Integration

1. Riemann integration and criteria

Exam focus: Lebesgue's criterion (bounded + discontinuities of measure zero) decides Riemann integrability instantly. Know Dirichlet, Thomae, and that composition can destroy integrability.

Lec-39 Integrability

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

The Riemann criterion; watch for the role of upper and lower sums.

Lec-39 Riemann/Riemann Stieltjes Integral

NPTEL · A Basic Course in Real Analysis (IIT Kharagpur)

Part C2023 Decembershow ▾

Previous-year question — unlock to view

Part C2023 Decembershow ▾

For a real number

λ

, consider the improper integrals

I_{λ} = \int_{0}^{1}

/ (1 - x)^{λ}

and

K_{λ} = \int_{1}^{\infty}

/ x^{λ}

. Which of the following statements are true?

A.There exists $λ$ such that $I_{λ}$ converges, but $K_{λ}$ does not converge.✓
B.There exists $λ$ such that $K_{λ}$ converges, but $I_{λ}$ does not converge.✓
C.There exists $λ$ such that $I_{λ}, K_{λ}$ both converge.
D.There exists $λ$ such that neither $I_{λ}$ nor $K_{λ}$ converges.✓

Solution

$I_{λ}$ converges iff $λ < 1; K_{λ}$ converges iff $λ > 1$ . So they never converge together, and at $λ = 1$ neither does.

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Part Cshow ▾

Which of the following functions on [0, 1] are Riemann integrable?

A.The Dirichlet function: f(x) = 1 for x rational, f(x) = 0 for x irrational.
B.Thomae's function: f(p/q) = 1/q in lowest terms, f(x) = 0 for x irrational, f(0) = 1.✓
C.f(x) = sin(1/x) for x ∈ (0, 1], f(0) = 0.✓
D.f(x) = 1/x for x ∈ (0, 1], f(0) = 0.

Solution

Lebesgue's criterion: a bounded function on [0,1] is Riemann integrable exactly when its set of discontinuities has measure zero. (1) Discontinuous everywhere — not integrable. (2) Continuous at every irrational and discontinuous only on $Q, a$ countable hence null set — integrable, with integral 0. (3) Bounded by 1 and discontinuous only at x = 0 — integrable. (4) Unbounded, so it is not Riemann integrable at all; the improper integral also diverges.

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Part Bshow ▾

The limit

l i m_{n \to \infty}

\sum_{k = 1}^{n}

k / n^{2}

equals

A.0
B.1/2✓
C.1
D.2

Solution

$\sum k / n^{2} = (1/ n) \sum (k / n)$ is a Riemann sum of $\int_{0}^{1} x$ dx = 1/2. Directly: $n (n + 1) / (2 n^{2}) \to 1/2$ .

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Part Bshow ▾

The improper integral

\int_{0}^{\infty}

x^{p - 1}

e^{- x}

dx converges if and only if

A.p > 0✓
B.p > 1
C.p ≥ 0
D.all real p

Solution

Near 0 the integrand behaves like $x^{p - 1}$ , integrable iff p > 0; at $\infty$ the exponential wins for every p. This is $Γ (p)$ .

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Part Bshow ▾

Let

f : [0, 1] \to R

be Riemann integrable and

g : R \to R

continuous. Which of the following is necessarily Riemann integrable on [0,1]?

A.f ∘ g (when defined)
B.g ∘ f✓
C.The function h with h(x) = 1 if $f (x) \in Q$ and 0 otherwise
D.f restricted to $Q \cap [0, 1]$ , extended by 0

Solution

Continuous outer function preserves integrability (discontinuities of g∘f ⊆ those of f).

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2. Improper integrals and convergence tests

Exam focus: Split at every singular point and at infinity, compare with 1/xᵖ, and distinguish convergence from absolute convergence — ∫ sin x/x is the standard example of the gap.

Lec-45 Improper Integrals

NPTEL · A Basic Course in Real Analysis (IIT Kharagpur)

Both kinds of improperness — unbounded interval and unbounded integrand.

Lec-46 Convergence Test for Improper Integrals

NPTEL · A Basic Course in Real Analysis (IIT Kharagpur)

Comparison and the p-thresholds you need at sight.

Improper integrals — thresholds and tests free

The two reference families

$\int_{1}^{\infty}$ dx/xᵖ converges $\Leftrightarrow * * p > 1 * * . \int_{0}^{1}$ dx/xᵖ converges ⇔ p < 1.

So $\int_{0}^{\infty}$ dx/xᵖ never converges — the exam uses this constantly. Similarly $\int_{0}^{1}$ dx $/ (1 - x)^{λ}$ converges iff $λ < 1$ , while $\int_{1}^{\infty}$ dx $/ x^{λ}$ converges iff $λ > 1$ : the two conditions are incompatible.

Logarithmic refinement: $\int_{2}^{\infty}$ dx/(x(log x)^p) converges ⇔ p > 1.

Tests

Comparison / limit comparison with xᵖ near each singular point.
Absolute convergence ⇒ convergence. The converse fails.
**Dirichlet $* * : \int f g$ with f monotone → 0 and $\int g$ bounded converges — this is how $\int_{1}^{\infty} sin x / x$ dx converges.
Split the integral at every singularity and at $\infty$ before testing; each piece must converge separately.

Standard values

Integral	Value
$\int_{0}^{\infty}$ $e^{- x^{2}}$ dx	$\sqrt π /2$
$\int_{0}^{\infty} sin x / x$ dx	$π /2 ($ conditionally)
$\int_{0}^{\infty}$ $x^{s - 1} e^{- x}$ dx	$Γ (s)$ , converges for s > 0
$\int_{0}^{1}$ $x^{a - 1} (1 - x)^{b - 1}$ dx	$B (a, b) = Γ (a) Γ (b) /Γ (a + b)$ , needs a, b > 0
$\int_{0}^{\infty}$ dx $/ (1 + x^{2})$	$π /2$
$\int_{0}^{\infty}$ $x^{a - 1} / (1 + x)$ dx	$π / sin (π a), 0 < a < 1$

The conditional-convergence gap

$\int_{1}^{\infty} sin x / x$ dx converges but $\int_{1}^{\infty}$ |sin x|/x dx $= \infty$ . The integral analogue of $\sum (- 1)^{n} / n$ .

Note also: convergence of $\int_{0}^{\infty} f$ does not force f(x) → 0. A function with tall thin spikes of area $2^{- n}$ at each integer converges yet is unbounded.

Key takeaways

p > 1 at infinity, p < 1 at the origin — and never both.
Conditional convergence is real for integrals too (sin x/x). $- \int f$ converging says nothing about f tending to 0.

Integration

1. Riemann integration and criteria

Riemann sums closing on the integralinteractivefree

Riemann integrability — the criterion and the zoo

Three equivalent criteria (f bounded on [a,b])

The zoo

2. Improper integrals and convergence tests

The two exponents that decide most improper integralsinteractive

Improper integrals — thresholds and tests free

The two reference families

Tests

Standard values

The conditional-convergence gap

Key takeaways