University Mathematics — Year 1 · Bachelor Year 1

15Integration on a Segment

The integral of the High School volume was founded on areas taken intuitively. This chapter constructs it: first for step functions, where the integral is a finite sum, then for continuous (and piecewise continuous) functions by uniform approximation — the place where Heine’s theorem (Theorem 13.22) earns its keep. The fundamental theorem of calculus then connects the construction to primitives, and Riemann sums connect it to discrete averages.

Throughout, $a < b$ are reals.

15.1 Step functions

Definition 15.1

$\varphi \colon \intcc{a}{b} \to \R$ is a step function when there is a subdivision $a = x_0 < x_1 < \dots < x_n = b$ such that $\varphi$ is constant, equal to $c_i$ , on each open interval $\intoo{x_{i-1}}{x_i}$ (values at the nodes are unconstrained). Its integral is

\int_a^b \varphi = \sum_{i=1}^{n} c_i\,(x_i - x_{i-1}),

independent of the chosen subdivision (refine two subdivisions by their common one: each side is unchanged under refinement).

Proposition 15.2

On step functions, the integral is linear, increasing ( $\varphi \leq \psi \implies \int\varphi \leq \int\psi$ ), and satisfies the Chasles relation $\int_a^b = \int_a^c + \int_c^b$ for $a < c < b$ .

Proof. The engine is refinement invariance, stated in the definition: inserting one extra node $t \in \intoo{x_{i-1}}{x_i}$ into a subdivision replaces the term $c_i(x_i - x_{i-1})$ by $c_i(t - x_{i-1}) + c_i(x_i - t)$ — the same number — so the integral is unchanged by any finite refinement. Now take $\varphi$ with subdivision $\sigma$ and $\psi$ with subdivision $\sigma'$ : on the common refinement $\sigma \cup \sigma'$ both are step functions with the same nodes, and on each piece $\varphi + \lambda\psi$ is constant equal to $c_i + \lambda d_i$ : linearity reduces to linearity of finite sums. Increase: $c_i \leq d_i$ on each piece gives $\sum c_i \Delta_i \leq \sum d_i \Delta_i$ (lengths $\Delta_i \geq 0$ ). Chasles: insert the node $c$ and split the sum at it. ∎

15.2 Integral of a continuous function

Theorem 15.3 (Uniform approximation)

Let $f$ be continuous on $\intcc{a}{b}$ . For every $\varepsilon > 0$ there are step functions $\varphi, \psi$ with

\varphi \leq f \leq \psi \qquad\text{and}\qquad \psi - \varphi \leq \varepsilon \text{ on } \intcc{a}{b}.

Proof. By Heine’s theorem (Theorem 13.22), $f$ is uniformly continuous: pick $\delta$ for $\varepsilon$ , and a subdivision of mesh $< \delta$ (equally spaced, say, with $n > \frac{b - a}{\delta}$ pieces). On each closed piece $\intcc{x_{i-1}}{x_i}$ , $f$ attains a minimum $m_i$ and a maximum $M_i$ (Theorem 13.13), and $M_i - m_i \leq \varepsilon$ (the two extremal points are within $\delta$ ). Define $\varphi = m_i$ and $\psi = M_i$ on $\intoo{x_{i-1}}{x_i}$ (and $\varphi = \psi = f$ at the nodes). ∎

Theorem 15.4 (Definition of the integral)

Let $f$ be continuous on $\intcc{a}{b}$ . The two numbers

I_-(f) = \sup\Bigl\{\int_a^b \varphi : \varphi \text{ step},\ \varphi \leq f\Bigr\}, \qquad I_+(f) = \inf\Bigl\{\int_a^b \psi : \psi \text{ step},\ \psi \geq f\Bigr\}

are equal; their common value is the integral $\int_a^b f$ (also written $\int_a^b f(t)\,\dd t$ ). It coincides with the previous notion on step functions, and extends to piecewise continuous functions by splitting $\intcc{a}{b}$ at the discontinuities (Chasles as a definition there).

Proof. Both sets are nonempty ( $f$ is bounded) and every lower step integral is $\leq$ every upper one (monotonicity on step functions): so $I_-(f) \leq I_+(f)$ . By Theorem 15.3, for every $\varepsilon$ there is a pair with $\int\psi - \int\varphi \leq \varepsilon(b - a)$ : the sup and inf are squeezed together, $I_- = I_+$ . ∎

Example 15.5 (Piecewise continuous, no drama)

The floor function on $\intcc{0}{3}$ is a step function in disguise: splitting at its jumps,

\int_0^3 \lfloor t \rfloor\,\dd t = \int_0^1 0 + \int_1^2 1 + \int_2^3 2 = 0 + 1 + 2 = 3 ,

and the values at the jump points $1, 2$ are irrelevant: changing a function at finitely many points changes no integral (the framing step functions are unaffected). This is the whole content of the “piecewise continuous” extension: cut at the finitely many discontinuities, integrate each continuous piece, add — Chasles as a definition.

Example 15.6 (The definition computes, once)

Let $f(x) = x$ on $\intcc{0}{1}$ and cut into $n$ equal pieces. The best step functions constant on the pieces are $\varphi = \frac{k-1}{n}$ and $\psi = \frac kn$ on the $k$ -th piece, with

\int_0^1 \varphi = \sum_{k=1}^{n} \frac{k-1}{n}\cdot\frac1n = \frac{n-1}{2n}, \qquad \int_0^1 \psi = \sum_{k=1}^{n} \frac{k}{n}\cdot\frac1n = \frac{n+1}{2n} .

Every lower integral is $\leq I_-(f) \leq I_+(f) \leq$ every upper one, so $\frac{n-1}{2n} \leq I_-(f) \leq I_+(f) \leq \frac{n+1}{2n}$ for all $n$ : both squeeze onto $\frac12$ , and $\int_0^1 x\,\dd x = \frac12$ straight from the definition. The closing insight: this is the first and last time we integrate from the definition — the fundamental theorem below replaces all such computations by one antiderivative lookup, which is the entire economic point of this chapter.

Theorem 15.7 (Properties)

For $f, g$ continuous (or piecewise continuous) on $\intcc{a}{b}$ and $\lambda \in \R$ :

linearity: $\int (f + \lambda g) = \int f + \lambda \int g$ ;
monotonicity: $f \leq g \implies \int_a^b f \leq \int_a^b g$ ; and $\bigl|\int_a^b f\bigr| \leq \int_a^b \abs f \leq (b - a)\, \sup\abs f$ ;
Chasles: $\int_a^b = \int_a^c + \int_c^b$ (with the convention $\int_b^a = -\int_a^b$ , valid for any order of the bounds);
strict positivity: if $f$ is continuous, $f \geq 0$ and $\int_a^b f = 0$ , then $f = 0$ everywhere on $\intcc{a}{b}$ .

Proof. (1)–(3) pass from step functions to the limit through the sup/inf definition. Linearity deserves the details once: given $\varepsilon > 0$ , frame $\varphi_f \leq f \leq \psi_f$ and $\varphi_g \leq g \leq \psi_g$ with gaps $\leq \varepsilon$ (Theorem 15.3). For $\lambda \geq 0$ , $\varphi_f + \lambda\varphi_g \leq f + \lambda g \leq \psi_f + \lambda\psi_g$ is a framing by step functions with gap $\leq (1 + \lambda)\varepsilon$ , and its step integrals equal $\int \varphi_f + \lambda\int\varphi_g$ etc. (Proposition 15.2): letting $\varepsilon \to 0$ squeezes $\int(f + \lambda g)$ onto $\int f + \lambda\int g$ . For $\lambda < 0$ , multiplying by $\lambda$ reverses the framing of $g$ — the lower step function of $\lambda g$ is $\lambda \psi_g$ — and the same squeeze runs with the roles swapped. The bound $\abs{\int f} \leq \int \abs f$ comes from $-\abs f \leq f \leq \abs f$ and monotonicity.

(4) Contrapositive: if $f(x_0) = m > 0$ , continuity provides a subinterval of length $\eta > 0$ on which $f \geq \frac m2$ ; the step function worth $\frac m2$ there and $0$ elsewhere is $\leq f$ , so $\int f \geq \frac{m\eta}{2} > 0$ . ∎

Example 15.8 (Chasles at work: integrals with absolute values)

To integrate an absolute value, cut where the sign changes.

\int_0^2 \abs{x - 1}\,\dd x = \int_0^1 (1 - x)\,\dd x + \int_1^2 (x - 1)\,\dd x = \frac12 + \frac12 = 1 ,

and, cutting $\intcc{0}{2\pi}$ at $\pi$ :

\int_0^{2\pi} \abs{\sin t}\,\dd t = \int_0^{\pi} \sin t\,\dd t - \int_{\pi}^{2\pi} \sin t\,\dd t = 2 + 2 = 4 ,

while $\int_0^{2\pi} \sin t\,\dd t = 0$ : cancellation is real, and this is why the strict positivity statement (Theorem 15.7 (4)) carries the hypothesis $f \geq 0$ — without it, a vanishing integral proves nothing about $f$ . The closing insight: $\int \abs f$ measures area, $\int f$ measures signed balance; the inequality $\abs{\int f} \leq \int\abs f$ is the exact record of what cancellation can destroy.

15.3 The fundamental theorem of calculus

Theorem 15.9 (Fundamental theorem of calculus)

Let $f$ be continuous on an interval $I$ and $a \in I$ . The function

F(x) = \int_a^x f(t)\, \dd t

is of class $C^1$ on $I$ , with $F' = f$ : every continuous function on an interval has primitives. Consequently, for any primitive $G$ of $f$ :

\int_a^b f(t)\,\dd t = G(b) - G(a) .

Proof. Fix $x_0 \in I$ and $\varepsilon > 0$ ; continuity at $x_0$ provides $\delta$ with $\abs{f(t) - f(x_0)} \leq \varepsilon$ for $\abs{t - x_0} \leq \delta$ . For $0 < \abs{h} \leq \delta$ (and $x_0 + h \in I$ ), Chasles gives

\frac{F(x_0 + h) - F(x_0)}{h} - f(x_0) = \frac 1h \int_{x_0}^{x_0+h} \bigl(f(t) - f(x_0)\bigr)\dd t ,

whose absolute value is $\leq \frac{1}{\abs h}\cdot \abs h\, \varepsilon = \varepsilon$ (bound (2), valid for either order of the bounds). So $F'(x_0) = f(x_0)$ ; $F' = f$ is continuous: $F$ is $C^1$ . If $G' = f$ too, then $(G - F)' = 0$ on the interval, so $G = F + c$ (Corollary 14.12), and $G(b) - G(a) = F(b) - F(a) = \int_a^b f$ . ∎

Example 15.10 (Symmetry before computation)

On a symmetric interval, parity does the work: if $f$ is odd, the substitution $t \mapsto -t$ sends $\int_{-a}^{0} f$ to $-\int_0^a f$ , so

\int_{-a}^{a} f(t)\,\dd t = 0 ; \qquad\text{if $f$ is even,}\quad \int_{-a}^{a} f = 2\int_0^a f .

Thus $\int_{-1}^{1} \frac{t^3\cos t}{1 + t^4}\,\dd t = 0$ with no primitive in sight (the integrand is odd), and $\int_{-\pi}^{\pi} t^2\cos t\,\dd t = 2\int_0^\pi t^2\cos t\, \dd t$ . Check symmetry before reaching for techniques: the fastest integral is the one never computed.

Example 15.11 (Recognizing a derivative on sight)

Compute $\displaystyle\int_0^{\pi/2} \frac{\dd x}{1 + \cos x}$ . The half-angle identity $1 + \cos x = 2\cos^2\frac x2$ turns the integrand into $\frac{1}{2}\bigl(1 + \tan^2\frac x2\bigr)$ , which is exactly the derivative of $\tan\frac x2$ :

\int_0^{\pi/2} \frac{\dd x}{1 + \cos x} = \Bigl[\tan\frac x2\Bigr]_0^{\pi/2} = \tan\frac\pi4 = 1 .

No substitution machinery was needed — only the reflex of reading an integrand as somebody’s derivative, the fundamental theorem doing the rest. (The systematic tool behind such trigonometric integrals, the substitution $t = \tan\frac x2$ , belongs to the standard toolkit built from Theorem 15.15 (2).)

Example 15.12 (Functions defined by integrals)

The fundamental theorem manufactures functions. Let

F(x) = \int_0^x \eu^{-t^2}\,\dd t .

No combination of classical functions has derivative $\eu^{-t^2}$ (a theorem of Liouville, admitted); yet $F$ exists, is $C^1$ with $F'(x) = \eu^{-x^2} > 0$ , strictly increasing, odd (substitute $t \mapsto -t$ ), and bounded: for $x \geq 1$ ,

F(x) - F(1) = \int_1^x \eu^{-t^2}\dd t \leq \int_1^x \eu^{-t}\dd t \leq \eu^{-1} ,

so $F \leq F(1) + \eu^{-1} \leq 1 + \eu^{-1}$ . (The exact limit, $\frac{\sqrt\pi}{2}$ , is computed with double integrals in the Year 3 volume.) Chain rule for moving bounds: $\frac{\dd}{\dd x}\int_x^{x^2} \eu^{-t^2}\dd t = 2x\,\eu^{-x^4} - \eu^{-x^2}$ . The closing insight: integration creates new functions from old ones, with all their properties readable from the integrand — the primitive you cannot write down is still a function you fully control.

Example 15.13 (Estimating without evaluating)

The integrals $R_n = \int_0^1 \frac{t^n}{1 + t}\,\dd t$ have no pleasant closed form, yet monotonicity pins them precisely: on $\intcc{0}{1}$ , $\frac12 \leq \frac{1}{1+t} \leq 1$ , so

\frac{1}{2(n+1)} = \frac12\int_0^1 t^n\,\dd t \;\leq\; R_n \;\leq\; \int_0^1 t^n \,\dd t = \frac{1}{n+1} :

the exact order of decay ( $R_n \sim$ a multiple of $\frac1n$ , in fact $R_n \sim \frac{1}{2n}$ ) with two lines and no antiderivative. The weekend problems of this chapter and the next run on exactly such framings — the analyst’s first instinct before an integral should be bound it, and only then, if needed, compute it.

Example 15.14 (Average values)

The mean of a continuous $f$ over $\intcc{a}{b}$ is $\frac{1}{b-a}\int_a^b f$ . For the arch of the sine:

\frac{1}{\pi}\int_0^\pi \sin t\,\dd t = \frac{1}{\pi}\bigl[-\cos t\bigr]_0^\pi = \frac{2}{\pi} \approx 0.637 :

a full positive arch averages not to $\frac12$ but to $\frac2\pi$ — the curve spends more time high than a triangle would. By Exercise 15.11 (mean value theorem for integrals, $g = 1$ ), the mean is a value: $\sin c = \frac2\pi$ for some $c \in \intoo{0}{\pi}$ . And by the Riemann sums of this chapter, the mean is the limit of ordinary averages of $n$ samples — the bridge between the discrete mean of data and the continuous mean of a signal, which is how the integral enters physics.

Theorem 15.15 (Integration by parts; substitution)

If $u, v$ are $C^1$ on $\intcc{a}{b}$ :

$\int_a^b u'v = \bigl[uv\bigr]_a^b - \int_a^b uv' .$
If $\varphi$ is $C^1$ on $\intcc{\alpha}{\beta}$ and $f$ continuous on $\varphi(\intcc{\alpha}{\beta})$ :

$\int_{\alpha}^{\beta} f\bigl(\varphi(t)\bigr)\,\varphi'(t)\, \dd t = \int_{\varphi(\alpha)}^{\varphi(\beta)} f(x)\, \dd x .$

Proof. (1) $(uv)' = u'v + uv'$ ; integrate over $\intcc{a}{b}$ and apply the fundamental theorem to the $C^1$ function $uv$ .

(2) Let $F$ be a primitive of $f$ on the image interval (Theorem 15.9). Then $(F \circ \varphi)' = (f \circ \varphi)\,\varphi'$ (chain rule), so both sides equal $F(\varphi(\beta)) - F(\varphi(\alpha))$ . ∎

Example 15.16

$\int_0^1 t\,\eu^t \dd t = \bigl[t\,\eu^t\bigr]_0^1 - \int_0^1 \eu^t\dd t = \eu - (\eu - 1) = 1$ . And with the substitution $x = \sin t$ ( $t \in \intcc{0}{\frac\pi2}$ ):

\int_0^1 \sqrt{1 - x^2}\, \dd x = \int_0^{\pi/2} \cos t \cdot \cos t \, \dd t = \int_0^{\pi/2} \frac{1 + \cos 2t}{2}\, \dd t = \frac\pi4 ,

— a quarter of the unit disk, as geometry demands. (Linearization from Method 3.11 at work.)

Remark 15.17 (Common pitfalls in integral calculus)

(i) Substitutions must be $C^1$ on the whole interval: the change $x = \frac1t$ is illegal across $0$ ; applied blindly to $\int_{-1}^{1}\frac{\dd x}{1 + x^2}$ it “proves” that the integral equals its own negative. When a substitution has a singularity, cut the interval first (Chasles), substitute on each piece, and only then recombine. (ii) Logarithmic primitives need absolute values: $\int \frac{\dd x}{x - 2} = \ln\abs{x - 2} + C$ on each side of $2$ separately — writing $\ln(x - 2)$ on $\intoo{0}{1}$ is writing the logarithm of a negative number; and the constant $C$ may differ on the two sides of the singularity. (iii) A vanishing integral does not kill the function: $\int_0^{2\pi}\sin = 0$ ; positivity of the integrand is required before concluding $f = 0$ (Example 15.8). (iv) Riemann sums must be calibrated: in $\frac{b-a}{n}\sum f\bigl(a + k\frac{b-a}{n}\bigr)$ , the step outside and the points inside must match the same subdivision — the frequent error is a sum $\sum_{k=1}^{n} f\bigl(\frac kn\bigr)$ without the factor $\frac1n$ , which diverges instead of converging to $\int_0^1 f$ . Checklist before invoking Theorem 15.20: factor out $\frac1n$ , rewrite the summand as $f$ of $\frac kn$ , name $f$ and check its continuity.

Example 15.18 (Guess, differentiate, adjust)

What is $\int_1^x (\ln t)^2\,\dd t$ ? Guess a primitive of the form $t\,P(\ln t)$ with $P$ polynomial and differentiate:

\bigl(t\,P(\ln t)\bigr)' = P(\ln t) + P'(\ln t) .

We need $P(u) + P'(u) = u^2$ : take $P(u) = u^2 - 2u + 2$ (matching coefficients downward from $u^2$ ). Hence

\int_1^x (\ln t)^2\,\dd t = \bigl[t\bigl((\ln t)^2 - 2\ln t + 2\bigr)\bigr]_1^x = x(\ln x)^2 - 2x\ln x + 2x - 2 ,

a result otherwise reached by two integrations by parts. The closing insight: for integrands of the form (polynomial in $\ln t$ ) or (polynomial times $\eu^{\lambda t}$ ), the primitive has the same shape — differentiating a shaped guess converts integration into linear algebra on coefficients, faster and less error-prone than iterated parts.

Example 15.19 (The boomerang integral)

Compute $I = \int_0^{\pi/2} \eu^x \cos x\,\dd x$ . Integrate by parts twice, differentiating the trigonometric factor each time:

I = \bigl[\eu^x \sin x\bigr]_0^{\pi/2} - \int_0^{\pi/2} \eu^x \sin x\,\dd x = \eu^{\pi/2} - J, \qquad J = \bigl[-\eu^x\cos x\bigr]_0^{\pi/2} + \int_0^{\pi/2} \eu^x\cos x\,\dd x = 1 + I .

The integral has returned to itself: $I = \eu^{\pi/2} - 1 - I$ , whence

I = \frac{\eu^{\pi/2} - 1}{2} .

The closing insight: when the integrand is a product of two functions that reproduce themselves under differentiation ( $\eu^{ax}$ , $\cos bx$ , $\sin bx$ ), two integrations by parts produce a linear equation for the unknown integral — solve it instead of integrating; equivalently, pass through $\eu^{(a + \iu b)x}$ (Chapter 3) and take real parts. Both roads give the same answer, and checking that they do is a free sanity test.

15.4 Riemann sums

Theorem 15.20 (Riemann sums)

Let $f$ be continuous on $\intcc{a}{b}$ . Then

S_n = \frac{b - a}{n} \sum_{k=0}^{n-1} f\Bigl(a + k\,\frac{b-a}{n}\Bigr) \xrightarrow[n \to \infty]{} \int_a^b f(t)\, \dd t ,

and likewise with any evaluation points inside the subintervals.

Proof. $S_n$ is the integral of the step function $\varphi_n$ equal to $f(a + k\frac{b-a}{n})$ on the $k$ -th subinterval. Given $\varepsilon > 0$ , uniform continuity (Heine) provides $\delta$ ; for $n > \frac{b-a}{\delta}$ , every point of a subinterval is within $\delta$ of its evaluation point, so $\abs{f - \varphi_n} \leq \varepsilon$ on $\intcc{a}{b}$ , whence

\Bigl| \int_a^b f - S_n \Bigr| = \Bigl| \int_a^b (f - \varphi_n) \Bigr| \leq (b-a)\,\varepsilon . \qedhere

∎

A left Riemann sum with n = 8 rectangles: as the mesh shrinks, uniform continuity forces the staircase area toward ∈t_ab f. — A left Riemann sum with $n = 8$ rectangles: as the mesh shrinks, uniform continuity forces the staircase area toward $\int_a^b f$ .

Example 15.21

$\displaystyle\sum_{k=1}^{n} \frac{1}{n + k} = \frac 1n \sum_{k=1}^{n} \frac{1}{1 + k/n} \xrightarrow[n\to\infty]{} \int_0^1 \frac{\dd x}{1 + x} = \ln 2$ : a limit invisible to elementary bounds, transparent as a Riemann sum.

Example 15.22 (A second Riemann sum, with calibration)

Find $\lim_{n\to\infty} \sum_{k=1}^{n} \dfrac{n}{(n+k)^2}$ . Calibrate:

\sum_{k=1}^{n} \frac{n}{(n+k)^2} = \frac{1}{n}\sum_{k=1}^{n} \frac{n^2}{(n+k)^2} = \frac1n \sum_{k=1}^{n} \frac{1}{\bigl(1 + \frac kn\bigr)^2} ,

a Riemann sum of the continuous $f(x) = \frac{1}{(1+x)^2}$ on $\intcc{0}{1}$ : the limit is

\int_0^1 \frac{\dd x}{(1 + x)^2} = \Bigl[-\frac{1}{1+x}\Bigr]_0^1 = \frac12 .

The closing insight: the whole art is the middle line — force the summand into the shape $f(\frac kn)$ at the cost of extracting exactly one factor $\frac1n$ ; once the shape is right, the theorem does the analysis and the fundamental theorem does the arithmetic.

Remark 15.23 (Where the integral works next)

The chapter’s constructions each have a sequel. Riemann sums return in Chapter 17 as the bridge between series and integrals (comparison of $\sum \frac{1}{n^\alpha}$ with $\int \frac{\dd t}{t^\alpha}$ ); the integral remainder is the sharpest form of Taylor’s formula in Chapter 16; the $\sup$ -based definition is the prototype for the Lebesgue integral of the Year 3 volume, where the same three properties (linearity, monotonicity, a convergence theorem) are rebuilt on a far larger class of functions. And the weekend problem below turns integration by parts into arithmetic: the irrationality of $\pi^2$ .

15.5 Exercises

Exercise 15.1 ★

Compute: $\displaystyle\int_0^1 \frac{\dd x}{x^2 - 4}$ (partial fractions, Chapter 9); $\displaystyle\int_1^{\eu} \ln t \,\dd t$ ; $\displaystyle\int_0^{\pi} t \sin t\, \dd t$ .

Solution

Solution of Exercise 15.1.

$\dfrac{1}{x^2-4} = \dfrac{1/4}{x - 2} - \dfrac{1/4}{x+2}$ (cover-up), so

\int_0^1 \frac{\dd x}{x^2 - 4} = \frac14\Bigl[\ln\abs{x-2} - \ln\abs{x+2}\Bigr]_0^1 = \frac14\Bigl(\ln\frac{1}{3} - \ln 1\Bigr) = -\frac{\ln 3}{4}.

By parts ( $u' = 1$ , $v = \ln t$ ): $\int_1^{\eu} \ln t\,\dd t = \bigl[t\ln t\bigr]_1^{\eu} - \int_1^{\eu} \dd t = \eu - (\eu - 1) = 1$ .

By parts ( $u' = \sin t$ , $v = t$ ): $\int_0^\pi t\sin t\,\dd t = \bigl[-t\cos t\bigr]_0^\pi + \int_0^\pi \cos t\,\dd t = \pi + 0 = \pi$ .

Exercise 15.2 ★

Compute $\displaystyle\int_0^{1} \frac{t}{(t^2+1)^2}\,\dd t$ (substitution) and $\displaystyle\int_0^{\pi/2} \cos^3 t\, \dd t$ (write $\cos^3 = \cos(1 - \sin^2)$ ).

Solution

Solution of Exercise 15.2.

Substitution $u = t^2 + 1$ , $\dd u = 2t\,\dd t$ :

\int_0^1 \frac{t\,\dd t}{(t^2+1)^2} = \frac12 \int_1^2 \frac{\dd u}{u^2} = \frac12\Bigl[-\frac1u\Bigr]_1^2 = \frac14 .

With $u = \sin t$ : $\int_0^{\pi/2} \cos^3 t\,\dd t = \int_0^{\pi/2} (1 - \sin^2 t)\cos t\,\dd t = \bigl[\sin t - \frac{\sin^3 t}{3}\bigr]_0^{\pi/2} = 1 - \frac13 = \frac23$ .

Exercise 15.3 ★

Find the limits, as Riemann sums:

u_n = \sum_{k=1}^{n} \frac{n}{n^2 + k^2}, \qquad v_n = \frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!\,n^n}} \ \emph{(take logarithms)}.

Solution

Solution of Exercise 15.3.

$u_n = \dfrac1n \sum_{k=1}^{n} \dfrac{1}{1 + (k/n)^2}$ : a Riemann sum of $x \mapsto \frac{1}{1+x^2}$ on $\intcc{0}{1}$ , so $u_n \to \int_0^1 \frac{\dd x}{1+x^2} = \arctan 1 = \dfrac\pi4$ .

$\ln v_n = \dfrac1n \sum_{k=1}^{n} \ln\dfrac{n+k}{n} = \dfrac 1n \sum_{k=1}^n \ln\Bigl(1 + \dfrac kn\Bigr) \to \int_0^1 \ln(1+x)\,\dd x = \bigl[(1+x)\ln(1+x) - x\bigr]_0^1 = 2\ln 2 - 1$ . Hence $v_n \to \eu^{2\ln 2 - 1} = \dfrac 4\eu$ . (Check of the identification: $\frac{(2n)!}{n!\,n^n} = \prod_{k=1}^{n} \frac{n+k}{n}$ .)

Exercise 15.4 ★

Let $f$ be continuous on $\intcc{0}{1}$ . Compute $\lim_{n\to\infty} \int_0^1 x^n f(x)\,\dd x$ . (Cut $\intcc{0}{1}$ at $1 - \delta$ .)

Solution

Solution of Exercise 15.4.

The limit is $0$ . Let $M = \sup \abs f$ and $\varepsilon \in \intoo{0}{1}$ . Cut at $1 - \varepsilon$ :

\Bigl| \int_0^1 x^n f \Bigr| \leq \int_0^{1 - \varepsilon} x^n \abs f + \int_{1-\varepsilon}^1 x^n \abs f \leq M\,(1-\varepsilon)^n + M\varepsilon .

Since $(1 - \varepsilon)^n \to 0$ (Exercise 11.3), the limsup of the left side is $\leq M\varepsilon$ for every $\varepsilon$ : the integral tends to $0$ .

Exercise 15.5 ★★

(Cauchy–Schwarz) For $f, g$ continuous on $\intcc{a}{b}$ , prove

\Bigl(\int_a^b fg\Bigr)^{\!2} \leq \int_a^b f^2 \cdot \int_a^b g^2 ,

by expanding $\int_a^b (f + \lambda g)^2 \geq 0$ as a quadratic in $\lambda$ . When is it an equality?

Solution

Solution of Exercise 15.5.

$Q(\lambda) = \int_a^b (f + \lambda g)^2 = \int f^2 + 2\lambda \int fg + \lambda^2 \int g^2 \geq 0$ for all $\lambda$ . If $\int g^2 = 0$ , then $g = 0$ (strict positivity, Theorem 15.7 (4)) and the inequality is $0 \leq 0$ . Otherwise $Q$ is a genuine quadratic, everywhere $\geq 0$ : its discriminant is $\leq 0$ , i.e. $\bigl(\int fg\bigr)^2 \leq \int f^2 \int g^2$ .

Equality iff the discriminant vanishes iff $Q(\lambda_0) = 0$ for some $\lambda_0$ , i.e. $\int (f + \lambda_0 g)^2 = 0$ , i.e. (strict positivity again) $f = -\lambda_0 g$ : equality holds exactly when $f$ and $g$ are proportional.

Exercise 15.6 ★★

Let $f$ be continuous on $\R$ , $T$ -periodic. Prove that $\int_a^{a+T} f$ does not depend on $a$ , and that $\frac1x \int_0^x f(t)\,\dd t \to \frac 1T \int_0^T f$ as $x \to +\infty$ .

Solution

Solution of Exercise 15.6.

Let $\Phi(a) = \int_a^{a+T} f$ . By the fundamental theorem (Theorem 15.9), $\Phi$ is differentiable with $\Phi'(a) = f(a + T) - f(a) = 0$ : constant.

For $x > 0$ , write $x = nT + r$ , $0 \leq r < T$ ( $n = \lfloor x/T \rfloor$ ). Chasles:

\int_0^x f = n \int_0^T f + \int_{nT}^{nT + r} f, \qquad \Bigl| \int_{nT}^{nT+r} f \Bigr| \leq T \sup_{\intcc{0}{T}} \abs f = C .

Then $\frac 1x \int_0^x f = \frac{nT}{x}\cdot\frac 1T \int_0^T f + O\bigl(\frac 1x\bigr)$ , and $\frac{nT}{x} \to 1$ : the limit is $\frac1T \int_0^T f$ .

Exercise 15.7 ★★

For $f$ continuous on $\intcc{0}{1}$ with $\int_0^1 f = \frac12$ , prove that $f$ has a fixed point in $\intcc{0}{1}$ . (Integrate $f(x) - x$ and use strict positivity, Theorem 15.7 (4), through its contrapositive combined with the intermediate value theorem.)

Solution

Solution of Exercise 15.7.

Let $g(x) = f(x) - x$ : continuous, with

\int_0^1 g = \int_0^1 f - \frac12 = 0 .

If $g$ never vanished, the intermediate value theorem would force a constant sign (a continuous function on an interval taking both signs vanishes); say $g > 0$ . Then, by strict positivity (Theorem 15.7 (4) applied to $g > 0$ , giving $\int g > 0$ ): contradiction with $\int g = 0$ . So $g(c) = 0$ for some $c$ : $f(c) = c$ .

Exercise 15.8 ★★★

(Wallis integrals) Let $W_n = \int_0^{\pi/2} \sin^n t\,\dd t$ .

Prove the recurrence $n W_n = (n-1) W_{n-2}$ ( $n \geq 2$ ) by parts, and compute $W_0, W_1$ , then $W_{2p}$ and $W_{2p+1}$ in closed form.
Prove that $(W_n)$ is decreasing with $\frac{W_{n+1}}{W_n} \to 1$ ; prove that the quantity $(n+1)\,W_{n+1} W_n$ is constant, equal to $\frac\pi2$ ; and deduce the equivalent $W_n \sim \sqrt{\dfrac{\pi}{2n}}$ .

Solution

Solution of Exercise 15.8.

By parts with $u' = \sin t$ , $v = \sin^{n-1} t$ :

$W_n = \bigl[-\cos t\sin^{n-1}t\bigr]_0^{\pi/2} + (n-1)\int_0^{\pi/2} \cos^2 t\,\sin^{n-2} t\,\dd t = (n-1)(W_{n-2} - W_n),$

so $nW_n = (n-1)W_{n-2}$ . From $W_0 = \frac\pi2$ , $W_1 = 1$ :

$W_{2p} = \frac{(2p-1)(2p-3)\cdots 1}{(2p)(2p-2)\cdots 2}\, \frac{\pi}{2} = \frac{(2p)!}{4^p (p!)^2}\,\frac\pi2, \qquad W_{2p+1} = \frac{(2p)(2p-2)\cdots 2}{(2p+1)(2p-1)\cdots 3} = \frac{4^p (p!)^2}{(2p+1)!} .$
On $\intoo{0}{\frac\pi2}$ , $0 < \sin t < 1$ , so $\sin^{n+1} < \sin^n$ and $(W_n)$ is (strictly) decreasing, positive. Sandwiching with the recurrence:

$\frac{n}{n+1} = \frac{W_{n+1}}{W_{n-1}} \leq \frac{W_{n+1}}{W_n} \leq 1 \quad\implies\quad \frac{W_{n+1}}{W_n} \to 1 .$

Invariant: $a_n = (n+1)W_{n+1}W_n$ satisfies $a_n = a_{n-1}$ by the recurrence $(n+1)W_{n+1} = nW_{n-1}$ , so $a_n = a_0 = 1 \cdot W_1 W_0 = \frac\pi2$ . Then

$n W_n^2 \sim (n+1) W_{n+1} W_n = \frac\pi2 \quad\implies\quad W_n \sim \sqrt{\frac{\pi}{2n}} .$

Exercise 15.9 ★★★

(Niven: $\pi$ is irrational) Suppose $\pi = \frac ab$ with $a, b \in \N^*$ , and set, for $n$ to be chosen,

P(x) = \frac{x^n (a - bx)^n}{n!}, \qquad I_n = \int_0^{\pi} P(x)\sin x\, \dd x .

Prove that $0 < I_n \leq \pi\,\frac{(\pi a)^n}{n!}$ , which is $< 1$ for $n$ large.
Prove that $P$ and all its derivatives take integer values at $0$ and at $\pi = \frac ab$ . (Binomial expansion: the coefficients of $P$ times $k!$ are integers for $k \geq n$ ; and $P(\pi - x) = P(x)$ .)
Set $Q = P - P'' + P^{(4)} - \dots$ (a finite sum). Check that $\bigl(Q'\sin x - Q\cos x\bigr)' = P \sin x$ , and deduce that $I_n = Q(\pi) + Q(0)$ is an integer.
Conclude.

Solution

Solution of Exercise 15.9.

On $\intoo{0}{\pi}$ : $x > 0$ , $a - bx = b(\frac ab - x) = b(\pi - x) > 0$ and $\sin x > 0$ , so the integrand is $> 0$ and $I_n > 0$ (strict positivity). Bound: on $\intcc{0}{\pi}$ , $x \leq \pi$ and $a - bx \leq a$ , so $P \leq \frac{\pi^n a^n}{n!}$ and $I_n \leq \pi\,\frac{(\pi a)^n}{n!}$ , which tends to $0$ (the factorial beats the geometric term: it is the general term of the convergent exponential series, cf. Example 11.12); in particular $I_n < 1$ for large $n$ .
Expand $x^n(a - bx)^n = \sum_{j=0}^{n} \binom nj a^{n-j} (-b)^j x^{n+j}$ : so $P = \frac{1}{n!}\sum_j c_j x^{n+j}$ with integer $c_j$ . Then $P^{(k)}(0) = 0$ for $k < n$ (valuation) and, for $n \leq k \leq 2n$ , $P^{(k)}(0) = \frac{k!}{n!} c_{k-n}$ , an integer since $n! \mid k!$ . Moreover $P(\pi - x) = P(x)$ (substitute: $\pi - x$ swaps the factors, using $a - b(\pi - x) = bx$ ), so $P^{(k)}(\pi) = \pm P^{(k)}(0)$ : integers as well.
With $Q = P - P'' + P^{(4)} - \dots$ (finite: $P$ has degree $2n$ ): $Q + Q'' = P$ , and

$\bigl(Q'\sin x - Q\cos x\bigr)' = (Q + Q'')\sin x = P\sin x .$

Hence $I_n = \bigl[Q'\sin x - Q\cos x\bigr]_0^{\pi} = Q(\pi) + Q(0)$ , a sum of values $P^{(2k)}$ at $0$ and $\pi$ : an integer by (2).
For $n$ large, $I_n$ is an integer with $0 < I_n < 1$ : impossible. The assumption $\pi = \frac ab$ fails: $\pi$ is irrational.

Exercise 15.10 ★★★

Let $f$ be $C^1$ on $\intcc{a}{b}$ . Prove the Riemann–Lebesgue-type limit

\int_a^b f(t)\sin(\lambda t)\,\dd t \xrightarrow[\lambda \to +\infty]{} 0

by integrating by parts. Then prove it again for $f$ merely continuous, by uniform approximation with step functions (Theorem 15.3).

Solution

Solution of Exercise 15.10.

$C^1$ case: by parts,

\int_a^b f(t)\sin\lambda t\,\dd t = \Bigl[-f(t)\frac{\cos\lambda t}{\lambda}\Bigr]_a^b + \frac{1}{\lambda}\int_a^b f'(t)\cos\lambda t\,\dd t,

bounded in absolute value by $\frac{2\sup\abs f + (b - a)\sup\abs{f'}} {\lambda} \to 0$ .

Continuous case: let $\varepsilon > 0$ and pick a step function $\varphi$ with $\abs{f - \varphi} \leq \varepsilon$ (Theorem 15.3 provides $\varphi \leq f \leq \psi$ with gap $\leq\varepsilon$ ; take $\varphi$ ). Then

\Bigl|\int_a^b f\sin\lambda t\Bigr| \leq \int_a^b \abs{f - \varphi} + \Bigl|\int_a^b \varphi \sin\lambda t\Bigr| \leq (b-a)\varepsilon + \sum_i \abs{c_i}\,\Bigl|\int_{x_{i-1}}^{x_i} \sin\lambda t\,\dd t\Bigr| ,

and each $\bigl|\int \sin \lambda t\bigr| = \bigl|\frac{\cos\lambda x_{i-1} - \cos\lambda x_i}{\lambda}\bigr| \leq \frac{2}{\lambda}$ : the second term tends to $0$ . Hence the limsup is $\leq (b-a)\varepsilon$ for every $\varepsilon$ : the limit is $0$ .

Exercise 15.11 ★★

(Mean value theorem for integrals) Let $f, g$ be continuous on $\intcc{a}{b}$ with $g \geq 0$ . Prove that there exists $c \in \intcc{a}{b}$ with

\int_a^b f(t)\,g(t)\,\dd t = f(c)\int_a^b g(t)\,\dd t ,

and show by an example that the hypothesis $g \geq 0$ cannot be dropped.

Solution

Solution of Exercise 15.11.

Let $m = \min f$ and $M = \max f$ , attained by the extreme value theorem. Since $g \geq 0$ : $m\,g \leq fg \leq M\,g$ , so by monotonicity

m \int_a^b g \;\leq\; \int_a^b fg \;\leq\; M \int_a^b g .

If $\int_a^b g = 0$ : strict positivity (Theorem 15.7 (4)) forces $g \equiv 0$ , both sides vanish, and any $c$ works. Otherwise $t = \frac{\int fg}{\int g}$ lies in $\intcc{m}{M} = f(\intcc{a}{b})$ (Theorems 13.13 and 13.10), so $t = f(c)$ for some $c$ .

Sign matters: on $\intcc{-1}{1}$ with $f(t) = g(t) = t$ : $\int fg = \int_{-1}^1 t^2 = \frac23$ , while $f(c)\int_{-1}^1 t\,\dd t = 0$ for every $c$ .

Exercise 15.12 ★★★

(Moments force zeros) Let $f$ be continuous on $\intcc{a}{b}$ with

\int_a^b f(t)\,t^k\,\dd t = 0 \qquad \text{for } k = 0, 1, \dots, n .

Prove that $f$ vanishes at $n + 1$ distinct points of $\intoo{a}{b}$ . (If $f$ changes sign only at $z_1 < \dots < z_m$ with $m \leq n$ , integrate $f$ against $P(t) = (t - z_1) \cdots (t - z_m)$ and use strict positivity.)

Solution

Solution of Exercise 15.12.

If $f \equiv 0$ the claim is vacuous (every point is a zero). So assume $f \not\equiv 0$ and suppose it has at most $n$ distinct zeros in $\intoo{a}{b}$ ; let $z_1 < \dots < z_m$ ( $m \leq n$ ) be those zeros where $f$ changes sign (possibly none). Set $P(t) = \prod_{i=1}^{m}(t - z_i)$ (empty product $= 1$ ), of degree $m \leq n$ . On each subinterval cut by the $z_i$ , both $f$ and $P$ have constant sign, and both flip sign when crossing some $z_i$ : the product $fP$ has one constant sign on all of $\intoo{a}{b}$ . Being continuous, not identically zero, of constant sign, it has $\bigl|\int_a^b fP\bigr| > 0$ (strict positivity applied to $\abs{fP}$ ). But $\int f P$ is a linear combination of the moments $\int f\,t^k$ , $k \leq n$ , all zero: contradiction. Hence $f$ has at least $n + 1$ distinct zeros in $\intoo{a}{b}$ .

Remark 15.24 (Perspectives inside this volume)

Three chapters ahead lean directly on this one. Chapter 16 carries the integral remainder — the sharpest of the three Taylor formulas is an integration by parts iterated $n$ times. Chapter 17 converts the framing of sums by integrals into the decisive test for $\sum n^{-\alpha}$ , and its weekend problem refines that framing into Euler’s constant. Chapter 24 makes the integral geometric: the length of a parametrized arc is $\int \sqrt{x'(t)^2 + y'(t)^2}\,\dd t$ , an integral of a continuous function on a segment — precisely the object built here, no improper theory needed. The single most reused fact will be the humblest: $\bigl|\int f\bigr| \leq (b - a)\sup\abs{f}$ , the inequality that turns every pointwise estimate into an integral estimate.

15.6 Problem: The integral irrationality machine

Problem 15.1

Weekend problem — $\eu$ and $\pi^2$ are irrational, $\eu$ to six decimals, and $\frac{22}{7} > \pi$ with proof

One mechanism powers this whole problem: an expression that must be a positive integer, yet is provably smaller than $1$ , cannot exist. Exercise 15.9 (Niven) ran it once to prove $\pi \notin \Q$ ; here we industrialize it. The machine needs three parts: an integrality input (endpoint values of well-chosen polynomials), a smallness input (a factor $\frac{1}{n!}$ crushing the integral), and a bridge (integration by parts) connecting them. We prove that $\eu$ is irrational and compute it with certified error, prove Legendre’s sharper theorem that $\pi^2$ is irrational, and end with the most charming integral in analysis: $\int_0^1 \frac{x^4(1 - x)^4}{1 + x^2}\dd x = \frac{22}{7} - \pi$ , which brackets $\pi$ by hand.

Part I — Fuel.

Prove that $\dfrac{c^{\,n}}{n!} \to 0$ for every fixed $c > 0$ (beyond $n \geq 2c$ , each step at least halves the term).
(Beta integrals) Prove, by induction on $m$ with integration by parts:

$\int_0^1 x^{\,k}\,(1 - x)^{\,m}\,\dd x = \frac{k!\,m!}{(k + m + 1)!} \qquad (k, m \in \N).$
Deduce $\displaystyle\int_0^1 \bigl(x(1-x)\bigr)^n \dd x = \frac{1}{(2n+1)\binom{2n}{n}}$ , and — comparing with the bound $x(1 - x) \leq \frac14$ — the estimate $\binom{2n}{n} \geq \dfrac{4^n}{2n+1}$ , matching $\binom{2n}{n}^{1/n} \to 4$ from Problem 11.1.
Prove the smallness lemma used twice below: for every continuous $g > 0$ on $\intoo{0}{1}$ ,

$0 < \int_0^1 \bigl(x(1-x)\bigr)^n g(x)\,\dd x \leq \frac{\sup_{\intcc{0}{1}}\abs g}{4^{\,n}} .$

Part II — $\eu$ : irrationality, then six decimals. Set $A_n = \displaystyle\int_0^1 x^n \eu^x \dd x$ .

Compute $A_0$ and $A_1$ , prove the recurrence $A_n = \eu - n A_{n-1}$ , and the bounds $0 < A_n \leq \dfrac{\eu}{n+1}$ .
Show by induction that $A_n = \alpha_n + \beta_n \eu$ with $\alpha_n, \beta_n \in \Z$ .
Deduce that $\eu$ is irrational (if $\eu = \frac pq$ , then $q A_n$ is an integer trapped in $\intoo{0}{1}$ for $n$ large). Compare with the proof of Exercise 11.9: same punchline, different fuel.
Prove, by induction and integration by parts, the exact remainder formula

$\eu = \sum_{k=0}^{n} \frac{1}{k!} + R_n, \qquad R_n = \frac{1}{n!}\int_0^1 (1 - t)^n\,\eu^{\,t}\,\dd t, \qquad \frac{1}{(n+1)!} \leq R_n \leq \frac{\eu}{(n+1)!} .$
Take $n = 9$ : bound $R_9$ using $\eu < 2.75$ (from $b_2 = 2.75$ in Example 11.12), evaluate the sum, and conclude the certified bracketing $2.7182818 \leq \eu \leq 2.7182823$ — six decimals, $\eu \approx 2.718282$ , with proof.

Part III — Legendre’s theorem: $\pi^2$ is irrational. Let $f(x) = \dfrac{x^n (1 - x)^n}{n!}$ , and suppose $\pi^2 = \frac ab$ with $a, b \in \N^*$ .

Show $f(1 - x) = f(x)$ and $0 < f \leq \dfrac{1}{4^n\,n!}$ on $\intoo{0}{1}$ .
Show that $f^{(k)}(0)$ and $f^{(k)}(1)$ are integers for every $k \geq 0$ (expand $x^n(1-x)^n$ with integer coefficients; $\frac{k!}{n!} \in \Z$ for $k \geq n$ ; then use the symmetry).
Define

$G = b^{\,n} \sum_{k=0}^{n} (-1)^k\, \pi^{2n - 2k} f^{(2k)} .$

Show that $G(0)$ and $G(1)$ are integers (each $b^n \pi^{2n-2k} = a^{\,n-k}\,b^{\,k}$ ).
Verify the telescoping $G'' + \pi^2 G = b^n \pi^{2n+2} f = \pi^2 a^n f$ , then

$\frac{\dd}{\dd x}\Bigl(G'(x)\sin \pi x - \pi\,G(x)\cos \pi x\Bigr) = \pi^2 a^n f(x)\sin \pi x .$
Integrate over $\intcc{0}{1}$ and conclude

$\pi\,a^n \int_0^1 f(x)\sin(\pi x)\,\dd x = G(0) + G(1) \in \Z ,$

a positive integer bounded by $\dfrac{\pi a^n}{4^n\,n!}$ .
Conclude with question 1 that $\pi^2$ is irrational (Legendre, 1794), and that this strengthens Exercise 15.9: why does the irrationality of $\pi^2$ imply that of $\pi$ , and not conversely?

Part IV — Understanding the machine.

Locate the two opposing forces (integrality of endpoint data; analytic smallness of the integral) and the bridge, in Parts II and III. Then explain why the factor $\frac{1}{n!}$ in $f$ is the crux: if it is removed, integrality survives, but which inequality dies, and for which claimed fractions $\frac ab$ does the proof then fail?
Effectivity: suppose someone claims $\pi^2 = \frac ab$ with $a \leq 10$ . Show that the contradiction already lands at $n = 7$ : compute $\pi\,(10/4)^7/7! \approx 0.38 < 1$ . The machine does not merely refute; it refutes by a fixed, computable stage.
Sanity check the bridge unconditionally: prove by two integrations by parts that

$\int_0^1 x(1 - x)\sin(\pi x)\,\dd x = \frac{4}{\pi^3},$

and reconcile with question 14 at $n = 1$ (keep $\pi^2$ symbolic: the telescoped identity reads $\pi^3 \int_0^1 f_1 \sin \pi x = -(f_1''(0) + f_1''(1)) = 4$ ).
What makes $\eu^x$ and $\sin \pi x$ eligible as the machine’s kernels? Identify the property (each satisfies a linear differential equation with constant coefficients, so repeated integration by parts cycles back to the start), and name the frontier: the same machine, refined by Hermite and Lindemann, proves $\eu$ and $\pi$ transcendental — beyond this volume.

Part V — $\frac{22}{7}$ versus $\pi$ , and the moral.

Establish the polynomial division

$\frac{x^4(1-x)^4}{1 + x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{1 + x^2},$

and deduce the celebrated identity

$\int_0^1 \frac{x^4 (1-x)^4}{1 + x^2}\,\dd x = \frac{22}{7} - \pi .$
The integrand is positive: conclude $\pi < \frac{22}{7}$ . Then, bounding $\frac{1}{1+x^2}$ between $\frac12$ and $1$ and using $\int_0^1 (x(1-x))^4 = \frac{1}{630}$ (question 3), prove

$\frac{22}{7} - \frac{1}{630} \;\leq\; \pi \;\leq\; \frac{22}{7} - \frac{1}{1260} ,$

i.e. $3.14126 \leq \pi \leq 3.14207$ : two correct decimals, by hand.
Generalize: dividing $x^{4m}(1-x)^{4m}$ by $1 + x^2$ , show the remainder is the constant $(-4)^m$ (work modulo $x^2 + 1$ : $(1-x)^2 \equiv -2x$ ), deduce rationals $r_m$ with

$\abs{\pi - r_m} \leq 4^{\,1 - 5m} ,$

and check that $m = 1$ reproduces question 20–21.
Confront these rationals with the approximation theory of Problem 14.1: compute $\abs{\pi - \frac{22}{7}} \approx 1.26\cdot10^{-3}$ against the Dirichlet guarantee $\frac{1}{49}$ , and quote $\abs{\pi - \frac{355}{113}} \approx 2.7\cdot10^{-7}$ against $\frac{1}{113^2} \approx 7.8\cdot10^{-5}$ : exceptionally good rational approximations exist for $\pi$ — consistent, since $\pi$ is not known to be badly approximable.
(The integer trap, abstracted) Prove the lemma that unifies everything: if $x \in \R$ and there exist integers $a_n, b_n$ with $0 < \abs{a_n + b_n x} \to 0$ , then $x$ is irrational. List its instances in this problem, in Exercise 15.9, in Exercise 11.9, and in Problem 14.1.
Synthesis, one sentence each: (i) the machine’s three parts and where each lives in the toolbox of this chapter; (ii) what the integral contributes that the mean value theorem of Problem 14.1 could not; (iii) the inventory of results extracted (two irrationalities, one six-decimal constant, one bracketing of $\pi$ , one binomial bound); (iv) the frontier (Hermite, Lindemann; and the same trap, run on $\zeta(2)$ and $\zeta(3)$ , in twentieth-century arithmetic).

Solution

Solution of Problem 15.1.

1. Let $N = \lceil 2c \rceil$ . For $n \geq N$ : $\frac{c^{n+1}/(n+1)!}{c^n/n!} = \frac{c}{n+1} \leq \frac12$ , so $0 < \frac{c^n}{n!} \leq \frac{c^N}{N!}\,2^{-(n - N)} \to 0$ : squeeze.

2. Fix $k$ ; induction on $m$ . For $m = 0$ : $\int_0^1 x^k = \frac{1}{k+1} = \frac{k!\,0!}{(k+1)!}$ . Step, by parts ( $u = (1-x)^m$ , $v' = x^k$ ):

\int_0^1 x^k (1-x)^m \dd x = \Bigl[\frac{x^{k+1}}{k+1}(1-x)^m\Bigr]_0^1 + \frac{m}{k+1}\int_0^1 x^{k+1}(1-x)^{m-1}\dd x = \frac{m}{k+1}\cdot\frac{(k+1)!\,(m-1)!}{(k+m+1)!} ,

which is $\frac{k!\,m!}{(k+m+1)!}$ .

3. $k = m = n$ : $\int_0^1 (x(1-x))^n = \frac{(n!)^2}{(2n+1)!} = \frac{1}{(2n+1)\binom{2n}{n}}$ . Since $x(1-x) \leq \frac14$ on $\intcc{0}{1}$ , the integral is $\leq 4^{-n}$ , whence $\binom{2n}{n} \geq \frac{4^n}{2n+1}$ — consistent with $\binom{2n}{n}^{1/n} \to 4$ (Problem 11.1).

4. The integrand is continuous, $\geq 0$ , and positive on $\intoo{0}{1}$ , hence not identically zero: its integral is $> 0$ (Theorem 15.7 (4)). Upper bound: $(x(1-x))^n \leq 4^{-n}$ and $g \leq \sup\abs g$ , then monotonicity.

5. $A_0 = \eu - 1$ ; $A_1 = [x\eu^x]_0^1 - \int_0^1 \eu^x = \eu - (\eu - 1) = 1$ . By parts: $A_n = [x^n \eu^x]_0^1 - n\int_0^1 x^{n-1}\eu^x = \eu - n A_{n-1}$ . Bounds: the integrand is positive, so $A_n > 0$ ; and $\eu^x \leq \eu$ gives $A_n \leq \eu\int_0^1 x^n = \frac{\eu}{n+1}$ .

6. $A_0 = -1 + 1\cdot\eu$ . If $A_{n-1} = \alpha_{n-1} + \beta_{n-1}\eu$ with integer entries, then

A_n = \eu - n\alpha_{n-1} - n\beta_{n-1}\eu = \underbrace{(-n\,\alpha_{n-1})}_{\alpha_n} + \underbrace{(1 - n\,\beta_{n-1})}_{\beta_n}\,\eu ,

both integers.

7. If $\eu = \frac pq$ : $q A_n = q\alpha_n + p\beta_n \in \Z$ , and $0 < qA_n \leq \frac{q\eu}{n+1} < 1$ for $n$ large: an integer strictly between $0$ and $1$ — impossible. So $\eu \notin \Q$ . In Exercise 11.9 the trapped integer was $q!\,\frac pq - q!\,a_q$ ; here it is $qA_n$ : same trap, integral fuel.

8. $n = 0$ : $R_0 = \int_0^1 \eu^t = \eu - 1$ , so $\eu = 1 + R_0$ . By parts ( $u = \eu^t$ , $v = -\frac{(1-t)^{n+1}}{n+1}$ ):

R_n = \frac{1}{n!}\Bigl(\Bigl[-\frac{(1-t)^{n+1}}{n+1} \eu^t\Bigr]_0^1 + \frac{1}{n+1}\int_0^1 (1-t)^{n+1}\eu^t\Bigr) = \frac{1}{(n+1)!} + R_{n+1} ,

so the formula propagates from $n$ to $n + 1$ . Bounds: $1 \leq \eu^t \leq \eu$ on $\intcc{0}{1}$ and $\int_0^1 (1-t)^n = \frac{1}{n+1}$ give $\frac{1}{(n+1)!} \leq R_n \leq \frac{\eu}{(n+1)!}$ .

9. $\sum_{k=0}^{9} \frac{1}{k!} = \frac{986410}{362880} = 2.71828152\dots$ , and

\frac{1}{10!} = 2.76\cdot10^{-7} \leq R_9 \leq \frac{2.75}{10!} = 7.58\cdot10^{-7} ,

so $2.7182818 \leq \eu \leq 2.7182823$ : with proof, $\eu = 2.718282$ to six decimals (true value $2.7182818\dots$ ).

10. $f(1 - x) = \frac{(1-x)^n x^n}{n!} = f(x)$ . On $\intoo{0}{1}$ : $0 < x(1-x) \leq \frac14$ , so $0 < f \leq \frac{1}{4^n\,n!}$ .

11. $x^n(1-x)^n = \sum_{j=0}^{n} (-1)^j\binom nj\,x^{n+j}$ , so $f = \frac{1}{n!}\sum_j c_j\,x^{n+j}$ with $c_j \in \Z$ . Hence $f^{(k)}(0) = 0$ for $k < n$ or $k > 2n$ , and for $n \leq k \leq 2n$ : $f^{(k)}(0) = \frac{k!}{n!}\, c_{k-n}$ , an integer because $n! \mid k!$ . The symmetry gives $f^{(k)}(1) = (-1)^k f^{(k)}(0) \in \Z$ .

12. $b^n \pi^{2n-2k} = b^n\bigl(\frac ab\bigr)^{\,n-k} = a^{\,n-k}\,b^{\,k} \in \Z$ , so $G(0) = \sum_k (-1)^k a^{n-k} b^k f^{(2k)}(0)$ and likewise $G(1)$ are integers by question 11.

13. In $\pi^2 G + G''$ , the term $k$ of $\pi^2 G$ carries $\pi^{2n-2k+2}f^{(2k)}$ and the term $j = k - 1$ of $G''$ carries $(-1)^{k-1}\pi^{2n-2k+2}f^{(2k)}$ : everything cancels except $k = 0$ in the first sum and $j = n$ in the second, i.e.

G'' + \pi^2 G = b^n\bigl(\pi^{2n+2} f + (-1)^n f^{(2n+2)}\bigr) = b^n \pi^{2n+2} f = \pi^2 a^n f

( $f$ has degree $2n$ , so $f^{(2n+2)} = 0$ ; and $b^n\pi^{2n} = a^n$ ). Then

\bigl(G'\sin\pi x - \pi G\cos\pi x\bigr)' = (G'' + \pi^2 G)\sin \pi x = \pi^2 a^n f\sin\pi x .

14. Integrating over $\intcc{0}{1}$ :

\pi^2 a^n \int_0^1 f\sin(\pi x)\,\dd x = \bigl[G'\sin\pi x - \pi G\cos\pi x\bigr]_0^1 = \pi\bigl(G(1) + G(0)\bigr) ,

so $\pi a^n \int_0^1 f\sin\pi x = G(0) + G(1) \in \Z$ . On $\intoo{0}{1}$ , $f > 0$ and $\sin\pi x > 0$ : the left side is positive, so $G(0) + G(1) \geq 1$ ; and $\sin \leq 1$ with question 10 bounds it by $\frac{\pi a^n}{4^n\,n!}$ .

15. By question 1 (with $c = \frac a4$ ), $\frac{\pi a^n}{4^n n!} \to 0$ : for large $n$ it is $< 1$ , contradicting $G(0) + G(1) \geq 1$ . So no fraction $\frac ab$ equals $\pi^2$ : Legendre’s theorem. If $\pi$ were rational, $\pi^2$ would be too: so $\pi \notin \Q$ — and the implication only runs this way ( $\sqrt2$ is irrational with rational square), which is why $\pi^2 \notin \Q$ is strictly stronger than Exercise 15.9.

16. Integrality: questions 11–12 (endpoint derivatives); smallness: questions 10 and 1; bridge: questions 13–14 (the telescoped double integration by parts). Without $\frac{1}{n!}$ , the endpoint data remain integers (even more easily), but the bound becomes $\frac{\pi a^n}{4^n}$ , which tends to $0$ only when $a < 4$ — and every candidate has $a = b\,\pi^2 > 9$ . The factorial is exactly what outruns the geometric growth $a^n$ : no factorial, no theorem.

17. For $a \leq 10$ , the integer $G(0) + G(1)$ is positive and at most $\pi\,(10/4)^n/n!$ . At $n = 7$ : $2.5^7 = 610.35\dots$ , so the bound is $\frac{\pi \times 610.35}{5040} \approx 0.38 < 1$ (at $n = 6$ it is still $1.07$ ): the contradiction lands at the seventh stage, explicitly.

18. Two integrations by parts:

\int_0^1 x(1-x)\sin\pi x\,\dd x = \frac1\pi\int_0^1 (1 - 2x)\cos\pi x\,\dd x = \frac{2}{\pi^2}\int_0^1 \sin\pi x\,\dd x = \frac{2}{\pi^2}\cdot\frac{2}{\pi} = \frac{4}{\pi^3}

(the bracket terms vanish: $x(1-x)$ at $0, 1$ , and $\sin\pi x$ at $0, 1$ ). Symbolically, the $n = 1$ telescope (no assumption on $\pi$ ) reads $\pi^3\int_0^1 f_1\sin\pi x = -(f_1''(0) + f_1''(1))$ with $f_1 = x(1 - x)$ , $f_1'' = -2$ : right side $4$ — the two computations agree.

19. $\eu^x$ solves $y' = y$ and $\sin\pi x$ solves $y'' = -\pi^2 y$ : linear equations with constant coefficients, so integration by parts cycles the kernel back to itself and keeps all boundary data inside $\Z + \Z\eu$ (resp. integer polynomials in $\pi^2$ ). That closure property is what the machine needs. Refined with kernels adapted to several points at once, the same mechanism yields Hermite’s theorem ( $\eu$ transcendental, 1873) and Lindemann’s ( $\pi$ transcendental, 1882) — beyond this volume.

20. Polynomial division (or multiply back and check):

x^4(1-x)^4 = (x^6 - 4x^5 + 5x^4 - 4x^2 + 4)(1 + x^2) - 4 .

Integrating the displayed identity divided by $1 + x^2$ :

\int_0^1 \frac{x^4(1-x)^4}{1+x^2}\dd x = \Bigl(\frac17 - \frac46 + 1 - \frac43 + 4\Bigr) - 4\arctan 1 = \frac{22}{7} - \pi ,

using $\frac17 - \frac23 + 1 - \frac43 + 4 = 3 + \frac17$ and $\arctan 1 = \frac\pi4$ .

21. The integrand is continuous, positive on $\intoo{0}{1}$ : the integral is $> 0$ , so $\pi < \frac{22}{7}$ . Moreover $\frac12 \leq \frac{1}{1+x^2} \leq 1$ on $\intcc{0}{1}$ and $\int_0^1 (x(1-x))^4 = \frac{(4!)^2}{9!} = \frac{1}{630}$ (question 3):

\frac{1}{1260} \leq \frac{22}{7} - \pi \leq \frac{1}{630} \quad\Longrightarrow\quad 3.14126 < \frac{22}{7} - \frac{1}{630} \leq \pi \leq \frac{22}{7} - \frac{1}{1260} < 3.14207 .

22. Modulo $x^2 + 1$ : $x^2 \equiv -1$ , so $x^{4m} = (x^2)^{2m} \equiv 1$ and $(1 - x)^2 = 1 - 2x + x^2 \equiv -2x$ , hence $(1-x)^{4m} \equiv (-2x)^{2m} = 4^m (x^2)^m \equiv (-4)^m$ : the remainder is the constant $(-4)^m$ , and the quotient $Q_m$ has integer coefficients (division by a monic integer polynomial). Dividing the identity by $1 + x^2$ and integrating:

J_m := \int_0^1 \frac{(x(1-x))^{4m}}{1+x^2}\dd x = s_m + (-4)^m\,\frac{\pi}{4}, \qquad s_m = \int_0^1 Q_m \in \Q .

Solving for $\pi$ : with $r_m = (-1)^{m+1}\,4^{\,1-m} s_m \in \Q$ , $\;\abs{\pi - r_m} = 4^{\,1-m} J_m \leq 4^{\,1-m}\cdot4^{-4m} = 4^{\,1-5m}$ . For $m = 1$ : $s_1 = \frac{22}{7}$ , $r_1 = \frac{22}{7}$ , bound $4^{-4} = \frac{1}{256}$ — questions 20–21 again.

23. $\bigl|\pi - \frac{22}{7}\bigr| = \frac{22}{7} - \pi \approx 1.26\cdot10^{-3}$ , sixteen times better than the order- $2$ benchmark $\frac{1}{7^2} \approx 2.0\cdot10^{-2}$ guaranteed by Dirichlet (Problem 14.1, question 4); and $\bigl|\pi - \frac{355}{113}\bigr| \approx 2.7\cdot10^{-7}$ beats $\frac{1}{113^2} \approx 7.8\cdot10^{-5}$ by a factor $\approx 300$ . No contradiction with anything proved: Liouville inequalities lower-bound approximation errors only for algebraic numbers, and no such bound for $\pi$ is available at this level — $\pi$ is free to be approximated spectacularly well.

24. Lemma: suppose $x = \frac pq$ and $0 < \abs{a_n + b_n x} \to 0$ . Then $\abs{a_n + b_n x} = \frac{\abs{q a_n + p b_n}}{q}$ , with $q a_n + p b_n$ a nonzero integer (nonzero because the absolute value is $> 0$ ): so $\abs{a_n + b_n x} \geq \frac1q$ for every $n$ , contradicting the convergence to $0$ . Instances: question 7 ( $x = \eu$ , $a_n = \alpha_n$ , $b_n = \beta_n$ ); Exercise 11.9 ( $x = \eu$ again, with $a_q = -q!\sum_{k \leq q}\frac{1}{k!}$ , $b_q = q!$ ); and Problem 14.1, question 1, is its geometric form. In Part III and in Exercise 15.9 the trap runs inside the contradiction: assuming rationality converts an expression into an integer, which the analysis then squeezes into $\intoo{0}{1}$ — the same principle, transposed.

25. (i) Integrality lives in the endpoint calculus of polynomials (questions 6, 11–12), smallness in the bounds $\sup$ -monotonicity gives (questions 4, 10), the bridge in integration by parts (questions 8, 13–14) — all three are theorems of this chapter. (ii) The integral supplies what the mean value theorem could not: an exact identity between the analytic object and the arithmetic data (equality, not just an inequality with an unknown $c$ ), which is why the machine reaches $\pi^2$ while Problem 14.1 reached only approximation exponents. (iii) Extracted: $\eu \notin \Q$ , $\pi^2 \notin \Q$ (hence $\pi \notin \Q$ ), $\eu = 2.718282$ certified, $\frac{22}{7} - \frac{1}{630} \leq \pi \leq \frac{22}{7} - \frac{1}{1260}$ , and $\binom{2n}{n} \geq \frac{4^n}{2n+1}$ . (iv) Frontier: Hermite and Lindemann push the same machine to transcendence; and Apéry (1979) ran the integer trap on $\zeta(3)$ — the machine is still producing twentieth-century mathematics.