This is not what it was meant to be, but my crash course in purgatory. I should have known better. I should have been wiser. The more things seems to change, the more they stay the same. Everything just keeps going round and round, people chasing the wind and clutching at clouds and shadows, an insane game of smoke and mirrors. Life is not just a mess mirage in a world ruled by chaos, lies, and deception, but a merry-go-round where we all go around in circles before the final curtains close. And there are never really happy endings, that’s not how the story goes, but love is always new, awesome, and makes the ride worthwhile, Desperate verses III, M. Anawim, #justtothepoint.

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:

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

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)] ^{c}⁄_{2}g(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 ⇒ x^{2}≥ x ⇒ -x^{2}≤ -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.

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.).

- $\int_{0}^{1} \frac{dx}{\sqrt{x}}$. We notice that there is a vertical asymptote at x = 0, so it is an integral improper.

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

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

- $\int_{0}^{1} \frac{dx}{x}$. We realize that there is an infinite discontinuity at x = 0, so we need to evaluate this improper integral by taking a limit as a parameter approaches the point of discontinuity.

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

$ = \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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