Proof of The Integral of 1/(1+s²) over ℝ

Routine facts about real numbers are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity, Toolkit for CA1c, Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (ii)). Riemann integrals are those of Single-Variable Calculus on an Interval: for $u<v$ and a function $g$ continuous on $[u,v]$, $\int_{u}^{v}g$ is the Riemann integral of $g|_{[u,v]}$, and $I_{n}(g)=\int_{-n}^{n}g$ for $n\in\mathbb{N}$. Let $\arctan:\mathbb{R}\to(-\pi/2,\pi/2)$ be the arctangent. The function $k$ is its derivative $\arctan'$ (The Arctangent: Continuity, Monotonicity, Derivative, Special Values and Oddness (ii)), since both are maps $\mathbb{R}\to\mathbb{R}$ with value $1/(1+y^{2})$ at every $y\in\mathbb{R}$. The Lebesgue integral of $k$ is computed as the supremum of its Riemann integrals over $[-n,n]$, which the arctangent evaluates.

Measurability of $k$. The function $k$ is positive and continuous on $\mathbb{R}$. Indeed, $s\mapsto1+s^{2}$ is the sum of a constant and the product of the identity map with itself, hence continuous (Continuity of Sums and Products of Real-Valued Functions on a Metric Space (i), Continuity of Identity Maps, Compositions and Reciprocals (i), Continuity of Sums and Products of Real-Valued Functions on a Metric Space (v)); it has no zeros, so its reciprocal $k$ is continuous (Continuity of Identity Maps, Compositions and Reciprocals (iii)). Continuous functions are Borel measurable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iv)), so $k$ is measurable, and being nonnegative it lies in $\mathcal{M}^{+}(\mathbb{R},\mathcal{B}(\mathbb{R}))$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)).

The Riemann integrals. Let $n\in\mathbb{N}$ be arbitrary. The restriction of $k$ to $[-n,n]$ is continuous (Restriction Stability of Continuity and of the Derivative (i)), hence Riemann integrable (A Continuous Function on a Closed Interval is Riemann Integrable (i)). The arctangent is an antiderivative of $k$ there: its restriction to $[-n,n]$ is continuous (The Arctangent: Continuity, Monotonicity, Derivative, Special Values and Oddness (i), Restriction Stability of Continuity and of the Derivative (i)) and differentiable with derivative $k$ (Restriction Stability of Continuity and of the Derivative (ii)). The fundamental theorem of calculus (Fundamental Theorem of Calculus, Part II: Evaluating an Integral by an Antiderivative (i)) and the oddness of the arctangent (The Arctangent: Continuity, Monotonicity, Derivative, Special Values and Oddness (iv)) therefore give

$$I_{n}(k)=\arctan n-\arctan(-n)=2\arctan n .$$

The supremum. By The Lebesgue Integral over ℝ of a Nonnegative Continuous Function as the Limit of Its Riemann Integrals over [−n, n] (ii) (a result resting on dependent choice), applied to the nonnegative continuous function $k$, and by The Riemann integrals,

$$\int_{\mathbb{R}}k\,d\lambda=\sup_{n\in\mathbb{N}}2\arctan n .$$

Since $\arctan$ takes values in $(-\pi/2,\pi/2)$ (The Arctangent Function (i)), each $2\arctan n$ is less than $\pi$. Conversely, let $\varepsilon$ with $0<\varepsilon<2\pi$ be given. First put $t=\pi/2-\varepsilon/2$, which lies in $(-\pi/2,\pi/2)$, and $y=\tan t$ (The Tangent Function on $(-\pi/2,\pi/2)$ (i)), so that $\arctan y=t$ (The Arctangent Function (i)); then choose $n\in\mathbb{N}$ with $n>y$ (The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (vii)). As the arctangent is increasing (The Arctangent: Continuity, Monotonicity, Derivative, Special Values and Oddness (i)), $y<n$ gives $\arctan y<\arctan n$, and so

$$2\arctan n>2t=\pi-\varepsilon .$$

Now let $s<\pi$. If $s\le0$, then $s<\pi/2=2\arctan1$ ($0=\arctan0$ lies in $(-\pi/2,\pi/2)$, and $\arctan1=\pi/4$, The Arctangent: Continuity, Monotonicity, Derivative, Special Values and Oddness (iii)), and $2\arctan1$ is one of the numbers $2\arctan n$. If $0<s<\pi$, then $\varepsilon=\pi-s$ lies in $(0,\pi)$, and the argument above gives an $n$ with $2\arctan n>\pi-\varepsilon=s$. Thus $\pi$ is an upper bound of the numbers $2\arctan n$ that no smaller number is, so the supremum is $\pi$ (Order and Suprema in the Extended Real Line (iv)). Together with Measurability of $k$, this proves (i).

Integrability. Since $\sup_{n}I_{n}(k)=\pi<\infty$, the integrability criterion for nonnegative continuous functions (The Lebesgue Integral over ℝ of a Nonnegative Continuous Function as the Limit of Its Riemann Integrals over [−n, n] (iii)) gives $k\in\mathcal{L}^{1}(\lambda)$, which is (ii).