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Improper integration.

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Improper integration

Improper integrals arise when the interval of integration is unbounded, $\int_{a}^{∞} f(x)dx = \lim_{N \to ∞}\int_{a}^{N} f(x)dx$.

An improper integral converges if the limit defining it exists -total area [a, ∞) under the curve f(x) is finite-. It is also possible for an improper integral to diverge to infinity. In that case, one may assign the value of ∞ (or −∞) to the integral. However, other improper integrals may simply diverge in no particular direction. This is called divergence by oscillation.

k > 0, $\int_{0}^{∞} e^{-kx}dx = \lim_{N \to ∞}\int_{0}^{N} e^{-kx}dx = \lim_{N \to ∞}\frac{-1}{k}e^{-kx}\bigg|_{0}^{N}$.

Therefore, $\int_{0}^{∞} e^{-kx}dx =\lim_{N \to ∞}\frac{-1}{k}e^{-kN}-(\frac{-1}{k}) = \frac{1}{k}$.

Example. $\int_{1}^{∞} \frac{dx}{x} = \lim_{N \to ∞}\int_{1}^{N} \frac{dx}{x} = \lim_{N \to ∞} ln(x)\bigg|_{1}^{N} = $. And now, it is very easy to show that it diverges.

$\lim_{N \to ∞} lnN -ln1 = \lim_{N \to ∞} lnN = ∞$

$\int_{1}^{∞} \frac{dx}{x^p} = \lim_{N \to ∞}\int_{1}^{N} \frac{dx}{x^p} = \lim_{N \to ∞} \frac{x^{-p+1}}{-p+1}\bigg|_{1}^{N}$

$= \lim_{N \to ∞} \frac{N^{-p+1}}{-p+1}-\frac{1^{-p+1}}{-p+1}$. Therefore, there are three cases:

  1. p = 1, $\int_{1}^{∞} \frac{dx}{x} = ∞$, it diverges.
  2. p < 1, $\int_{1}^{∞} \frac{dx}{x^p} = ∞$, it diverges.
  3. p > 1, $\int_{1}^{∞} \frac{dx}{x^p} = -\frac{1}{-p+1} = \frac{1}{p-1}$, so it converges.

Limit Comparison

Limit or asymptotic comparison. Assume f, g are positive functions. If f(x) ≈ g(x) as x → ∞, that is, $\lim_{x \to ∞} \frac{f(x)}{g(x)} = c$, c ≠ 0 then $\int_{a}^{∞} f(x)dx$ and $\int_{a}^{∞} g(x)dx$ both improper integrals converge or both diverge.

Proof. $\lim_{x \to ∞} \frac{f(x)}{g(x)} = c$ ⇒ ∃a: |f(x)g(x)-c| < c/2 ∀x>a ⇒ -c/2 < f(x)g(x)-c < c/2 ⇒ -cg(x)/2 < f(x)-cg(x) < cg(x)/2 ⇒[Adding +cg(x)] c2g(x) < f(x) < 3cg(x)/2 ∀x>a ⇒ $\frac{c}{2}\int_{a}^{∞} g(x)dx < \int_{a}^{∞} f(x)dx < \frac{3c}{2}\int_{a}^{∞} g(x)dx$. This shows both integrals converge or both diverge, QED.

Example. $\int_{0}^{∞} \frac{dx}{\sqrt{x^2+10}}$. Consider that $\sqrt{x^2+10}≈\sqrt{x^2}=x$ as x → ∞, $\int_{0}^{1} \frac{dx}{\sqrt{x^2+10}}$ is a certain (finite) number, and $\int_{1}^{∞} \frac{dx}{x}$ diverges, so our original improper integral diverges.

$\int_{0}^{∞} \frac{dx}{\sqrt{x^3+7}}$ [$\frac{1}{\sqrt{x^3+7}}≈\frac{1}{\sqrt{x^3}}=\frac{1}{x^{3/2}}$] ≈ $\int_{0}^{∞} \frac{dx}{x^{3/2}}$ convergent.

$\int_{-∞}^{∞} e^{-x^2}dx =$[By symmetry] $2\int_{0}^{∞} e^{-x^2}dx ≤$ [x ≥ 1 ⇒ x2≥ x ⇒ -x2≤ -x ⇒ $e^{-x^2}≤e^{-x}$] $2\int_{0}^{1} e^{-x^2}dx +2\int_{1}^{∞} e^{-x}dx$, and we know that $2\int_{0}^{1} e^{-x^2}dx$ is a finite number and $+2\int_{1}^{∞} e^{-x}dx$ converges, so the original improper integrate converges.

Improper integrals of second type

One may be tempted to calculate $\int_{-1}^{1} \frac{dx}{x^2}=-x^{-1}\bigg|_{-1}^{1} = -2$, but this is wrong because 😞😞😞

$\int_{0}^{1} \frac{dx}{x^2}$ diverges.

An improper integral of the second kind arises when the integrand becomes unbounded or approaches infinity within the interval of integration. The integral $\int_{0}^{1} f(x)dx = \lim_{a \to 0^+} \int_{a}^{1} f(x)dx$ converges if the limit exists, and diverges if the limit does not exist (Figure 1.a.).  

$\int_{0}^{1} x^{\frac{-1}{2}}dx = \lim_{a \to 0^+} \int_{a}^{1} x^{\frac{-1}{2}}dx = \lim_{a \to 0^+} 2x^{1/2}\bigg|_{a}^{1}$

$= \lim_{a \to 0^+} 2 -2\sqrt{a} = 2$, so the improper integral $\int_{0}^{1} \frac{dx}{\sqrt{x}}$ converges and is equal to 2 (Figure 1.b.).  

$\int_{0}^{1} \frac{dx}{x} = \lim_{a \to 0^+} \int_{a}^{1} \frac{dx}{x} = \lim_{a \to 0^+} lnx\bigg|_{a}^{1}$

$ = \lim_{a \to 0^+} ln1 -lna = \lim_{a \to 0^+} -lna = -(-∞) = ∞$, so this improper integral diverges (Figure 1.c.).  

The idea is quite simple, $\frac{1}{x^{1/2}} « \frac{1}{x}« \frac{1}{x^2}$ as x → 0+, the first integral is convergent (small enough), but the other two improper integral diverge. Similarly, $\frac{1}{x^{1/2}} » \frac{1}{x} » \frac{1}{x^2}$ as x → ∞, the only one that is convergent is the last one.

$\int_{0}^{∞} \frac{dx}{(x-3)^2} = \int_{0}^{5} \frac{dx}{(x-3)^2} + \int_{5}^{∞} \frac{dx}{(x-3)^2}$. The integral $\int_{0}^{∞} \frac{dx}{(x-3)^2}$ is improper because there is an infinite discontinuity (the function spikes up to infinity at 3 from both sides) at x = 3.

$\int_{5}^{∞} \frac{dx}{(x-3)^2} = \lim_{b \to ∞} \int_{5}^{b} \frac{dx}{(x-3)^2} = \lim_{b \to ∞} \frac{-1}{x-3}\bigg|_{5}^{b} = $

$\lim_{b \to ∞} (\frac{-1}{b-3}+\frac{1}{5-3}) = \frac{1}{2}$. Hence, $\int_{5}^{∞} \frac{dx}{(x-3)^2}$ converges, and its value is 1/2. However,

$\int_{0}^{3} \frac{dx}{(x-3)^2} = \lim_{c \to 3^-} \int_{0}^{c} \frac{dx}{(x-3)^2} = \lim_{c \to 3^-} \frac{-1}{x-3}\bigg|_{0}^{c} = $

$\lim_{c \to 3^-} (\frac{-1}{c-3}+\frac{1}{0-3}) = ∞$. So, the improper integral $\int_{0}^{3} \frac{dx}{(x-3)^2}$ diverges to infinite.

Similarly, $\int_{3}^{5} \frac{dx}{(x-3)^2} = \lim_{c \to 3^+} \int_{c}^{5} \frac{dx}{(x-3)^2} = \lim_{c \to 3^+} \frac{-1}{x-3}\bigg|_{c}^{5} = $

$\lim_{c \to 3^+} (\frac{-1}{5-3}+\frac{1}{c-3}) = ∞$. So, the improper integral $\int_{3}^{5} \frac{dx}{(x-3)^2}$ diverges to infinite ⇒ $\int_{0}^{∞} \frac{dx}{(x-3)^2}$ diverges (Figure 1.d).  


This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
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