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# Week 4 Tuesday 7/11 brief notes.
## Improper integrals (7.8)
So far the integrals we have been dealing with are all definite integrals, or finding general expressions of antiderivatives. The definite integrals exist when we integrate a continuous function over a closed and bounded interval.
Note. If a function $f(x)$ is **continuous** on the closed and bounded interval $[a,b]$, the value $\displaystyle\int_{a}^{b}f(x)dx$ **always exist** and is finite.
What if the interval is unbounded, or the function we are integrating has a discontinuity on the interval of integration? These are called **improper integrals**, and we will need to **define** how we should deal with them.
# Type 1: Unbounded interval.
This is the situation where we get unbounded interval of integration, like: $$
\begin{align*}
\int_{a}^{+\infty} f(x)dx \\
\int_{-\infty}^{a} f(x)dx \\
\int_{-\infty}^{\infty}f(x)dx
\end{align*}
$$
What do we mean by these? Since a definite integral returns a value, and geometrically is the (signed) area under the curve $f(x)$, we would like to interpret an improper integral like that as well. So how do we speak of this "infinite area"?
> **Definition.**
> (A) Given a function $f(x)$ and let $a$ be a real number. We define $$
\int_{a}^{+\infty} f(x)dx = \lim_{t\to+\infty} \int_{a}^{t}f(x)dx
$$If this limit exists, then we say the integral **converges**. Otherwise **diverges** (an all encompassing term for "does not converge").
> (B) Similarly, we define $$
\int_{-\infty}^{a}f(x)dx = \lim_{t\to-\infty} \int_{t}^{a}f(x)dx
$$ (C) Lastly, if both $\displaystyle\int_{a}^{\infty}f(x)dx$ and $\displaystyle\int_{-\infty}^{a}f(x)dx$ exist, for any choice of $a$, then we can define $$
\int_{-\infty}^{\infty} f(x)dx = \int_{a}^{+\infty}f(x)dx +\int_{-\infty}^{a}f(x)dx
$$ and say the integral converges. Otherwise we say it diverges if we can find some $a$ such that either $\int_{a}^{\infty}f(x)dx$ or $\int_{-\infty}^{a}f(x)dx$ diverges.
**Example.** Determine whether $\displaystyle\int_{0}^{+\infty} \frac{1}{1+x^{2}}dx$ converges or not. If converges, determine its value.
$\blacktriangleright$ This is an improper integral. To examine its convergence we need to examine its limit definition: $$
\begin{align*}
\int_{0}^{+\infty} \frac{1}{1+x^{2}}dx & = \lim_{t\to\infty} \int_{0}^{t} \frac{1}{1+x^{2}}dx \\
& = \lim_{t\to\infty} [\arctan(x)\bigg|_{0}^{t} \\
& =\lim_{t\to\infty} \arctan(t) \\
& = \frac{\pi}{2}
\end{align*}
$$So the integral converges, and has value $\displaystyle\frac{\pi}{2}$. $\blacklozenge$
**Example.** Determine whether $\displaystyle\int_{-\infty}^{\infty} \frac{1}{1+x^{2}}dx$ converges or not. If so determine its value.
$\blacktriangleright$ To examine its convergence by definition, we take any convenient point to break it into two improper integrals. Taking $a=0$ is a good one. We already showed $\displaystyle\int_{0}^{\infty} \frac{1}{1+x^{2}}dx = \frac{\pi}{2}$. As for $\displaystyle\int_{-\infty}^{0} \frac{1}{1+x^{2}}dx$, note by definition $$
\begin{align*}
\int_{-\infty}^{0} \frac{1}{1+x^{2}}dx & = \lim_{t\to-\infty}\int_{t}^{0} \frac{1}{1+x^{2}}dx \\
& = \lim_{t\to-\infty} [\arctan(x)\bigg|_{t}^{0} \\
& = \lim_{t\to-\infty} -\arctan(t) \\
& =- \left( -\frac{\pi}{2} \right) = \frac{\pi}{2}
\end{align*}
$$Of course, we could have obtained this by symmetry as well.
In any case, since both these improper integrals converge, we say $\displaystyle\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} dx$ converges and has value $\pi$. $\blacklozenge$
**Example.** Determine whether the integral $\displaystyle\int_{0}^{\infty} \frac{1}{1+x}dx$ converges or not. If so determine its value.
$\blacktriangleright$ Again by definition, $$
\begin{align*}
\int_{0}^{\infty} \frac{1}{1+x}dx & = \lim_{t\to\infty} \int_{0}^{t} \frac{1}{1+x}dx \\
& = \lim_{t\to\infty} \ln(1+x)\bigg|_{0}^{t} \\
& = \lim_{t\to\infty} \ln(1+t) \\
& = +\infty
\end{align*}
$$Hence we say $\displaystyle\int_{0}^{\infty} \frac{1}{1+x} dx$ **diverges**. (And diverges to $+\infty$, if we care about what kind.) $\blacklozenge$
**Example.** Determine whether the integral $\displaystyle\int_{0}^{\infty}\sin(x)dx$ converges or not. If so determine its value.
$\blacktriangleright$ By definition, $$
\begin{align*}
\int_{0}^{\infty}\sin(x)dx & = \lim_{t\to\infty}\int_{0}^{t} \sin(x)dx \\
& = \lim_{t\to\infty} [-\cos(t)+1]
\end{align*}
$$which diverges (does not converge to any one value). Hence the integral $\displaystyle\int_{0}^{\infty}\sin(x)dx$ diverges. $\blacklozenge$
**Example.** Examine the convergence of $\displaystyle\int_{1}^{\infty} \frac{1}{x^{2}}dx$. If converges, find its value.
$\blacktriangleright$ By definition, $$
\begin{align*}
\int_{1}^{\infty} \frac{1}{x^{2}}dx & = \lim_{t\to +\infty} \int_{1}^{t} \frac{1}{x^{2}}dx \\
& = \lim_{t\to +\infty} \left[ -\frac{1}{x} \right|_{1}^{t} \\
& = \lim_{t\to +\infty} \left( -\frac{1}{t} +1\right) \\
& = 1
\end{align*}
$$So $\displaystyle\int_{1}^{\infty} \frac{1}{x^{2}}dx = 1$ converges. $\blacklozenge$
**Slight remark.** The integral type $\int_{-\infty}^{\infty}f(x)dx$ needs good care to define. And there are various other ways to define it. Let us call our definitions **classical**. Sometimes one would like to assign a value to these divergent improper integrals. One way is called **Cauchy principal value**, where it is defined $\displaystyle\text{P.V.}\int_{-\infty}^{\infty}f(x) dx = \lim_{t\to\infty}\int_{-t}^{t}f(x)dx$. This can often give a value to an otherwise divergent classical improper integral. For instance $\displaystyle\int_{-\infty}^{\infty}xdx$ diverges classically, but $\displaystyle\text{P.V.}\int_{-\infty}^{\infty}xdx = 0$. When using Cauchy principal value, it is common and polite to denote $\text{P.V.}$ in front. In any case, we won't be using this that often in this class.
# Type 2: Discontinuity on the interval.
If the integral $\displaystyle\int_{a}^{b}f(x)dx$ is such that $f$ has a discontinuity on the boundary $x=a$ or $x=b$, or discontinuous in the interior at $c\in(a,b)$, then this is also improper. Here is how we define them and their convergence
>**Definition.**
>(A) If $f(x)$ is not continuous at $x=b$, then we define $$
\int_{a}^{b}f(x)dx = \lim_{t\to b^{-}} \int_{a}^{t}f(x)dx
$$(B) if $f(x)$ is not continuous at $x=a$, then we define $$
\int_{a}^{b}f(x)dx = \lim_{t\to a^{+}}\int_{t}^{b}f(x)dx
$$ (C) If $f(x)$ has a discontinuity in the interior at some $c\in(a,b)$, then we define $$
\int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx
$$where the convergence depends on the two improper integrals.
> In any of these cases, if the limit exists then we say the improper integral **converges** and take on the value of the limit. Otherwise we say they **diverge**.
**Example.** Determine whether $\displaystyle\int_{0}^{1} \frac{1}{x} dx$ converges or not.
$\blacktriangleright$ The integrand $\frac{1}{x}$ has a discontinuity at $x=0$. So by definition, $$
\begin{align*}
\int_{0}^{1} \frac{1}{x}dx & = \lim_{t\to 0^{+}} \int_{t}^{1} \frac{1}{x}dx \\
& = \lim_{t\to 0^{+}} \ln(x) \bigg|_{t}^{1} \\
& = \lim_{t\to 0^{+}} - \ln(t) \\
& = +\infty
\end{align*}
$$So $\displaystyle\int_{0}^{1} \frac{1}{x} dx$ diverges. $\blacklozenge$
**Example.** Determine whether $\displaystyle\int_{0}^{1} \frac{1}{x^{2}}dx$ converges or not. If so, find its value.
$\blacktriangleright$ By definition $$
\begin{align*}
\int_{0}^{1} \frac{1}{x^{2}}dx & = \lim_{t\to 0^{+}} \int_{t}^{1} \frac{1}{x^{2}}dx \\
& = \lim_{t\to 0^{+}} \left[ -\frac{1}{x} \right|_{t}^{1} \\
& = \lim_{t\to 0^{+}} \left(-1 + \frac{1}{t}\right) \\
& = +\infty
\end{align*}
$$Whence $\displaystyle\int_{0}^{1} \frac{1}{x^{2}}dx$ diverges! $\blacklozenge$
**Example.** Determine whether $\displaystyle\int_{0}^{1} \frac{1}{x^{1/3}}dx$ converges or not. If so, find its value.
$\blacktriangleright$ By definition $$
\begin{align*}
\int_{0}^{1} \frac{1}{x^{1/3}}dx & = \lim_{t\to 0^{+}} \int_{t}^{1} \frac{1}{x^{1/3}}dx \\
& = \lim_{t\to 0^{+}} \left[ \frac{3}{2} x^{2/3} \right|_{t}^{1} \\
& = \lim_{t\to 0^{+}} \left(\frac{3}{2} - \frac{3}{2} t^{2/3}\right) \\
& = \frac{3}{2}
\end{align*}
$$So here $\displaystyle\int_{0}^{1} \frac{1}{x^{1/3}}dx$ converges! $\blacklozenge$
**Example.** Does $\displaystyle\int_{0}^{1}\ln(x)dx$ converge or diverge? If converges, find its value.
$\blacktriangleright$ Here $\ln(x)$ is discontinuous at $x=0$. Using its definition, $$
\begin{align*}
\int_{0}^{1}\ln(x)dx & = \lim_{t\to 0^{+}} \int_{t}^{1}\ln(x)dx \\
& = \lim_{t\to 0 ^{+}} [x\ln(x)-x\bigg|_{t}^{1} \\
& =\lim_{t\to 0 ^{+}} (-1-t\ln(t)+t) \\
\end{align*}
$$
Here we have a limit but one term is indeterminant, $\lim_{t\to 0^{+}} t\ln(t)$, of type $0\cdot\infty$. But using a bit algebra and L'Hospital rule we get $\lim_{t\to 0^{+}} \frac{\ln(t)}{\frac{1}{t}} \stackrel{\text{L'H}}{=} \lim_{t\to 0^{+}} \frac{(\frac{1}{t})}{(-\frac{1}{t^{2}})} = \lim_{t\to 0^{+}} -t =0$.
So we see $\displaystyle\int_{0}^{1} \ln(x)dx = -1$ converges! $\blacklozenge$
**Example.** Determine whether $\displaystyle \int_{0}^{2} \frac{1}{x-1}dx$ converges or diverges.
$\blacktriangleright$ Note that the integrand has a discontinuity at $x=1$. So we need to break it into two parts:$$
\int_{0}^{2}= \int_{0}^{1}+\int_{1}^{2}
$$ We need both to converge in order to declare this improper integral converges. If any one part diverges, the this improper integral diverges.
The antiderivative $\displaystyle\int \frac{1}{x-1}dx =\ln|x-1|+C$, so for the part $\int_{0}^{1}$, we get$$
\begin{align*}
\int_{0}^{1} \frac{1}{x-1}dx & = \lim_{t\to 1^{-}} \ln|x-1|\bigg|_{0}^{t} \\
& =\lim_{t\to1^{-}} \ln|t-1| = -\infty
\end{align*}
$$Since one part of the integral diverges, the original improper integral diverges. $\blacklozenge$
# An important class of improper integral: $p$-integrals.
The function $\displaystyle f(x)=\frac{1}{x^{p}}$ has a natural discontinuity at $x=0$. So we should investigate the various types of improper integrals relating to these types. As it turns out we have the following:
> **Theorem. $p$-integrals.**
> **("Tail" case)** The improper integral $$
\int_{1}^{\infty} \frac{1}{x^p}dx
$$converges if $p > 1$, and diverges if $p\le 1$.
>**("Pole" case)** Also, the improper integral $$
\int_{0}^{1} \frac{1}{x^{p}}dx
$$ converges if $p < 1$, and diverges if $p \ge 1$.
> The result remains the same if we replace the $1$ in the integration bound with any positive $a > 0$.
**Example.** The integral $\displaystyle\int_{0}^{3} \frac{1}{x^{2.3}} dx$ diverges by $p$-integral theorem. And the integral $\displaystyle\int_{5}^{\infty} \frac{1}{x^{2.3}}dx$ converges by $p$-integrals theorem. $\blacklozenge$
These can quickly help us determine convergence of various improper integrals if used correctly.
**Example.** Let us look at the improper integral $\displaystyle\int_{1}^{2} \frac{1}{\sqrt{x-1}}dx$. Apply a simple translation $u=x-1$, shifting the bounds as well, we get $$
\int_{x=1}^{2} \frac{1}{\sqrt{x-1}}dx = \int_{u=0}^{1} \frac{1}{\sqrt{u}}du
$$This gives a pole type $p$-integral of $p=\frac{1}{2}$, which we known convergences. Hence $\displaystyle\int_{1}^{2} \frac{1}{\sqrt{x-1}}dx$ converges! (If we want what the actual value is, we need to do more work.) $\blacklozenge$
## Comparison test for integrals.
Quite often we just want to know whether an improper integral converges or not. The idea is to compare with known improper integrals (such at the $p$-integrals!)
Since if the integral is easier to compare when the integrand is positive (or the same sign through out), we will focus on positive integrands.
The general mantra is simple. If a function is bigger than some function with divergent improper integral, then the corresponding improper integral will also be divergent. And if a function is smaller than some function with convergent improper integral, then the corresponding improper integral will also be convergent.
> **Theorem. Comparison test**
> Let $f(x)$ and $g(x)$ both be positive functions.
> If $f(x) \le g(x)$ and the improper integral $\int_{a}^{b} g(x)dx$ converges, then $\int_{a}^{b}f(x)dx$ also converges.
> If $f(x)\ge g(x)$ and the improper integral $\int_{a}^{b}g(x)dx$ diverges, then $\int_{a}^{b}f(x)dx$ also diverges.
Often, $p$-integrals are the ones that are useful to compare to.
**Example.** Determine convergence of $$
\int_{0}^{\infty} \frac{x}{x^{3}+1}dx
$$
$\blacktriangleright$ Note the integral can be chopped into two pieces, $\int_{0}^{\infty} =\int_{0}^{1}+\int_{1}^{\infty}$. The part $\int_{0}^{1}$ is just a simple definite integral that has a finite value, as the function $\displaystyle$ is continuous on $[0,1]$. We should worry about the improper $\int_{1}^{\infty}$ part.
Notice when $x \ge 1$, we have $$
\frac{x}{x^{3}+1}= \frac{1}{x^{2} + \frac{1}{x}} \le \frac{1}{x^{2}}
$$
So $$
\int_{1}^{\infty} \frac{x}{x^{3}+1}dx =\int_{1}^{\infty} \frac{1}{x^{2}+\frac{1}{x}}dx \le \int_{1}^{\infty} \frac{1}{x^{2}}dx
$$
But the improper integral $\displaystyle\int_{1}^{\infty} \frac{1}{x^{2}}dx$ converges by $p$-integrals! So the improper integral $\displaystyle\int_{1}^{\infty} \frac{x}{x^{3}+1}dx$ converges, and hence the original improper integral $\displaystyle\int_{0}^{1} \frac{x}{x^{3}+1}dx$ converges! $\blacklozenge$
**Example.** Determine the convergence of $$
\int_{1}^{\infty} \frac{1+\sin^{2}(x)}{\sqrt{x}}dx
$$
$\blacktriangleright$ Observe the integrand $$
\frac{1+\sin^{2}(x)}{\sqrt{x}} \ge \frac{1}{\sqrt{x}}
$$and so $$
\int_{1}^{\infty} \frac{1+\sin^{2}(x)}{\sqrt{x}}dx \ge \int_{1}^{\infty} \frac{1}{\sqrt{x}}dx
$$But by $p$-integrals, we know the tail integral with $p=\frac{1}{2} \le 1$ diverges. So the original improper integral $\displaystyle\int_{1}^{\infty} \frac{1+\sin^{2}(x)}{\sqrt{x}}dx$ diverges. $\blacklozenge$
## Using limits and continuity to help us analyze improper integrals.
Sometimes use properties of the integrand to help make comparisons.
**Example.** Does the integral $\int_{0}^{\infty} \arctan(x)dx$ converge or diverge?
$\blacktriangleright$ Note that $\arctan(x) \to \frac{\pi}{2}$ as $x\to\infty$. That is $\arctan(x)$ is close to $\frac{\pi}{2}$ for all large $x$. Note $\frac{\pi}{2} > 1$. So at some point, say $x=a$, $x > a$ would imply $\arctan(x) > 1$. But this mean the improper integral $\int_{0}^{\infty} \arctan(x) dx > \int_{a}^{\infty}\arctan(x) > \int_{a}^{\infty} 1dx = \infty$.
So $\int_{0}^{\infty}\arctan(x)dx$ diverges. $\blacklozenge$
Sometimes we exploit continuity of the parts that don't blow up to help us analyze.
**Example.** Does the integral $\displaystyle\int_{2}^{3} \frac{x^{2}+3}{\sqrt{x^{2}-4}}dx$ converge or diverge?
$\blacktriangleright$ Observe the integral has a singularity at $x=2$. Let us try to isolate that part:$$
\frac{x^{2}+3}{\sqrt{x^{2}-4}}=\frac{x^{2}+3}{\sqrt{x+2}}\cdot \underbrace{\frac{1}{\sqrt{x-2}}}_{\text{blows up}}
$$But the factor $\displaystyle \frac{x^{2}+3}{\sqrt{x+2}}$ is continuous at $x=2$, and has value $\frac{7}{2}$. This means when $x\approx 2$, we have $\displaystyle \frac{x^{2}+3}{\sqrt{x+2}}\approx \frac{7}{2}$. In other words, when $x\approx 2$, say on the interval $(2,2+\epsilon)$ we have $$
\frac{x^{2}+3}{\sqrt{x^{2}-4}}=\frac{x^{2}+3}{\sqrt{x+2}}\cdot \underbrace{\frac{1}{\sqrt{x-2}}}\approx \frac{7}{2} \frac{1}{\sqrt{x-2}} \le \frac{4}{\sqrt{x-2}}
$$
Now the improper integral (with $u=x-2$) $$
\int_{2}^{2+\epsilon} \frac{4}{\sqrt{x-2}}dx =\int_{u=0}^{\epsilon} \frac{4}{\sqrt{u}}du
$$converges by $p$-integral theorem. So by comparison test, we deduce our improper integral $\displaystyle\int_{2}^{3} \frac{x^{2}+3}{\sqrt{x^{2}-4}}dx$ converges! $\blacklozenge$
Please read sections 7.7 and 7.8 for more examples and their exercises to gain more understanding.
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