Proof of Logarithmic Integrals of Reciprocal Linear Functions over Bounded and Unbounded Intervals

Each result cited is universally quantified over the data in its own statement. Throughout, $\mathrm{Leb}$ is Lebesgue measure on $\mathcal{B}(\mathbb{R})$; for a Borel set $B$ and $f:B\to\mathbb{R}$, $\tilde f:\mathbb{R}\to\mathbb{R}$ is the extension of $f$ by $0$, as in Rays, and Integrability and the Integral of a Real Function on a Borel Subset of the Real Line §integral. Integrals of nonnegative measurable functions are those of Lebesgue Integral of a Nonnegative Measurable Function, with values in $[0,\infty]$; integrals of integrable functions are those of Integrable Function and the Lebesgue Integral, which also records that a measurable $f:\mathbb{R}\to\mathbb{R}$ is integrable if and only if $\int_{\mathbb{R}}|f|\,d\mathrm{Leb}<\infty$. For a real number $p$, the singleton $\{p\}$ is the closed interval $[p,p]$, so $\mathrm{Leb}(\{p\})=p-p=0$ by claim 4 of Existence of Lebesgue Measure on the Real Line. Every interval is a Borel set by Borel Sigma-Algebra on the Real Line. The real line $(\mathbb{R},d_{\mathbb{R}})$ is that of The Absolute Value Metric on the Real Line.

Part 1 (bounded intervals). Let $a,s,b$ be real with $a<s$ and $a+b>0$.

Positivity. For $t\in[a,s]$ we have $a\le t$, hence $b+a\le b+t$, and $0<b+a$; so $0<b+t$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, and moreover $(b+t)^{-1}\le(b+a)^{-1}$ and $0<(b+t)^{-1}$ by Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal. Hence $u:[a,s]\to\mathbb{R}$, $u(t)=(b+t)^{-1}$, is well defined, and $0<u(t)\le(b+a)^{-1}$ for every $t\in[a,s]$.

Continuity of $u$. As maps from subsets of the real line to the real line: the map $t\mapsto t$ on $[a,s]$ is continuous by Continuity of the Identity Map, of Powers, and of Polynomial Functions on a Subset of the Real Line §identity; the constant map $t\mapsto b$ is continuous at every point by claim 1 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space; so $t\mapsto b+t$ is continuous on $[a,s]$ by claim 5 of the same theorem. It does not vanish on $[a,s]$, so $u$ is continuous on $[a,s]$ by claim 2 of Continuity of the Reciprocal of a Nonvanishing Real-Valued Function on a Metric Space.

An antiderivative. Let $I=\{t\in\mathbb{R}:-b<t\}$ and $J=\{y\in\mathbb{R}:0<y\}=(0,\infty)$. Both are intervals, and every point of each is an interior point: for $t_{0}\in I$ the points $(t_{0}-b)/2$ and $t_{0}+1$ belong to $I$ and satisfy $(t_{0}-b)/2<t_{0}<t_{0}+1$ (because $-b<t_{0}$, using claims 1, 6 and 8 of Elementary Order Arithmetic in an Ordered Field), and for $y_{0}\in J$ the points $y_{0}/2$ and $y_{0}+1$ do the same job (claims 1, 6 and 8 of the same lemma). By the positivity step, $0<b+t$ for $t\in[a,s]$, which by claim 1 of Elementary Order Arithmetic in an Ordered Field (adding $-b$) is $-b<t$; so $[a,s]\subseteq I$. Let $\gamma:I\to\mathbb{R}$, $\gamma(t)=b+t$; then $\gamma(t)\in J$ for every $t\in I$. At every $t_{0}\in I$ the map $\gamma$ is differentiable with $\gamma'(t_{0})=1$: the restriction to $I$ of $x\mapsto x^{1}=x$ has derivative $1$ at $t_{0}$ by claim 1 of Derivative of a Polynomial Function on the Real Line, the constant map with value $b$ has derivative $0$ by claim 1 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, and the derivative of a sum is the sum of the derivatives by claim 2 of that lemma. Next, $\log:J\to\mathbb{R}$ (The Natural Logarithm) is differentiable at every $y\in J$ in the sense of Derivative at an Interior Point, with derivative $1/y$; we check this directly from the definition. Fix $y\in J$ and a real $\varepsilon>0$. By claims 5, 7 and 8 of Elementary Order Arithmetic in an Ordered Field, $y/2$ and $\varepsilon y^{2}/2$ are positive, and by claim 9 of that lemma there is a positive real $\eta$ equal to the smaller of them. Let $h\in\mathbb{R}$ satisfy $0<|h|<\eta$ and $y+h\in J$. Since $|h|<y/2$, claim 9 of Properties of the Absolute Value in an Ordered Field gives $-y/2<h$, hence $y/2<y+h$ by claim 1 of Elementary Order Arithmetic in an Ordered Field (as $y-y/2=y/2$). Thus $r=(y+h)/y$ is a positive real number (claims 2, 5 and 7 of Elementary Order Arithmetic in an Ordered Field), and $\log(y+h)=\log(r\,y)=\log(r)+\log(y)$ by the product rule for $\log$ recorded in The Natural Logarithm. By The Function $s\log s$: Continuity, Young's Inequality and Lower Bounds, with the Elementary Bounds for the Exponential and the Logarithm §log, $1-r^{-1}\le\log(r)\le r-1$, and by field arithmetic $r-1=h/y$ and $1-r^{-1}=h/(y+h)$. Adding $-h/y$ to both inequalities (compatibility of $\le$ with addition in Ordered Field) gives

$$-\frac{h^{2}}{y(y+h)}=\frac{h}{y+h}-\frac{h}{y}\le\log(y+h)-\log(y)-\frac{h}{y}\le0 ,$$

so, by claims 1, 4 and 6 of Properties of the Absolute Value in an Ordered Field (note $h^{2}=|h|^{2}$), $\bigl|\log(y+h)-\log(y)-h/y\bigr|\le|h|^{2}\,\bigl(y(y+h)\bigr)^{-1}$. From $y/2<y+h$ and $0<y$ we get $y^{2}/2<y(y+h)$ by claim 10 of Elementary Order Arithmetic in an Ordered Field, so $\bigl(y(y+h)\bigr)^{-1}\le2/y^{2}$ by Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal. Dividing by $|h|>0$ (which is legitimate by claim 7 of Elementary Order Arithmetic in an Ordered Field; claim 4 of Properties of the Absolute Value in an Ordered Field gives $|x/h|=|x|/|h|$, multiplication by the nonnegative $|h|$ preserves $\le$ by Ordered Field, and multiplication by the positive $2/y^{2}$ preserves $<$ by claim 10 of Elementary Order Arithmetic in an Ordered Field),

$$\left|\frac{\log(y+h)-\log(y)}{h}-\frac{1}{y}\right|\le\frac{|h|}{y(y+h)}\le\frac{2|h|}{y^{2}}<\frac{2\eta}{y^{2}}\le\varepsilon .$$

As $\varepsilon>0$ was arbitrary, $\log$ is differentiable at $y$ with $\log'(y)=1/y$. Put $F=\log\circ\gamma:I\to\mathbb{R}$, so $F(t)=\log(b+t)$. By Chain Rule for One-Dimensional Derivatives, $F$ is differentiable at every $t_{0}\in I$ with

$$F'(t_{0})=\frac{1}{b+t_{0}}\cdot1=\frac{1}{b+t_{0}} .$$

By Differentiability at an Interior Point Implies Continuity There, $F$ is continuous at every point of $I$ relative to $I$, that is, continuous on $I$ as a map from a subset of the real line to the real line; and by claim 1 of Restriction Stability of Continuity and of the Derivative its restriction $F|_{[a,s]}$ is continuous on $[a,s]$. Every $x$ with $a<x<s$ is an interior point of $[a,s]$, so by claim 2 of Restriction Stability of Continuity and of the Derivative the restriction $F|_{[a,s]}$ is differentiable at $x$ with derivative $F'(x)=u(x)$.

The Riemann integral. By claim 3 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval, the continuous function $u$ is Riemann integrable on $[a,s]$. Hence Fundamental Theorem of Calculus, Part II, on a Closed Real Interval, applied on $[a,s]$ to $u$ and $F|_{[a,s]}$, gives

$$\int_{a}^{s}u(t)\,dt=F(s)-F(a)=\log(b+s)-\log(b+a).$$

The Lebesgue integral over $[a,s]$. Let $([a,s],\mathcal{B}_{[a,s]},\lambda_{[a,s]})$ be the restricted measure space of claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval. By claim 3 of that lemma, $u$ is a random variable on the associated probability space, hence $\mathcal{B}_{[a,s]}$-measurable; so by claim 2 of that lemma its zero extension $w:\mathbb{R}\to\mathbb{R}$ (equal to $u$ on $[a,s]$ and to $0$ elsewhere) is measurable. By the positivity step, $0\le w\le(b+a)^{-1}\mathbf{1}_{[a,s]}$ pointwise, so by claim 1 of Linearity and Monotonicity of the Lebesgue Integral and The Integral of an Indicator Function is the Measure of the Set,

$$\int_{\mathbb{R}}|w|\,d\mathrm{Leb}\le(b+a)^{-1}\,\mathrm{Leb}([a,s])=(b+a)^{-1}(s-a)<\infty ,$$

using claim 4 of Existence of Lebesgue Measure on the Real Line. Thus $w$ is integrable. By claim 2 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (real-valued case), $u$ is integrable with respect to $\lambda_{[a,s]}$ and $\int_{\mathbb{R}}w\,d\mathrm{Leb}=\int_{[a,s]}u\,d\lambda_{[a,s]}$, and by claim 3 of that lemma the latter equals the Riemann integral computed above. Hence

$$\int_{\mathbb{R}}w\,d\mathrm{Leb}=\log(b+s)-\log(b+a).$$

Passage to the open interval. Let $v:(a,s)\to\mathbb{R}$, $v(t)=1/(b+t)$, and let $\tilde v$ be its extension by $0$. Then $\tilde v=w\,\mathbf{1}_{(a,s)}$ pointwise, and $(a,s)$ is a Borel set, so $\tilde v$ is measurable by claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. The functions $\tilde v$ and $w$ agree outside $\{a\}\cup\{s\}$, which is null: apply The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-union to the sequence $N_{1}=\{a\}$, $N_{2}=\{s\}$ and $N_{m}=\varnothing$ for $m\ge3$, all null by the preliminary remark (and $\mathrm{Leb}(\varnothing)=0$), whose union is $\{a\}\cup\{s\}$. By The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison, $\tilde v$ is integrable and $\int_{\mathbb{R}}\tilde v\,d\mathrm{Leb}=\int_{\mathbb{R}}w\,d\mathrm{Leb}$. By Rays, and Integrability and the Integral of a Real Function on a Borel Subset of the Real Line §integral, $t\mapsto1/(b+t)$ is integrable on $(a,s)$ and

$$\int_{(a,s)}\frac{dt}{b+t}=\log(b+s)-\log(b+a).$$

Part 2 (half-lines). Let $a,b,c$ be real with $a+b>0$ and $a+c>0$. For $t>a$ we have $b+t>b+a>0$ and $c+t>c+a>0$ by claims 1 and 2 of Elementary Order Arithmetic in an Ordered Field, so $(b+t)^{-1}$ and $(c+t)^{-1}$ exist by claim 7 of that lemma and $h:(a,\infty)\to\mathbb{R}$, $h(t)=\frac{1}{b+t}-\frac{1}{c+t}$, is well defined. Put $\delta=c-b$.

The case $\delta=0$. Then $c=b$, so $h(t)=\frac{1}{b+t}-\frac{1}{b+t}=0$ for every $t>a$; and $\tilde h(t)=0$ for $t\le a$ by definition of the extension by $0$. Hence $\tilde h$ is identically $0$ on $\mathbb{R}$. It is measurable by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, its positive part $\tilde h^{+}$ and negative part $\tilde h^{-}$ are both the zero function, and each has integral $0$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-integral (with the null set $\varnothing$). By Integrable Function and the Lebesgue Integral, $\tilde h$ is integrable with $\int_{\mathbb{R}}\tilde h\,d\mathrm{Leb}=0-0=0$. Moreover $a+c=a+b$, so $\log(a+c)-\log(a+b)=0$. By Rays, and Integrability and the Integral of a Real Function on a Borel Subset of the Real Line §integral, $h$ is integrable on $(a,\infty)$ with integral $0=\log(a+c)-\log(a+b)$.

The case $\delta\ne0$: a positive factor. Let $k:(a,\infty)\to\mathbb{R}$, $k(t)=\bigl((b+t)(c+t)\bigr)^{-1}$. By claims 5 and 7 of Elementary Order Arithmetic in an Ordered Field, $k(t)>0$, and by field arithmetic

$$h(t)=\frac{(c+t)-(b+t)}{(b+t)(c+t)}=\delta\,k(t)\qquad(t>a).$$

Truncations. For $N\in\mathbb{N}$, regarded as a real number $N\ge1$, put $s_{N}=a+N$, so $a<s_{N}$. Let $k_{N}:\mathbb{R}\to\mathbb{R}$ equal $k$ on $(a,s_{N})$ and $0$ elsewhere, and let $k_{\infty}=\tilde k$. By Part 1, applied to $(a,s_{N},b)$ and to $(a,s_{N},c)$ (note $a+b>0$ and $a+c>0$), the functions $t\mapsto1/(b+t)$ and $t\mapsto1/(c+t)$ are integrable on $(a,s_{N})$; let $p_{N},q_{N}$ be their extensions by $0$. Then $k_{N}=\delta^{-1}(p_{N}-q_{N})$ pointwise on $\mathbb{R}$ (on $(a,s_{N})$ by the identity above, elsewhere both sides vanish). By claim 2 of Linearity and Monotonicity of the Lebesgue Integral and Part 1, $k_{N}$ is integrable and

$$I_{N}:=\int_{\mathbb{R}}k_{N}\,d\mathrm{Leb}=\delta^{-1}\Bigl(\bigl[\log(b+s_{N})-\log(b+a)\bigr]-\bigl[\log(c+s_{N})-\log(c+a)\bigr]\Bigr).$$

Since $k_{N}\ge0$, its positive part is $k_{N}$ and its negative part is $0$, whose integral is $0$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-integral; so, by Integrable Function and the Lebesgue Integral, $I_{N}$ is also the integral of $k_{N}$ as a nonnegative measurable function.

The limit of $I_{N}$. Let $r_{N}=(b+s_{N})/(c+s_{N})$, a positive real number. By the product rule for $\log$, $\log(b+s_{N})=\log\bigl(r_{N}(c+s_{N})\bigr)=\log(r_{N})+\log(c+s_{N})$, so

$$I_{N}=\delta^{-1}\bigl(\log(r_{N})+\log(c+a)-\log(b+a)\bigr).$$

By The Function $s\log s$: Continuity, Young's Inequality and Lower Bounds, with the Elementary Bounds for the Exponential and the Logarithm §log, $1-r_{N}^{-1}\le\log(r_{N})\le r_{N}-1$, and by field arithmetic

$$r_{N}-1=\frac{-\delta}{c+s_{N}},\qquad 1-r_{N}^{-1}=\frac{-\delta}{b+s_{N}} .$$

Since $b+s_{N}=(a+b)+N>N>0$ and likewise $c+s_{N}>N$, Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal gives $(b+s_{N})^{-1}\le N^{-1}$ and $(c+s_{N})^{-1}\le N^{-1}$. Hence both endpoints $1-r_{N}^{-1}$ and $r_{N}-1$ have absolute value at most $|\delta|N^{-1}$, and so $|\log(r_{N})|\le|\delta|N^{-1}$. Given real $\varepsilon>0$, claim 2 of The Archimedean Property of the Real Numbers gives $n\in\mathbb{N}$ with $|\delta|<n\varepsilon$; then for every $N\ge n$, again by Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal, $|\log(r_{N})|\le|\delta|N^{-1}\le|\delta|n^{-1}<\varepsilon$. Thus $\log(r_{N})\to0$ in the sense of Limit of a Sequence of Real Numbers. The constant sequence with value $\log(c+a)-\log(b+a)$ converges to that value, since every term is at distance $0<\varepsilon$ from it. Hence, by claims 1 and 3 of Arithmetic of Limits of Real Sequences,

$$I_{N}\to L:=\delta^{-1}\bigl(\log(c+a)-\log(b+a)\bigr)\qquad(N\to\infty).$$

Monotone convergence. For $N\in\mathbb{N}$ we have $(a,s_{N})\subseteq(a,s_{N+1})$ and $k>0$, so $k_{N}\le k_{N+1}\le k_{\infty}$ pointwise. For $t\le a$, $k_{N}(t)=0=k_{\infty}(t)$ for every $N$; for $t>a$, claim 1 of The Archimedean Property of the Real Numbers gives $n\in\mathbb{N}$ with $t-a<n$, and then $k_{N}(t)=k(t)=k_{\infty}(t)$ for every $N\ge n$. So at every $t\in\mathbb{R}$ the nondecreasing sequence $(k_{N}(t))_{N}$ is eventually equal to $k_{\infty}(t)$; hence it converges to $k_{\infty}(t)$ and its least upper bound is $k_{\infty}(t)$. By claim 5 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, $k_{\infty}$ is measurable, and by Monotone Convergence Theorem, in $[0,\infty]$,

$$\int_{\mathbb{R}}k_{\infty}\,d\mathrm{Leb}=\sup_{N}I_{N}.$$

The real sequence $(I_{N})_{N}$ is nondecreasing by claim 1 of Linearity and Monotonicity of the Lebesgue Integral and converges to $L$. So $L$ is an upper bound of it: if $I_{N_{0}}>L$ for some $N_{0}$, then $|I_{N}-L|\ge I_{N_{0}}-L>0$ for all $N\ge N_{0}$, contradicting convergence with $\varepsilon=I_{N_{0}}-L$. And no smaller number is an upper bound: for real $\varepsilon>0$ some $I_{N}$ satisfies $|I_{N}-L|<\varepsilon$, hence $I_{N}>L-\varepsilon$. Therefore $\sup_{N}I_{N}=L$, and $\int_{\mathbb{R}}k_{\infty}\,d\mathrm{Leb}=L$ is finite.

Conclusion. As $k_{\infty}\ge0$ is measurable with $\int_{\mathbb{R}}|k_{\infty}|\,d\mathrm{Leb}=L<\infty$, it is integrable by Integrable Function and the Lebesgue Integral, with integral $L$ (its negative part being $0$). Moreover $\tilde h=\delta\,k_{\infty}$ pointwise on $\mathbb{R}$ (on $(a,\infty)$ by the identity $h=\delta k$, elsewhere both sides vanish). By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, $\tilde h$ is integrable and

$$\int_{\mathbb{R}}\tilde h\,d\mathrm{Leb}=\delta L=\log(a+c)-\log(a+b).$$

By Rays, and Integrability and the Integral of a Real Function on a Borel Subset of the Real Line §integral, $h$ is integrable on $(a,\infty)$ with $\int_{(a,\infty)}h(t)\,dt=\log(a+c)-\log(a+b)$.