Improper integration.

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 messy 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.

Recall

A definite integral is a mathematical concept used to find the area under a curve between two fixed limits. It is represented as $\int_{a}^{b}f(x)dx$, where a and b are the lower and upper limits, and f(x) is the integrand.

A definite integral of a function is defined as the area of the region bounded by the function's graph or curve between two points, say a and b, and the x-axis. The integral of a real-valued function f(x) on an interval [a, b] is written or expressed as $\int_{a}^{b} f(x)dx$ and it is illustrated in Figure 1.a.

The Fundamental Theorem of Calculus states roughly that the integral of a function f over an interval is equal to the change of any antiderivate F (F'(x) = f(x)) between the ends of the interval, i.e., $\int_{a}^{b} f(x)dx = F(b)-F(a)=F(x) \bigg|_{a}^{b}$

Improper integration

Definite integrals have specific limits of integration which determine the net area under the curve of the function between these limits. Improper integrals arise when the interval of integration is unbounded, that is, $\int_{a}^{∞} f(x)dx = \lim_{b \to ∞}\int_{a}^{b} f(x)dx$, or when the integrand is undefined at certain points within the interval of integration. Techniques for evaluating improper integrals involve taking limits as the integration bounds approach infinity or as singularities are approached.

An improper integral converges if the limit defining it exists, that is, the total area [a, ∞) under the curve f(x) is finite. In other words, improper integrals are said to be convergent if the limit is finite and that limit is the value of the improper integral.

An improper integral diverges if the limit defining it does not exist 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.

Solved exercises

• $\int_{-∞}^{∞} \frac{1}{1+x^2}dx = \int_{-∞}^{0} \frac{1}{1+x^2}dx + \int_{0}^{∞} \frac{1}{1+x^2}dx$.

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

$lim_{a \to -∞} arctan(0)-arctan(a) = lim_{a \to -∞} 0-arctan(a) = \frac{π}{2} ⇒$[By symmetry] $\int_{-∞}^{∞} \frac{1}{1+x^2}dx =\frac{π}{2}+\frac{π}{2} = π.$

• $\int_{1}^{∞} \frac{e^{\frac{1}{x}}}{x^2}dx$ [$u = \frac{1}{x}, du = \frac{-1}{x^2}dx,$ x = 1 → u = 1, x = ∞ → u = 0] $\int_{1}^{0} -e^udu$ =[Properties of integrals] $\int_{0}^{1} e^udu = e^u\bigg|_{0}^{1} = e-e^0 = e -1.$

• $\int_{1}^{∞} \frac{ln(x)}{x^2}dx$ [Integration by parts, u = ln(x), $v’ = \frac{1}{x^2}dx, u’ = \frac{1}{x}dx, v = -x^{-1}$] $\lim_{b \to ∞} ln(x)(-x^{-1})-\int_{1}^{b} -x^{-1}\frac{1}{x}dx = \lim_{b \to ∞} -\frac{ln(x)}{x}+\int_{1}^{b}\frac{1}{x^2}dx = \lim_{b \to ∞} -\frac{ln(x)}{x}-\frac{1}{x}\bigg|_{1}^{b}$

$\lim_{b \to ∞} (-\frac{ln(b)}{b}-\frac{1}{b}) - (-\frac{ln(1)}{1}-\frac{1}{1}) = \lim_{b \to ∞} (-\frac{ln(b)}{b}-\frac{1}{b})-(-0-1)=0-(0-1)=1.$

By L’Hospital Rule, $\lim_{b \to ∞} (\frac{ln(b)}{b}) = \lim_{b \to ∞} \frac{\frac{1}{b}}{1} = \lim_{b \to ∞} \frac{1}{b} = 0$

• $\int_{0}^{2} \frac{2x}{x^2-4}dx$. The reader should realize that the function is not even defined when x = 2 because the denominator of $\frac{2x}{x^2-4}$ is zero when x = 2, $\int_{0}^{2} \frac{2x}{x^2-4}dx = \lim_{b \to 2⁻} \int_{0}^{b} \frac{2x}{x^2-4}dx =$[Substitution u=x²-4, 2xdx = du] $\lim_{b \to 2⁻} \int_{0}^{b} \frac{du}{u} = \lim_{b \to 2⁻} ln|u| = \lim_{b \to 2⁻} ln|x^2-4|\bigg|_{0}^{b}$ =

$\lim_{b \to 2⁻} ln|b^2-4| -ln(4)=-∞,$ so the integral diverges.

• $\int_{0}^{∞} \frac{x^2}{\sqrt{x^3+4}}dx = $[U-substitution, u = x³+4, du = 3x²dx, x = 0 ⇒ u = 4, x = ∞ ⇒ u = ∞] $\int_{4}^{∞} \frac{1}{3}\frac{du}{\sqrt{u}}du = \lim_{b \to ∞} \int_{4}^{b} \frac{1}{3}\frac{du}{\sqrt{u}}du = \lim_{b \to ∞} \frac{1}{3}·2 u^{\frac{1}{2}}\bigg|_{4}^{b} =$

$= \frac{2}{3} \lim_{b \to ∞} (\sqrt{b}-{\sqrt{4}})= \frac{2}{3}(∞ -2) = ∞.$

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

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

• $\int_{1}^{∞} \frac{dx}{x} = \lim_{b \to ∞}\int_{1}^{b} \frac{dx}{x} = \lim_{b \to ∞} ln(x)\bigg|_{1}^{b} = $.

$\lim_{b \to ∞} lnb -ln1 = \lim_{b \to ∞} lnb = ∞$, so our original integral diverges.

• $\int_{0}^{∞} sin(x)dx = \lim_{b \to ∞} \int_{0}^{b}sin(x)dx = \lim_{b \to ∞} (-cos(x))\bigg|_{0}^{b}$

However, this is equal to $= \lim_{b \to ∞} (-cos(b)+cos(0)) = \lim_{b \to ∞} -cos(b)+1$, this limit does not exist, this improper integral is divergent.

• $\int_{-∞}^{0} xe^xdx = \lim_{a \to -∞} \int_{a}^{0} xe^xdx$ =[Integration by parts, u = x, v’=e^xdx, u’ = dx, v = e^x] $\lim_{a \to -∞} xe^x-\int_{a}^{0} e^xdx = \lim_{a \to -∞} (xe^x-e^x)dx = \lim_{a \to -∞} e^x(x-1)dx\bigg|_{a}^{0} = $

$ = \lim_{a \to -∞} (e^0(0-1))-(e^a(a-1)) = \lim_{a \to -∞} -1-ae^a+e^a = -1 -\lim_{a \to -∞} ae^a +0$. However, $\lim_{a \to -∞} ae^a = \lim_{a \to -∞} \frac{a}{e^{-a}}$[L’Hôpital’s, -∞/∞] $\lim_{a \to -∞} \frac{1}{-e^{-a}} = 0$, hence $\int_{-∞}^{0} xe^xdx = -1 -0 + 0 = -1.$

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

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

1. p = 1, $\int_{1}^{∞} \frac{dx}{x} = ∞$, it diverges (it was solved previously).
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.

• $\int_{0}^{∞} \frac{x}{1+x^4}dx$ =[U-Substitution, u = x², du = 2xdx] $\frac{1}{2}·\int_{0}^{∞}\frac{du}{1+u^2} = \frac{1}{2}·\lim_{b \to ∞} \int_{0}^{b} \frac{du}{1+u^2} = \frac{1}{2}·\lim_{b \to ∞} tan^{-1}(u)\bigg|_{0}^{b} = $

$= \frac{1}{2}·\lim_{b \to ∞}(tan^{-1}(b)-tan^{-1}(0)) = \frac{1}{2}·\lim_{b \to ∞}(tan^{-1}(b)-0) = \frac{1}{2}·tan^{-1}(∞) = \frac{1}{2}·\frac{π}{2} = \frac{π}{4}.$

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
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6. Contemporary Abstract Algebra, Joseph, A. Gallian.
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9. SFU, Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.