c>0$

We know ${\displaystyle\int_0^1\frac{1}{x^{p}}\,dx=\begin{cases}<\infty & \text{if }p<1\\ =\infty & \text{if }p\geq1\end{cases}}$ and ${{{\displaystyle\int_1^{\infty}\frac{1}{x^{p}}\,dx=\begin{cases}<\infty & \text{if }p>1\\ =\infty & \text{if }p\leq1\end{cases}}}}$

We use ratio test, let $g(x)=\frac{1}{x^p}$, then $\lim_{x\to\infty}\frac{x^{p}}{\left(ax+b\right)^{p}}=\lim_{x\to\infty}\frac{1}{\left(a+\frac{b}{x}\right)^{p}} =\frac{1}{a^{p}}=M$

Now, $ax+b=0\iff x=-\frac{b}{a}$, then we take $c>-\frac{b}{a}$, we have ${{{\displaystyle\int_{c}^{\infty}\frac{1}{\left(ax+b\right)^{p}}\,dx=\begin{cases}<\infty & \text{if }p>1\\ =\infty & \text{if }p<1\end{cases}}}}$
Let $p, q \in \mathbb{Q}$, analyze the convergence of $f(x)=\frac{1}{x^p(x+1)^q}$ in $(0, \infty)$.
We analyze first the problem at 0. Take $g(x) = \frac{1}{x^p}$ and consider the limit $\lim_{x \to 0} \frac{\frac{1}{x^p(1+x)^q}}{\frac{1}{x^p}} = \lim_{x \to 0} \frac{1}{(1+x)^q} = 1 = M$

Hence $\displaystyle\int_{0}^{1}\frac{1}{x^{p}(1+x)^{q}}d x = \begin{cases} \infty & \text{if } p>1 \\ <\infty & \text{if } p<1 \end{cases}$
Since $\displaystyle\int_{0}^{\infty}\frac{1}{x^{p}(1+x)^{q}}d x = \int_{0}^{1}\frac{1}{x^{p}(1+x)^{q}} d x + \int_{1}^{\infty}\frac{1}{x^{p}(1+x)^{q}}d x$
$I_1$ $I_2$

Now we study $I_2$
We consider $g(x) = \frac{1}{x^{p+q}}$, then $\lim_{x \to \infty} \frac{\frac{1}{x^{p}(1+x)^{q}}}{\frac{1}{x^{p+q}}} = \lim_{x \to \infty} \frac{x^{p+q}}{x^{p}(1+x)^{q}} = \lim_{x \to \infty} \frac{x^{q}}{(1+x)^{q}} = \lim_{x \to \infty} (\frac{x}{1+x})^q = 1^q = 1$
Thus $\displaystyle\int_{1}^{\infty}\frac{1}{x^{p}(1+x)^{q}}dx= \begin{cases} <\infty & \text{if }p+q>1 \\ \infty & \text{if }p+q<1 \end{cases}$

Thus $p<1$ and $p+q>1$ are the condition of convergence

To analyze integrals that are not non-negative, we introduce the notion of absolute convergence

Definition

Let $f$ be a locally integrable function on $[a,b)$. We say that the improper integral of $f$ converges absolutely or it is absolutely convergent if ${\displaystyle\int_{a}^{b}|f(x)|\,dx<\infty}$

Theorem

Let $f$ be a locally integrable function on $[a, b)$.

If the improper integral of $f$ is absolutely convergent, then the improper integral of $f$ is convergent.

Main idea: $g(x)=|f(x)|-f(x)$ and comparison test to get $g(x)$ is local integrable, then discuss $f(x)$

Proof

Since $f$ is integrable, then $|f|$ is also integrable. Since $|f(x)|\geq f(x)$, then $g(x)=|f(x)|-f(x)\geq 0$

Thus $g$ is also integrable, and we know that $\displaystyle\int_{a}^{b}\left|f(x)\right|\,dx<\infty$

Since $0\leq g(x)\leq 2|f(x)|$, $\forall x\in [a,b)$. By comparison test, since ${{{\displaystyle\int_{a}^{b}2\left|f(x)\right|\,dx}}}<\infty$, then ${{\displaystyle\int_{a}^{b}g(x)\,dx}}<\infty$

Now, $f(x) = |f(x)| - g(x)$, $\forall x \in (a, b)$

$\displaystyle\int_{a}^{b} f(x) dx = \lim_{c \to b^-}\int_{a}^{c} f(x) dx =$$\displaystyle \lim_{c \to b^-}\int_{a}^{c} (|f(x)| - g(x)) dx$ $\displaystyle= \lim_{c \to b^-}(\int_{a}^{c} |f(x)| dx - \int_{a}^{c} g(x) dx)$
Both limit exist, then $\displaystyle= \lim_{c \to b^-}\int_{a}^{c} |f(x)| dx - \lim_{c \to b^-}\int_{a}^{c} g(x) dx < \infty$

Remark

Kindly reminder: Do not use the methods of integration directly on the improper integrals. Use first the definition with the limit.

Remark

In calculus, the function $\arctan(y)$ is defined to return the unique angle $\theta$ in the principal range $(-\pi/2, \pi/2)$ such that $\tan(\theta) = y$ , and $-\frac{\pi}{4}$ is that unique angle for $y=-1$ .

Remark

Convergent $\not\Rightarrow $ absolutely Convergent

Example: $f(x) = \frac{\sin(x)}{x}$ in $[1, \infty)$

Since $|f(x)| = |\frac{\sin(x)}{x}|$ in $[1, \infty)$

Let us take a look at ${\displaystyle\int_1^{c}\frac{\sin(x)}{x}dx=\int_1^{c}\frac{\sin(x)}{x}dx}$

$u(x) = \frac{1}{x}$, $u'(x) = -\frac{1}{x^2}$ and $v'(x) = \sin(x)$, $v(x) = -\cos(x)$

$\displaystyle= [-\frac{\cos(x)}{x}]_{1}^{c}- \int_{1}^{c}-\frac{\cos(x)}{x^{2}}dx$

Now $| \frac{\cos(x)}{x^2} | = \frac{|\cos(x)|}{x^2} \leq \frac{1}{x^2}$, then$\displaystyle\int_{1}^{\infty}\frac{1}{x^{2}}dx < \infty \implies \int_{1}^{\infty}|\frac{\cos(x)}{x^{2}} | dx < \infty$

Then $\displaystyle\int_{1}^{\infty}\frac{\cos(x)}{x^{2}}dx<\infty$

Then $\displaystyle\lim_{c\to\infty}\int_{1}^{c}\frac{\sin(x)}{x}dx$ depends on $\lim_{c \to \infty}[-\frac{\cos(x)}{x}]_{1}^{c}= \lim_{c \to \infty}(-\frac{\cos(c)}{c} ) - (-\frac{\cos(1)}{1})$ $= 0 + \cos(1)$ $= \cos(1)$

Thus $\displaystyle\int_{1}^{\infty}\frac{\sin(x)}{x}dx<\infty$

Now, $|\frac{\sin(x)}{x}| = \frac{|\sin(x)|}{x}$ $\ge \frac{\sin^2(x)}{x}$

Since $\cos(2x) = \cos(x+x) = \cos^2(x) - \sin^2(x)$ $\implies \sin^2(x) = \cos^2(x) - \cos(2x)$
$2\sin^2(x) = \sin^2(x) + \cos^2(x) - \cos(2x)$ $= 1 - \cos(2x)$ $\implies \sin^2(x) = \frac{1-\cos(2x)}{2}$

Then $\frac{|\sin(x)|}{x} \ge \frac{1-\cos(2x)}{2x}$ $\forall x \in (1, \infty)$ $\displaystyle\implies \int_{1}^{c}\frac{|\sin(x)|}{x}dx \ge \int_{1}^{c}\frac{1-\cos(2x)}{2x} dx$
$\displaystyle= \int_{1}^{c}\frac{1}{2x}dx - \int_{1}^{c}\frac{\cos(2x)}{2x}dx \to \infty$ as $c \to \infty$.

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