Improper integrals — Type I (infinite limits), Type II (discontinuous integrand)

If $f$ is continuous on $[a,\infty)$ : $\int_a^{\infty} f(x)\,dx \;=\; \lim_{t\to\infty}\int_a^{t} f(x)\,dx$ If $f$ is continuous on $(-\infty,b]$ : $\int_{-\infty}^{b} f(x)\,dx \;=\; \lim_{t\to-\infty}\int_{t}^{b} f(x)\,dx$ If both limits are infinite, split at any convenient point $c$ : $\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^{c} f(x)\,dx + \int_{c}^{\infty} f(x)\,dx$ The integral converges if the limit exists and is finite; otherwise it diverges. For the double-infinite case, both pieces must converge independently.

HOW to derive the master result $\int_1^\infty \frac{1}{x^p}\,dx$

This single result is the 80/20 powerhouse — it underlies almost every convergence test.

Derivation from scratch.

Step 1 — replace $\infty$ with $t$ . $\int_1^{\infty}\frac{dx}{x^p}=\lim_{t\to\infty}\int_1^{t}x^{-p}\,dx.$ Why this step? We're forbidden from integrating to $\infty$ ; $t$ is a real upper limit we can handle with FTC.

Step 2 — antiderivative (case $p\ne1$ ). Power rule: $\int x^{-p}dx=\dfrac{x^{-p+1}}{-p+1}=\dfrac{x^{1-p}}{1-p}$ . $\int_1^t x^{-p}dx=\frac{t^{1-p}-1}{1-p}.$ Why this step? Apply FTC on the finite interval $[1,t]$ — totally legal now.

Step 3 — take the limit. As $t\to\infty$ , behaviour of $t^{1-p}$ decides everything:

If $p>1$ : exponent $1-p<0$ , so $t^{1-p}\to 0$ . Result $=\dfrac{0-1}{1-p}=\dfrac{1}{p-1}$ ✓ converges.
If $p<1$ : exponent $1-p>0$ , so $t^{1-p}\to\infty$ → diverges.

Step 4 — the borderline $p=1$ . Power rule fails; use $\int x^{-1}dx=\ln x$ : $\lim_{t\to\infty}\big[\ln t-\ln 1\big]=\lim_{t\to\infty}\ln t=\infty \Rightarrow \textbf{diverges}.$ Why this matters: $1/x$ is the knife-edge — it just barely loses. This is why $p=1$ is the famous boundary.

Figure — Improper integrals — Type I (infinite limits), Type II (discontinuous integrand)

Type II — Discontinuous integrand (blow-up)

Derivation (case $p\ne1$ ). Blow-up at left end $0$ : $\int_0^1 x^{-p}dx=\lim_{t\to0^+}\frac{x^{1-p}}{1-p}\Big|_t^1=\lim_{t\to0^+}\frac{1-t^{1-p}}{1-p}.$ Why this step? The spike is at $0$ , so we creep up from $t>0$ .

$p<1$ : $1-p>0$ → $t^{1-p}\to0$ → answer $\dfrac{1}{1-p}$ converges.
$p>1$ : $1-p<0$ → $t^{1-p}\to\infty$ → diverges.
$p=1$ : $\int_0^1\frac{dx}{x}=\lim_{t\to0^+}(\ln1-\ln t)=+\infty$ diverges.

Worked examples

Recall Feynman: explain it to a 12-year-old

Imagine painting under a curve. Normally the strip of paint has a left and right wall, so the paint is finite. Now I knock down the right wall — the strip goes on forever. Surprisingly, sometimes you still need only a finite bucket of paint, because the strip gets thinner and thinner so fast. Other times the bucket never fills — it leaks to infinity. The trick to decide: paint up to a movable wall at position $t$ , see how much paint that needs, then slide the wall away to infinity and watch whether the paint total settles on a number or keeps growing. Type II is the same game, but instead of a far-away wall it's a spot where the curve shoots up to the sky — we tiptoe toward that spot and see if the skinny-but-tall sliver still holds finite paint.

Flashcards

What makes an integral "Type I" improper?

At least one limit of integration is infinite (

\pm\infty

What makes an integral "Type II" improper?

The integrand is unbounded (has a vertical asymptote) somewhere on the interval.

How do you handle

\int_a^\infty f\,dx

Replace

\infty

with

t

\lim_{t\to\infty}\int_a^t f\,dx

, then check if the limit is finite.

For

\int_1^\infty x^{-p}dx

, the convergence condition is?

Converges iff

p>1

; value is

\frac{1}{p-1}

For

\int_0^1 x^{-p}dx

, the convergence condition is?

Converges iff

p<1

; value is

\frac{1}{1-p}

At which

p

does the

1/x^p

integral diverge in BOTH the Type I and Type II

p

-cases?

p=1

(the borderline; gives

\ln

, which diverges).

For

\int_{-\infty}^\infty f

, what must hold to "converge"?

Split at any

c

; BOTH halves must converge independently (not just a symmetric limit).

Why is the symmetric-limit answer not "convergence"?

It's the Cauchy principal value; it can be finite even when the true improper integral diverges (e.g.

\int_{-\infty}^\infty x\,dx

Value of

\int_0^\infty e^{-x}dx

1

Value of

\int_0^1 \frac{dx}{\sqrt x}

2

(Type II,

p=1/2<1

, converges).

What is the red-flag that you missed an interior singularity?

A positive integrand giving a negative or impossible answer (e.g.

\int_{-1}^1 x^{-2}dx

"="

-2

State the Comparison Test idea for improper integrals.

0\le f\le g

and

\int g

converges, then

\int f

converges; if

\int f

diverges, then

\int g

diverges.

Connections

Fundamental Theorem of Calculus — the legality limit we extend with $t\to$ bad point.
Limits at Infinity — every Type I reduces to $\lim_{t\to\infty}$ .
Comparison Test for Integrals — built directly on the $p$ -integral.
p-series and Integral Test — $\sum 1/n^p$ converges by the same $p>1$ rule.
Laplace Transform & Gamma Function — defined by convergent Type I integrals.
Probability Density Functions — require $\int_{-\infty}^\infty f=1$ to converge.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, normal definite integral $\int_a^b f\,dx$ tabhi kaam karta hai jab interval finite ho aur function bounded ho. Improper integral wahan aata hai jab ya to limit infinite ho (Type I), ya function kisi point pe phat jaaye yaani vertical asymptote ho (Type II). Trick simple hai: hum kabhi seedha infinity tak ya blow-up ke through integrate nahi karte. Hum us "kharab" point ko ek variable $t$ se replace karte hain, normal integrate karte hain, phir $t$ ko us kharab point ki taraf limit me le jaate hain. Agar answer ek finite number pe settle ho gaya — converge, warna diverge.

Sabse important cheez ek hi hai, jise yaad kar lo to 80% sawaal ban jaate hain: $\int_1^\infty \frac{1}{x^p}dx$ converge karta hai sirf jab $p>1$ , aur $\int_0^1 \frac{1}{x^p}dx$ converge karta hai sirf jab $p<1$ . Yaad rakhne ka mantra: "infinity pe bada $p$ chahiye, zero pe chhota $p$ chahiye". Aur $p=1$ dono jagah akela loser hai — wahan $\ln$ aata hai jo infinity tak chala jaata hai.

Ek common galti: $\int_{-1}^1 \frac{1}{x^2}dx$ me log seedha antiderivative laga ke $-2$ nikaal dete hain. Galat! Kyunki $x=0$ pe function phat raha hai aur wo interval ke andar hai — FTC wahan apply nahi hota. Positive function ka area kabhi negative nahi ho sakta, yahi red flag hai. Sahi tareeka: $0$ pe split karo aur dono halves alag-alag check karo (yahan dono diverge karte hain).

Jab integrate karna mushkil ho (jaise $\frac{1}{x^3+1}$ ), to comparison test lagao: agar $0\le f\le g$ aur $\int g$ converge karta hai, to $\int f$ bhi zaroor karega. Yeh poora concept Laplace transform, Gamma function aur probability density (jahan total area $=1$ chahiye) me directly use hota hai — isliye yeh foundation pakka karna zaroori hai.

Improper integrals — Type I (infinite limits), Type II (discontinuous integrand)

WHY do we even need this?

Type I — Infinite limits

HOW to derive the master result $\int_1^\infty \frac{1}{x^p}\,dx$

Type II — Discontinuous integrand (blow-up)

Worked examples

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus II — Integration

Connections