151 lines
5 KiB
Markdown
151 lines
5 KiB
Markdown
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# Improper integrals
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Proper integrals are [definite integrals](integration.md/#the-definite-integral) where the integrand $f$ is *continuous* on a *closed, finite* interval $[a,b]$. For positive $f$ it corresponds to the area of a **bounded region** of the plane, a region contained inside some disk of finite radius with centre at the origin. To extend the definite integral by allowing for two possibilities excluded in the situation described above.
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1. There may exist $a=-\infty$ or $b=\infty$ or both.
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2. $f$ may be unbounded as $x$ approaches $a$ or $b$ or both.
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Integrals satisfying 1. are called **improper integrals of type I.** and integrals satisfying 2. are called **improper integrals of type II**.
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## Improper integrals of type I
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If $f$ is continuous on $[a,\infty]$, the improper integral of $f$ over $[a,\infty]$ is defined as a limit of proper integrals:
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$$
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\int_a^\infty f(x)dx = \lim_{R \to \infty} \int_a^R f(x)dx.
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$$
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Similarly, if $f$ is continuous on $[-\infty,b]$, then the improper integrals is defined as:
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$$
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\int_{-\infty}^b f(x)dx = \lim_{R \to -\infty} \int_R^b f(x)dx.
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$$
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In either case, if the limit exists, the improper integral **converges**. If the limit does not exist, the improper integral **diverges**. If the limit is $\infty$ or $-\infty$, the proper integral **diverges to (negative) infinity**.
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## Improper integrals of type II
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If $f$ is continuous on the interval $(a,b]$ and is possibly unbounded near $a$, the improper integral may be defined as:
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$$
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\int_a^b f(x)dx = \lim_{c \downarrow a} \int_c^b f(x)dx.
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$$
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Similarly, if $f$ is continuous on $[a,b)$ and is possibly unbounded near $b$, the improper integral may be defined as:
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$$
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\int_a^b f(x)dx = \lim_{c \uparrow b} \int_a^c f(x)dx.
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$$
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These improper integrals may also converge, diverge or diverge to (negative) infinity.
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## p-integrals
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Summerizing the behaviour of improper integrals of types I and II for powers of $x$ if $0 < a< \infty$, then:
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1.
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$$
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\int_a^\infty x^{-p}dx =
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\begin{cases}
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\text{converges to } \frac{a^{1-p}}{p-1} \quad \text{if } p > 1, \\
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\text{diverges to } \infty \quad \text{if } p \leq 1.
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\end{cases}
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$$
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2.
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$$
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\int_0^a x^{-p}dx =
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\begin{cases}
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\text{converges to } \frac{a^{1-p}}{1-p} \quad \text{if } p < 1, \\
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\text{diverges to } \infty \quad \text{if } p \geq 1.
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\end{cases}
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$$
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**Proof of 1:**
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For $p=1$:
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$$
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\int_a^\infty x^{-1}dx = \lim_{R \to \infty} \int_a^R x^{-1}dx = \lim_{R \to \infty} (\ln R - \ln a) = \infty.
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$$
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For $p < 1$:
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$$
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\begin{array}{ll}
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\int_a^\infty x^{-p}dx &= \lim_{R \to \infty} \int_a^R x^{-p}dx, \\
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&= \lim_{R \to \infty} [\frac{x^{-p+1}}{-p+1}]_a^R, \\
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&= \lim_{R \to \infty} \frac{R^{1-p}-a^{1-p}}{1-p} = \infty.
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\end{array}
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$$
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For $p > 1$
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$$
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\begin{array}{ll}
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\int_a^\infty x^{-p}dx &= \lim_{R \to \infty} \int_a^R x^{-p}dx, \\
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&= \lim_{R \to \infty} [\frac{x^{-p+1}}{-p+1}]_a^R, \\
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&= \lim_{R \to \infty} \frac{a^{-(p-1)}-R^{-(p-1)}}{p-1} = \frac{a^{1-p}}{p-1}.
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\end{array}
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$$
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**Proof of 2:**
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For $p=1$:
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$$
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\int_0^a x^{-1}dx = \lim_{c \space\downarrow\space 0} \int_c^a x^{-1}dx = \lim_{c \space\downarrow\space 0} (\ln a - \ln c) = \infty.
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$$
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For $p > 1$
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$$
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\begin{array}{ll}
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\int_0^a x^{-p}dx &= \lim_{c \space\downarrow\space 0} \int_c^a x^{-p}dx, \\
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&= \lim_{c \space\downarrow\space 0} [\frac{x^{-p+1}}{-p+1}]_c^a, \\
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&= \lim_{c \space\downarrow\space 0} \frac{c^{-(p-1)} - a^{-(p-1)}}{p-1} = \infty.
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\end{array}
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$$
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For $p < 1$:
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$$
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\begin{array}{ll}
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\int_0^a x^{-p}dx &= \lim_{c \space\downarrow\space 0} \int_c^a x^{-p} dx, \\
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&= \lim_{c \space\downarrow\space 0} [\frac{x^{-p+1}}{-p+1}]_c^a, \\
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&= \lim_{c \space\downarrow\space 0} \frac{a^{1-p}-c^{1-p}}{1-p} = \frac{a^{1-p}}{1-p}.
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\end{array}
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$$
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## Comparison theorem for integrals
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Let $-\infty \leq a < b \leq \infty$, and suppose that functions $f$ and $g$ are continuous on the interval $(a,b)$ and satisfy $0 \leq f(x) \leq g(x)$. If $\int_a^b g(x)dx$ converges, then so does $\int_a^b f(x)dx$, and:
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$$
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\int_a^b f(x)dx \leq \int_a^b g(x)dx.
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$$
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Equivalently, if $\int_a^b f(x)dx$ diverges to $\infty$, then so does $\int_a^b g(x)dx$.
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**Proof:** Since both integrands are nonnegative, there are only two possibilities for each integral: it can either converge to a nonnegative number or diverge to $\infty$. Since $f(x) \leq g(x)$ on $(a,b)$, it follows by the [properties of the definite integral](integration.md/#properties) that if $a < r < s < b$, then:
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$$
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\int_r^s f(x)dx \leq \int_r^s g(x)dx.
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$$
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By taking limits as $r \space\downarrow\space a$ and $s \space\uparrow\space b$.
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### To prove convergence
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To find a function $g$, such that
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1. $\forall x \in [a,\infty], \space 0 \leq f(x) \leq g(x)$.
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2. $\int_0^\infty g(x)dx$ is convergent.
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### To prove divergence
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To find a function $f$ such that
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1. $\forall x \in [a,\infty], \space g(x) \geq f(x) \geq 0$.
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2. $\int_0^\infty f(x)dx$ is divergent.
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