24 Integration of Convergent Series and Swapping Limit and Integrals, Improper Integrals

Lecture from: 27.05.2024 | Video: Video ETHZ

Review: Partial Fraction Decomposition – Breaking Down Complex Fractions

Quick Recap: The Big Idea of Partial Fractions When we have a rational function, which is a ratio of two polynomials $\frac{P(x)}{Q(x)}$, and we want to integrate it, the strategy is to break it down into a sum of simpler fractions. These simpler pieces are ones we usually know how to integrate (like $\frac{A}{x-a}$ or $\frac{B}{(x-a{)}^2}$, etc.).

Key Assumptions:

1. The degree of the numerator $P(x)$ is strictly less than the degree of the denominator $Q(x)$. (If not, we do polynomial long division first to get a polynomial part plus a new fraction that does satisfy this).
2. We assume $Q(x)$ is monic (its leading coefficient is 1). (If not, we can divide both $P(x)$ and $Q(x)$ by that leading coefficient).

Case 1: “All roots of $Q(x)$ are real”

If all the roots $x_i$ of the denominator $Q(x)$ are real numbers, we can factor $Q(x)$ like this: $Q(x)={\prod }_{i=1}^k(x-x_i{)}^{n_i}$ Here:

• $x_i$ are the distinct real roots of $Q(x)$.
• $n_i$ is the multiplicity of the root $x_i$ (how many times $(x-x_i)$ appears as a factor).

The Setup for Partial Fractions: The theory tells us that we can then write our original fraction $\frac{P(x)}{Q(x)}$ as a sum of terms, where for each distinct root $x_i$ with multiplicity $n_i$, we include terms with denominators $(x-x_i{)}^1,(x-x_i{)}^2,\dots ,(x-x_i{)}^{n_i}$: $\frac{P(x)}{Q(x)}={\sum }_{i=1}^k\left({\sum }_{j=1}^{n_i}\frac{C_{ij}}{(x-x_i{)}^j}\right)$ The $C_{ij}$ are constants that we need to find. The inner sum means that if a root $x_i$ has multiplicity $n_i$, we’ll have $n_i$ terms in our partial fraction expansion corresponding to that root. For example, if $Q(x)=(x-2{)}^3(x-5)$, we’d have terms for $\frac{A}{x-2}$, $\frac{B}{(x-2{)}^2}$, $\frac{D}{(x-2{)}^3}$, and $\frac{E}{x-5}$.

Clicker Question: Setting up a Partial Fraction Decomposition

What is the correct setup for the partial fraction decomposition of $\frac{1}{x^2(x+1)}$? Here, $Q(x)=x^2(x+1)$.

The roots are:

• $x_1=0$ with multiplicity $n_1=2$.
• $x_2=-1$ with multiplicity $n_2=1$.

So the setup is: $\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$ (We need terms for $(x-0{)}^1$ and $(x-0{)}^2$ because of the $x^2$ factor).

Finding the Coefficients: Multiply both sides by the common denominator $x^2(x+1)$: $1=A\cdot x(x+1)+B\cdot (x+1)+C\cdot x^2$ $1=A(x^2+x)+Bx+B+C{x}^2$ $0x^2+0x+1=(A+C)x^2+(A+B)x+B$

Now, we equate the coefficients of the powers of $x$:

• Coefficient of $x^0$ (constant term): $B=1$.
• Coefficient of $x^1$: $A+B=0$. Since $B=1$, $A+1=0⟹A=-1$.
• Coefficient of $x^2$: $A+C=0$. Since $A=-1$, $-1+C=0⟹C=1$.

So, $A=-1,B=1,C=1$.

Integrating: $\int \frac{1}{x^2(x+1)}dx=\int \left(-\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\right)dx$ $=-\int \frac{1}{x}dx+\int x^{-2}dx+\int \frac{1}{x+1}dx$ $=-\ln |x|+\frac{x^{-1}}{-1}+\ln |x+1|+K$ $=-\ln |x|-\frac{1}{x}+\ln |x+1|+K$ (Using $K$ for the constant of integration to avoid confusion with $C$ from the coefficients).

Case 2: ”$Q(x)$ also has non-real (complex) roots”

What if the Denominator Doesn’t Factor Nicely into Real Linear Terms? If $Q(x)$ (our denominator polynomial with real coefficients) has complex roots, they always show up in conjugate pairs. For example, if $\alpha +i\beta$ is a root, then $\alpha -i\beta$ must also be a root. When we multiply the factors corresponding to these conjugate pairs, we get an irreducible quadratic with real coefficients: $(x-(\alpha +i\beta ))(x-(\alpha -i\beta ))=((x-\alpha )-i\beta )((x-\alpha )+i\beta )=(x-\alpha {)}^2-(i\beta {)}^2=(x-\alpha {)}^2+{\beta }^2$. This kind of quadratic (like $x^2+1$ or $x^2-2x+5$) doesn’t have real roots.

If $Q(x)$ has non-real roots, they always appear in complex conjugate pairs (since $Q(x)$ is a real polynomial).

A pair of complex conjugate roots, say $\alpha +i\beta$ and $\alpha -i\beta$ (where $\beta \ne 0$), corresponds to an irreducible quadratic factor in the real factorization of $Q(x)$: $(x-(\alpha +i\beta ))(x-(\alpha -i\beta ))=((x-\alpha )-i\beta )((x-\alpha )+i\beta )=(x-\alpha {)}^2-(i\beta {)}^2=(x-\alpha {)}^2+{\beta }^2$ This quadratic term $(x-\alpha {)}^2+{\beta }^2$ has no real roots (its discriminant is negative).

So, the general real factorization of $Q(x)$ (monic) will look like: $Q(x)=\left({\prod }_{l=1}^e((x-{\alpha }_l{)}^2+{\beta }_l^2{)}^{m_l}\right)\cdot \left({\prod }_{i=1}^k(x-x_i{)}^{n_i}\right)$ where:

• $(x-{\alpha }_l{)}^2+{\beta }_l^2$ are the distinct irreducible quadratic factors, each with multiplicity $m_l$.
• $(x-x_i)$ are the distinct real linear factors, each with multiplicity $n_i$.

The General Setup for Partial Fractions (including complex roots)

The partial fraction decomposition for $\frac{P(x)}{Q(x)}$ (with $\mathrm{deg}(P)<\mathrm{deg}(Q)$) will now include:

1. Terms for each real root $x_i$ with multiplicity $n_i$: ${\sum }_{j=1}^{n_i}\frac{C_{ij}}{(x-x_i{)}^j}$
2. Terms for each irreducible quadratic factor $(x-{\alpha }_l{)}^2+{\beta }_l^2$ with multiplicity $m_l$: ${\sum }_{j=1}^{m_l}\frac{A_{lj}x+B_{lj}}{((x-{\alpha }_l{)}^2+{\beta }_l^2{)}^j}$ (Notice the numerator for quadratic factors is linear: $A_{lj}x+B_{lj}$).

The overall setup is: $\frac{P(x)}{Q(x)}={\sum }_{l=1}^e{\sum }_{j=1}^{m_l}\frac{A_{lj}x+B_{lj}}{((x-{\alpha }_l{)}^2+{\beta }_l^2{)}^j}+{\sum }_{i=1}^k{\sum }_{j=1}^{n_i}\frac{C_{ij}}{(x-x_i{)}^j}$ The coefficients $A_{lj},B_{lj},C_{ij}$ are found by multiplying by the common denominator $Q(x)$ and comparing coefficients of powers of $x$, just like before. It leads to a system of linear equations.

How do we integrate terms like $\frac{Ax+B}{((x-\alpha {)}^2+{\beta }^2{)}^j}$?

First, we try to make the numerator related to the derivative of the “inside” of the quadratic part, or a constant. Write $Ax+B=A(x-\alpha )+(A\alpha +B)$. (This is like completing the square for the linear term).

So the term becomes: $\frac{A(x-\alpha )}{((x-\alpha {)}^2+{\beta }^2{)}^j}+\frac{A\alpha +B}{((x-\alpha {)}^2+{\beta }^2{)}^j}$

1. Integrating the first part: $\int \frac{A(x-\alpha )}{((x-\alpha {)}^2+{\beta }^2{)}^j}dx$. Use substitution: $u=(x-\alpha {)}^2+{\beta }^2$. Then $du=2(x-\alpha )dx$. The integral becomes $\frac{A}{2}\int \frac{1}{u^j}du$, which we know how to integrate (it’s $u^{-j}$, leading to $\frac{u^{-j+1}}{-j+1}$ if $j\ne 1$, or $\ln |u|$ if $j=1$).

2. Integrating the second part: $\int \frac{A\alpha +B}{((x-\alpha {)}^2+{\beta }^2{)}^j}dx=(A\alpha +B)\int \frac{1}{((x-\alpha {)}^2+{\beta }^2{)}^j}dx$. Use substitution: $x-\alpha =\beta u$. Then $dx=\beta du$. The integral becomes $(A\alpha +B)\int \frac{1}{({\beta }^2{u}^2+{\beta }^2{)}^j}(\beta du)=(A\alpha +B)\int \frac{\beta }{{\beta }^{2j}(u^2+1{)}^j}du$ $=\frac{A\alpha +B}{{\beta }^{2j-1}}\int \frac{1}{(u^2+1{)}^j}du$. The integral $\int \frac{1}{(u^2+1{)}^j}du$ is $J_j$ from our earlier discussion (Example 5.2.7 (5)), which can be solved using reduction formulas (often involving integration by parts or other tricks) to eventually get arctangents and rational functions of $u$.

It’s a systematic process, though it can be algebraically intensive!

Example: A Denominator with an Irreducible Quadratic

Let’s integrate $\int \frac{1-x}{x^3+x}dx=\int \frac{1-x}{x(x^2+1)}dx$. Here $Q(x)=x(x^2+1)$.

Roots: $x_1=0$ (real, multiplicity 1). The factor $x^2+1$ is an irreducible quadratic (like $(x-0{)}^2+1^2$, so $\alpha =0,\beta =1$). Multiplicity 1.

Setup: $\frac{1-x}{x(x^2+1)}=\frac{Ax+B}{x^2+1}+\frac{C}{x}$ Multiply by $x(x^2+1)$: $1-x=(Ax+B)x+C(x^2+1)$ $1-x=A{x}^2+Bx+C{x}^2+C$ $0x^2-1x+1=(A+C)x^2+Bx+C$.

Compare coefficients:

• $x^0$: $C=1$.
• $x^1$: $B=-1$.
• $x^2$: $A+C=0$. Since $C=1$, $A+1=0⟹A=-1$.

So, $\frac{1-x}{x(x^2+1)}=\frac{-x-1}{x^2+1}+\frac{1}{x}=-\frac{x+1}{x^2+1}+\frac{1}{x}$. $\int \left(-\frac{x}{x^2+1}-\frac{1}{x^2+1}+\frac{1}{x}\right)dx$

• $\int -\frac{x}{x^2+1}dx$: Let $u=x^2+1,du=2xdx$. This is $-\frac{1}{2}\int \frac{1}{u}du=-\frac{1}{2}\ln |u|=-\frac{1}{2}\ln (x^2+1)$.
• $\int -\frac{1}{x^2+1}dx=-\arctan (x)$.
• $\int \frac{1}{x}dx=\ln |x|$.

So, the integral is $-\frac{1}{2}\ln (x^2+1)-\arctan (x)+\ln |x|+K$.

Integration of Convergent Series

The Big Question: Can we swap the order of limit and integration? That is, if $f_n(x)\to f(x)$, is it true that ${\lim }_{n\to \infty }\int f_n(x)dx=\int ({\lim }_{n\to \infty }{f}_n(x))dx=\int f(x)dx$?

Example: When Swapping Fails with Pointwise Convergence

Consider a sequence of “tent” functions $f_n:[0,2]\to \mathbb{R}$. $f_n(x)=\{\begin{array}{ll}{n}^{2}x& 0\le x\le 1/n\\ 2n-n^{2}x& 1/n\le x\le 2/n\\ 0& 2/n\le x\le 2\end{array}$

Each $f_n(x)$ forms a triangle with base $2/n$ and height $n$. The area of this triangle is $\frac{1}{2}\cdot \text{base}\cdot \text{height}=\frac{1}{2}\cdot \frac{2}{n}\cdot n=1$. So, ${\int }_0^2{f}_n(x)dx=1$ for all $n\ge 1$. Therefore, ${\lim }_{n\to \infty }{\int }_0^2{f}_n(x)dx=1$.

Now, what about the pointwise limit of $f_n(x)$? For any fixed $x\in (0,2]$, if $n$ is large enough such that $2/n<x$, then $f_n(x)=0$. If $x=0$, $f_n(0)=0$ for all $n$. So, $f_n(x)\to 0$ for all $x\in [0,2]$ as $n\to \infty$. The pointwise limit function is $f(x)=0$. Then ${\int }_0^2({\lim }_{n\to \infty }{f}_n(x))dx={\int }_0^{2}0dx=0$.

We have $1=\lim \int f_n\ne \int \lim f_n=0$. So, in this case, we cannot swap the limit and the integral. The problem here is that while $f_n\to 0$ pointwise, the “mass” of the function $f_n$ (its integral) doesn’t go to zero; it just gets squeezed into a narrower and taller spike. This is not uniform convergence.

Theorem: Swapping Limit and Integral with Uniform Convergence

Uniform Convergence Saves the Day! If a sequence of integrable functions $f_n$ converges uniformly to a function $f$, then this “bad behavior” doesn’t happen. The limit function $f$ will also be integrable, and we can swap the limit and integral.

Let $f_n:[a,b]\to \mathbb{R}$ be a sequence of bounded, Riemann integrable functions such that $(f_n{)}_{n\ge 1}$ converges uniformly to $f:[a,b]\to \mathbb{R}$. Then $f$ is also bounded and Riemann integrable, and: ${\lim }_{n\to \infty }{\int }_a^b{f}_n(x)dx={\int }_a^{b}f(x)dx$ In other words: “You can swap $\lim$ and $\int$ under uniform convergence.” ${\lim }_{n\to \infty }{\int }_a^b{f}_n(x)dx={\int }_a^b\left({\lim }_{n\to \infty }{f}_n(x)\right)dx$ (Proof: see script. It uses the $ϵ-\delta$ definition of uniform convergence and properties of integrals.)

Clicker Question: Pointwise Convergence and Integrability

Let $f_n:[a,b]\to \mathbb{R}$ be bounded, integrable, and $f_n\to f$ pointwise as $n\to \infty$, where $f:[a,b]\to \mathbb{R}$ is bounded.

• Is $f$ then integrable? No! Consider the sequence of functions $(f_n)$ where $f_n(x)=1$ if $x$ is one of the first $n$ rational numbers in $[0,1]$ (in some fixed enumeration of $\mathbb{Q}\cap [0,1]$), and $f_n(x)=0$ otherwise. Each $f_n$ is integrable (it’s non-zero at only finitely many points, so $\int f_n=0$). The pointwise limit is $f(x)=\{\begin{array}{ll}1& x\in \mathbb{Q}\cap [0,1]\\ 0& x\in [0,1]\setminus \mathbb{Q}\end{array}$ (the Dirichlet function). We know the Dirichlet function is not integrable. (Compare with Example 5.1.6).

• If $f$ is integrable, does it then hold that ${\lim }_{n\to \infty }{\int }_a^b{f}_n(x)dx={\int }_a^{b}f(x)dx$? No! See the “tent functions” example at the beginning of this note. There, $f_n\to f\equiv 0$ pointwise, $f$ is integrable ($\int 0dx=0$), but $\lim \int f_n=1\ne 0$.

• If $f$ is integrable and $\lim \int f_n=\int f$, does $f_n\to f$ uniformly? No! Consider $f_n(x)=x^n$ on $[0,1]$. Pointwise limit $f(x)=\{\begin{array}{ll}0& 0\le x<1\\ 1& x=1\end{array}$. This $f$ is integrable, ${\int }_0^{1}f(x)dx=0$. ${\int }_0^1{f}_n(x)dx={\int }_0^1{x}^{n}dx=\frac{x^{n+1}}{n+1}{|}_0^1=\frac{1}{n+1}$. ${\lim }_{n\to \infty }{\int }_0^1{f}_n(x)dx={\lim }_{n\to \infty }\frac{1}{n+1}=0$. So, $\lim \int f_n=\int f$ holds. However, $f_n(x)=x^n$ does not converge uniformly to $f(x)$ on $[0,1]$.

• If $f_n\to f$ uniformly, is $f$ integrable and $\lim \int f_n=\int f$? Yes! This is exactly what Theorem 5.5.1 states.

Application: Integration of Power Series Term by Term

Integrating Infinite Polynomials Piece by Piece Since power series converge uniformly on any closed interval strictly inside their radius of convergence, this theorem allows us to integrate a power series term by term, just like we do with finite polynomials.

Corollary: Term-by-Term Integration of Power Series

Let $f(x)={\sum }_{k=0}^{\infty }{c}_k{x}^k$ be a convergent power series with a positive radius of convergence $\rho >0$. Then for any $X\in (-\rho ,\rho )$: ${\int }_0^{X}f(t)dt={\sum }_{k=0}^{\infty }{\int }_0^X{c}_k{t}^{k}dt={\sum }_{k=0}^{\infty }{c}_k{\left[\frac{t^{k+1}}{k+1}\right]}_0^X={\sum }_{k=0}^{\infty }\frac{c_k}{k+1}{X}^{k+1}$ The resulting series for the integral also has the same radius of convergence $\rho$.

Example: The Series for $\ln (1+X)$ (Value of the Alternating Harmonic Series)

We know the geometric series $\frac{1}{1+t}=\frac{1}{1-(-t)}=1-t+t^2-t^3+t^4-\cdots ={\sum }_{k=0}^{\infty }(-1{)}^k{t}^k$. This converges for $|-t|<1$, so for $t\in (-1,1)$. Radius of convergence is $\rho =1$.

Integrate from $0$ to $X$, where $0\le X<1$: ${\int }_0^X\frac{1}{1+t}dt=\ln (1+t){|}_0^X=\ln (1+X)-\ln (1)=\ln (1+X)$.

Using term-by-term integration: $\ln (1+X)={\int }_0^X\left({\sum }_{k=0}^{\infty }(-1{)}^k{t}^k\right)dt={\sum }_{k=0}^{\infty }(-1{)}^k{\int }_0^X{t}^{k}dt$ $={\sum }_{k=0}^{\infty }(-1{)}^k{\left[\frac{t^{k+1}}{k+1}\right]}_0^X={\sum }_{k=0}^{\infty }(-1{)}^k\frac{X^{k+1}}{k+1}$ $=X-\frac{X^2}{2}+\frac{X^3}{3}-\frac{X^4}{4}+\frac{X^5}{5}-\dots$

This series for $\ln (1+X)$ also has radius of convergence 1.

What happens at the boundary $X=1$?

The series becomes $1-1/2+1/3-1/4+\dots$, which is the alternating harmonic series.

We know $\ln (1+X)$ is continuous at $X=1$ (from the left), so ${\lim }_{X\to 1^{-}}\ln (1+X)=\ln (2)$.

Abel’s theorem on power series states that if a power series converges at an endpoint of its interval of convergence, then the function defined by the series is continuous up to that endpoint.

The alternating harmonic series converges (by Leibniz criterion).

Therefore, we can say: $\ln (2)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots ={\sum }_{k=0}^{\infty }(-1{)}^k\frac{1}{k+1}={\sum }_{j=1}^{\infty }(-1{)}^{j-1}\frac{1}{j}$

Example: The Series for $\arctan (X)$

We know $\frac{1}{1+t^2}=1-t^2+t^4-t^6+t^8-\cdots ={\sum }_{k=0}^{\infty }(-1{)}^k{t}^{2k}$. This converges for $|t^2|<1$, so for $t\in (-1,1)$. Radius $\rho =1$.

Integrate from $0$ to $X$, where $0\le X<1$: ${\int }_0^X\frac{1}{1+t^2}dt=\arctan (t){|}_0^X=\arctan (X)-\arctan (0)=\arctan (X)$.

Using term-by-term integration: $\arctan (X)={\sum }_{k=0}^{\infty }(-1{)}^k{\int }_0^X{t}^{2k}dt={\sum }_{k=0}^{\infty }(-1{)}^k{\left[\frac{t^{2k+1}}{2k+1}\right]}_0^X$ $={\sum }_{k=0}^{\infty }(-1{)}^k\frac{X^{2k+1}}{2k+1}=X-\frac{X^3}{3}+\frac{X^5}{5}-\frac{X^7}{7}+\dots$

This series for $\arctan (X)$ also has $\rho =1$. At $X=1$, the series is $1-1/3+1/5-1/7+\dots$. This is Leibniz’s formula for $\pi /4$.

Since $\arctan (X)$ is continuous at $X=1$ and the series converges at $X=1$ (by Leibniz criterion), we have: $\frac{\pi }{4}=\arctan (1)=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$

Improper Integrals

Extending the Idea of Area: Unbounded Regions or Functions So far, we’ve only defined integrals ${\int }_a^{b}f(x)dx$ where the interval $[a,b]$ is finite and the function $f(x)$ is bounded on that interval. “Improper integrals” allow us to consider two new scenarios:

1. Integrating over an unbounded interval (like $[a,\infty )$, $(-\infty ,b]$, or $(-\infty ,\infty )$).
2. Integrating a function that becomes unbounded at one or more points within a finite interval.

Goal: Extension of the integral to unbounded domains or unbounded functions.

Example: How can we understand ${\int }_1^{\infty }\frac{1}{x^2}dx$?

Intuition: An antiderivative of $1/x^2=x^{-2}$ is $-1/x=-x^{-1}$. So maybe ${\int }_1^{\infty }\frac{1}{x^2}dx={\left[-\frac{1}{x}\right]}_1^{\infty }={\lim }_{x\to \infty }(-\frac{1}{x})-(-\frac{1}{1})=0-(-1)=1$?

This looks promising!

Definition: Improper Integrals

Let $J\subseteq \mathbb{R}$ be an interval, and $f:J\to \mathbb{R}$ be a function that is bounded and Riemann integrable on every compact (closed and bounded) subinterval $[a,b]\subseteq J$. (Often, $f$ is simply continuous on $J$).

Let $A=\inf (J)$ and $B=\sup (J)$ (where $A$ can be $-\infty$ and $B$ can be $\infty$).

• Case 1: “J is open or unbounded to the right” (e.g., $J=[1,\infty )$, $J=[0,1)$, $J=(0,1)$). This means $A\in J$ but $B\notin J$ (if $J$ is $(A,B)$ or $[A,B)$) or $B=\infty$. $f$ is (improperly) integrable on $J$ if the limit ${\lim }_{b\to B^{-}}{\int }_A^{b}f(x)dx$ exists in $\mathbb{R}$ (i.e., is a finite real number). If it exists, we define ${\int }_{J}f(x)dx:={\lim }_{b\to B^{-}}{\int }_A^{b}f(x)dx$.

• Case 2: “J is open or unbounded to the left” (e.g., $J=(-\infty ,0]$, $J=(0,1]$, $J=(0,1)$). This means $A\notin J$ (if $J$ is $(A,B)$ or $(A,B]$) or $A=-\infty$, but $B\in J$. $f$ is (improperly) integrable on $J$ if the limit ${\lim }_{a\to A^{+}}{\int }_a^{B}f(x)dx$ exists in $\mathbb{R}$. If it exists, we define ${\int }_{J}f(x)dx:={\lim }_{a\to A^{+}}{\int }_a^{B}f(x)dx$.

• Case 3: “J is open or unbounded on both sides” (e.g., $J=(-\infty ,\infty )$, $J=\mathbb{R}$, $J=(0,1)$). This means $A\notin J$ and $B\notin J$. $f$ is (improperly) integrable on $J$ if for some (any) $c\in J$, both improper integrals ${\int }_A^{c}f(x)dx$ (as in Case 2) and ${\int }_c^{B}f(x)dx$ (as in Case 1) exist in $\mathbb{R}$. In this case, we define: ${\int }_{J}f(x)dx:={\int }_A^{c}f(x)dx+{\int }_c^{B}f(x)dx$ (The value of this improper integral is then independent of the choice of $c\in J$).

Convergence/Divergence: If an improper integral ${\int }_{J}f(x)dx$ exists in $\mathbb{R}$ (i.e., is finite), we say the improper integral converges. Otherwise, we say it diverges.

Examples of Improper Integrals

1. ${\int }_0^{\infty }{e}^{-x}dx$: This is Case 1 with $A=0,B=\infty$. ${\int }_0^{\infty }{e}^{-x}dx={\lim }_{b\to \infty }{\int }_0^b{e}^{-x}dx$ ${\int }_0^b{e}^{-x}dx=[-e^{-x}{]}_0^b=(-e^{-b})-(-e^0)=-e^{-b}+1=1-e^{-b}$. ${\lim }_{b\to \infty }(1-e^{-b})=1-0=1$ So, ${\int }_0^{\infty }{e}^{-x}dx=1$ (it converges).

2. ${\int }_1^{\infty }\frac{1}{x^{\alpha }}dx$ for $\alpha \in \mathbb{R}$. This is Case 1 with $A=1,B=\infty$. Consider ${\int }_1^b\frac{1}{x^{\alpha }}dx={\int }_1^b{x}^{-\alpha }dx$.

   • If $\alpha =1$: ${\int }_1^b{x}^{-1}dx=[\ln |x|{]}_1^b=\ln (b)-\ln (1)=\ln (b)$. ${\lim }_{b\to \infty }\ln (b)=\infty$. So for $\alpha =1$, the integral diverges.
   • If $\alpha \ne 1$: ${\int }_1^b{x}^{-\alpha }dx={\left[\frac{x^{-\alpha +1}}{-\alpha +1}\right]}_1^b=\frac{b^{1-\alpha }}{1-\alpha }-\frac{1^{1-\alpha }}{1-\alpha }=\frac{1}{1-\alpha }(b^{1-\alpha }-1)$. Now we need to look at ${\lim }_{b\to \infty }{b}^{1-\alpha }$.
     • If $1-\alpha >0$ (i.e., $\alpha <1$): ${\lim }_{b\to \infty }{b}^{1-\alpha }=\infty$. Integral diverges.
     • If $1-\alpha <0$ (i.e., $\alpha >1$): ${\lim }_{b\to \infty }{b}^{1-\alpha }={\lim }_{b\to \infty }\frac{1}{b^{\alpha -1}}=0$. In this case, the limit of the integral is $\frac{1}{1-\alpha }(0-1)=\frac{-1}{1-\alpha }=\frac{1}{\alpha -1}$.

   Conclusion for ${\int }_1^{\infty }\frac{1}{x^{\alpha }}dx$:

   • Converges to $\frac{1}{\alpha -1}$ if $\alpha >1$.
   • Diverges if $\alpha \le 1$.

3. ${\int }_0^1\frac{1}{x^{\alpha }}dx$ for $\alpha \in \mathbb{R}$. Here the potential problem is at $x=0$ if $\alpha >0$. This is Case 2 with $A=0,B=1$. Consider ${\lim }_{ϵ\to 0^{+}}{\int }_{ϵ}^1\frac{1}{x^{\alpha }}dx$.

   • If $\alpha =1$: ${\int }_{ϵ}^1{x}^{-1}dx=[\ln |x|{]}_{ϵ}^1=\ln (1)-\ln (ϵ)=-\ln (ϵ)$. ${\lim }_{ϵ\to 0^{+}}(-\ln (ϵ))=\infty$. So for $\alpha =1$, the integral diverges.
   • If $\alpha \ne 1$: ${\int }_{ϵ}^1{x}^{-\alpha }dx={\left[\frac{x^{1-\alpha }}{1-\alpha }\right]}_{ϵ}^1=\frac{1^{1-\alpha }}{1-\alpha }-\frac{{ϵ}^{1-\alpha }}{1-\alpha }=\frac{1}{1-\alpha }(1-{ϵ}^{1-\alpha })$. Now we need ${\lim }_{ϵ\to 0^{+}}{ϵ}^{1-\alpha }$.
     • If $1-\alpha >0$ (i.e., $\alpha <1$): ${\lim }_{ϵ\to 0^{+}}{ϵ}^{1-\alpha }=0$. The limit of the integral is $\frac{1}{1-\alpha }(1-0)=\frac{1}{1-\alpha }$.
     • If $1-\alpha <0$ (i.e., $\alpha >1$): ${\lim }_{ϵ\to 0^{+}}{ϵ}^{1-\alpha }={\lim }_{ϵ\to 0^{+}}\frac{1}{{ϵ}^{\alpha -1}}=\infty$. Integral diverges.

   Conclusion for ${\int }_0^1\frac{1}{x^{\alpha }}dx$:

   • Converges to $\frac{1}{1-\alpha }$ if $\alpha <1$.
   • Diverges if $\alpha \ge 1$.

Continue here: 25 Integral Test for Series Convergence, Stirling’s Formula to Approximate Factorial, Summary