4.21 Properties of improper integrals
In the last lecture we defined integrals for unbounded functions on finite intervals or for bounded functions on infinite intervals.
These are called improper integrals and were defined using limits. If such a limit exists, we have for \(b = \infty\) or \(a = -\infty\)
How do we define the case when \(b = \infty\) and \(a = -\infty\)?
Definition
Let f be a locally integrable function on the open interval \((a, b)\).
We define \(\displaystyle\int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx\) where \(a < c < b\), provided both improper integrals on the right hand side converge.
Remark
The existence of the integral in the definition above does not depend on the particular choice of the point \(c \in (a, b)\). (Exercise)
Main Idea: separate local integral to finite integral and use limit to approach
Then use the integral on right hand side to express left hand side
Proof
Let choose any \(c,d\in(a,b)\) and \(d<c\), then \(\displaystyle\int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx\) and \(\displaystyle\int_{a}^{b}f(x)dx = \int_{a}^{d}f(x)dx + \int_{d}^{b}f(x)dx\)
Then assume \(\displaystyle\int_{a}^{b}f(x)dx\) exists, then NTP: \(\displaystyle \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx = \int_{a}^{d}f(x)dx + \int _{d}^{b}f(x)dx\)
First, for \(\displaystyle \int_{a}^{c}f(x)dx\), let choose \(t\in(a,d)\), then \(\displaystyle \int_{t}^{c}f(x)dx=\int_{t}^{d}f(x)dx+\int_{d}^{c}f(x)dx\)
Then let \(t\to a^+\), we get \({\displaystyle\int_{a}^{c}f(x)dx=\int_{a}^{d}f(x)dx+\int_{d}^{c}f(x)dx}\) since assumption and definition of local integrable
Then \({\displaystyle\int_{a}^{d}f(x)dx=\int_{a}^{c}f(x)dx-\int_{d}^{c}f(x)dx}\)
Similarly for \(\displaystyle\int_{c}^{b}f(x)dx\), let choose \(t\in(c,b)\), then \(\displaystyle\int_{d}^{t}f(x)dx=\int_{d}^{c}f(x)dx+\int_{c}^{t}f(x)dx\)
Let \(t\to b^-\), then \(\displaystyle\int_{d}^{b}f(x)dx=\int_{d}^{c}f(x)dx+\int_{c}^{b}f(x)dx\)
Thus \(\displaystyle \int_{a}^{d}f(x)dx + \int_{d}^{b}f(x)dx=\int_{a}^{c}f(x)dx-\int_{d} ^{c}f(x)dx+\int_{d}^{c}f(x)dx+\int_{c}^{b}f(x)dx= \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx\)
Example
Let \(p \in \mathbb{Q}, p \ne 1\). Consider the function \(f(x) = \frac{1}{x^p} = x^{-p}\)
In \((0, 1]\), \(f\) is locally integrable with primitive \(F(x) = \frac{x^{-p+1}}{-p+1}\)
\(\displaystyle\int_0^1\frac{1}{x^{p}}dx=\lim_{c\to0^{+}}\int_{c}^1\frac{1}{x^{p}}dx=\lim_{c\to0^{+}}\left[\frac{x^{-p+1}}{-p+1}\right]_{c}^1=\lim_{c\to0^{+}}\frac{1}{-p+1}-\frac{1}{(-p+1)c^{p-1}}\)
If \(p > 1\): \(p-1 > 0\), then \(= \infty\). If \(p-1 < 0\), then \(= \frac{1}{-p+1}\)
In \([1, \infty)\), \(f\) is locally integrable
\(\displaystyle\int_{1}^{\infty} \frac{1}{x^{p}}dx = \lim_{c \to \infty}\int_{1}^{c} \frac{1}{x^{p}} dx\) \(= \lim_{c \to \infty} \left[ \frac{x^{-p+1}}{-p+1} \right]_1^c\) \(= \lim_{c \to \infty} \frac{c^{-p+1}}{-p+1} - \frac{1}{-p+1}\)
\(= \lim_{c \to \infty} \frac{1}{(-p+1)c^{p-1}} - \frac{1}{-p+1}\)
If \(p > 1\): \(p-1 > 0\), then \(= 0 - \frac{1}{-p+1} = \frac{1}{p-1}\)
If \(p < 1\), then \(p-1 < 0\), then \(= \infty\)
Thus if \(p > 1\)\(\displaystyle\implies \int_{0}^{\infty} \frac{1}{x^{p}}dx\) diverges. If \(p < 1\):\(\displaystyle\int_{0}^{\infty}\frac{1}{x^{p}}dx\) diverges.
Conclusion: \(\displaystyle\int_{0}^{\infty} \frac{1}{x^{p}}dx\) diverges \(\forall p \ne 1\).
Exercise What happens with \(p=1\)? Diverges both sides
Remark
Observe that, by definition, we ask that both improper integrals \(\displaystyle\int_{a}^{c}f(x)dx\) and \(\displaystyle\int_{c}^{b}f(x)dx\) must converge separately.
When studying \(\displaystyle\int_{-\infty}^{\infty}f(x) dx\) the convergence
One may consider the Principal Value \(\displaystyle\lim_{R \to \infty}\int_{-R}^{R}f(x) dx\)
Caution: The existence of this limit does not imply the convergence.
For example, \(f(x)=x, F(x)=\frac{x^2}{2}\), then \(\displaystyle\lim_{R \to \infty}\int_{-R}^{R}x dx = \lim_{R \to \infty}\left[ \frac{x^{2}}{2} \right]_{-R}^{R}\)
\(= \lim_{R \to \infty} \frac{R^2}{2} - \frac{(-R)^2}{2} = 0\). But \(\displaystyle\int_{-\infty}^{\infty}x dx\) is not convergent.(split into sum of two integral and each is divergent)
Sometimes it is difficult to determine if a given improper integral converges or not.
The idea is then to compare it with an integral that we already know its behaviour.
This will help us to determine the convergence of the integral, but rarely its values in case it converges.
To compare integrals, we assume that the functions are non-negative.
Theorem
Let \(f\) be a locally integrable function on the interval \([a, b)\). We define and assume \(f(x) \geq 0 ,\forall x \in [a, b)\).
Then \(\displaystyle\int_{a}^{b}f(x)dx\) converges if the function \(\displaystyle F(x) = \int_{a}^{x}f(t)dt\) is bounded in \([a, b)\) and diverges if the function \(F(x)\) is not bounded.
These are the only possibilities and \(\displaystyle\int_{a}^{b}f(x)dx = \sup_{a \le x < b}F(x)\) in either case.
Proof
Since \(f(x)\geq 0\), \(\forall x\in[a,b)\), then \(F(x)\) is increasing(Easy proof)
Thus \(\displaystyle\lim_{c\to b^{-}}\int^c_af(t)dt=\lim_{c\to b^-}F(c)=\sup_{x\in[a,b)}F(x)\)
Then if \(F\) is bounded, then it is converges. If \(F\) is unbounded, then it is diverges