A simple regulariation

(This is a slight modification of my old post.)

Proposition. Let $f : (0,\infty) \to \Bbb{C}$ be a locally integrable function that satisfies the following conditions:

1. For each $\epsilon > 0$ the following limit converges: $$I(\epsilon) = \lim_{R\to\infty} \int_{\epsilon}^{R} \frac{f(x)}{x} \, \mathrm{d}x$$
2. There exist constants $m = m(f)$ and $c = c(f)$ such that $$I(\epsilon) = -m\log\epsilon + c + o(1) \qquad \text{as }\epsilon \to 0^+.$$

Then the Laplace transform $\mathcal{L}f(x)$ is a well-defined continuous function on $(0,\infty)$ and we have $$c = \lim_{\substack{\epsilon & \to 0^+ \\ R&\to\infty}} \left( \int_{\epsilon}^{R} \mathcal{L}f(s) \, \mathrm{d}s - m \log R \right) - m\gamma, \tag{1}$$ where $\gamma$ is the Euler-Mascheroni constant.

Remark. Before the actual proof, we make some remarks.

1. If in addition that $\lim_{x\to 0^+} f(x)$ converges, then this limit is exactly $m(f)$: $$m(f) = \lim_{x\to 0^+} f(x)$$ This renders $m(f)$ uninteresting in most applications.
2. $c(f)$ can be represented by the integral $$\begin{align*} c(f) &= \int_{0}^{\infty} \frac{f(x) - m(f) \mathbf{1}_{(0,1)}(x)}{x} \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{f(x) - m(f)}{x} \, \mathrm{d}x + \int_{1}^{\infty} \frac{f(x)}{x} \, \mathrm{d}x. \end{align*}$$
3. If we impose a strong assumption on $f$, the proof greatly simplifies. The following lengthy proof is only meaningful when a full generality is needed.

Proof. Let $g : (0,\infty) \to \Bbb{C}$ and $J : (0,\infty) \to \Bbb{C}$ be defined by $$g(x) = \frac{f(x) - m\mathbf{1}_{(0,1)}(x)}{x}, \qquad J(\epsilon) = \lim_{R\to\infty} \int_{\epsilon}^{R} g(x) \, \mathrm{d}x.$$ By the assumption, $J$ is well-defined and satisfies $$c = \lim_{x \to 0^+} J(x), \qquad 0 = \lim_{x\to\infty} J(x).$$ In particular, $J$ extends to a continuous function on $[0,\infty]$ and hence bounded. Now using the identity $J'(x) = -g(x)$ on $(0,\infty)$, for $0 \lt \epsilon \lt 1 \lt R$ we have $$\begin{align*} &\int_{\epsilon}^{R} f(x) e^{-sx} \, \mathrm{d}x \\ &= m \int_{\epsilon}^{1} e^{-sx} \, \mathrm{d}x - \int_{\epsilon}^{R} x J'(x) e^{-sx} \, \mathrm{d}x \\ &= m \int_{\epsilon}^{1} e^{-sx} \, \mathrm{d}x - \left[ xJ(x)e^{-sx} \right]_{\epsilon}^{R} + \int_{\epsilon}^{R} J(x)(1 - sx)e^{-sx} \, \mathrm{d}x. \end{align*}$$ For $s \gt 0$, taking $\epsilon \to 0^+$ and $R \to \infty$ shows that the expression above converges to $$\begin{align*} \mathcal{L}f(s) &:= \lim_{\substack{\epsilon & \to 0^+ \\ R&\to\infty}} \int_{\epsilon}^{R} f(x) e^{-sx} \, \mathrm{d}x \\ &= m\cdot\frac{1 - e^{-s}}{s} + \int_{0}^{\infty} J(x)(1 - sx)e^{-sx} \, \mathrm{d}x. \end{align*}$$ An easy but important remark is that the last integral converges absolutely for $s \gt 0$. Now by the simple estimate $$\int_{\epsilon}^{R} \int_{0}^{\infty} (sx)^\alpha e^{-sx} \, \mathrm{d}x \, \mathrm{d}s = \int_{\epsilon}^{R} \frac{\Gamma(\alpha+1)}{s} \, \mathrm{d}s \lt \infty$$ that holds true for $\alpha \gt -1$, we can apply Fubini's theorem to write $$\begin{align*} &\int_{\epsilon}^{R} \mathcal{L}f(s) \, \mathrm{d}s \\ &= m \int_{\epsilon}^{R} \frac{1 - e^{-s}}{s} \, \mathrm{d}s + \int_{0}^{\infty} J(x) \left( \int_{\epsilon}^{R} (1 - sx)e^{-sx} \, \mathrm{d}s \right) \, \mathrm{d}x \\ &= m \left[ (1-e^{-s})\log s \right]_{\epsilon}^{R} - m \int_{\epsilon}^{R} e^{-s}\log s \, \mathrm{d}s \\ &\qquad + \int_{0}^{\infty} J(x) \left( Re^{-Rx} - \epsilon e^{-\epsilon x} \right) \, \mathrm{d}x. \end{align*}$$ From this computation, it follows that $$\lim_{\substack{\epsilon & \to 0^+ \\ R&\to\infty}} \left( \int_{\epsilon}^{R} \mathcal{L}f(s) \, \mathrm{d}s - m\log R \right) = m \gamma + c.$$ This proves $\text{(1)}$ as expected. ////

In many applications, the logarithmic term is cancelled out and thus the primary interest lies in the value of $c = c(f)$. An easy observation is that both $m$ and $c$ are linear. A less obvious properties are summarized in the following table.

Transformation	Relation	Conditions
$g(x) = f(x^p)$	$c(g) = \frac{1}{p}c(f)$	$p > 0$
$g(x) = f(px)$	$c(g) = c(f) - m(f)\log p$	$p > 0$
$g(x) = f(x)e^{-\alpha x}$	$c(g) = c(f) - \int_{0}^{\alpha} \mathcal{L}f(s) \, ds$	$\Re(\alpha) > 0$

It is worth it noting that $m(g) = m(f)$ in all of the transformations listed above. Next, the following table summaries some well-known values of $c$.

Function $f(x)$	Value of $m(f)$	Value of $c(f)$	Conditions
$e^{-x}$	$f(0) = 1$	$-\gamma$	—
$\cos x$	$f(0) = 1$	$-\gamma$	—
$\dfrac{1}{(1+x)^p}$	$f(0) = 1$	$-H_{p-1}$	$p > 0$
$\dfrac{x}{e^x - 1}$	$f(0) = 1$	$0$	—

Here are some easy examples.

Example 1. Let $p, q \gt 0$. Then $$\begin{align*} &\int_{0}^{\infty} \frac{\cos (x^p) - \exp(-x^q)}{x} \, \mathrm{d}x \\ &= c\{\cos(x^p) - \exp(-x^q)\} = \frac{1}{p} c\{\cos x\} - \frac{1}{q} c\{e^{-x}\} \\ & = \gamma \left( \frac{1}{q} - \frac{1}{p} \right). \end{align*}$$

Example 2. Let $\alpha, \beta \gt 0$. Then $$\begin{align*} &\int_{0}^{\infty} \frac{1}{x} \left( \frac{1}{1 + \alpha^2 x^2} - \cos (\beta x) \right) \, \mathrm{d}x \\ &= c\left\{ \frac{1}{1 + \alpha^2 x^2} - \cos (\beta x) \right\} \\ &= \frac{1}{2} c\left\{ \frac{1}{1 + x} \right\} - c \{\cos x\} - \log \left( \frac{\alpha}{\beta} \right) \\ &= \gamma - \log \left( \frac{\alpha}{\beta} \right). \end{align*}$$

Example 3. The Laplace transform of the Bessel function $J_0$ of the first kind and order $0$ is given by^[1] $$\mathcal{L} \{J_0\} (s) = \frac{1}{\sqrt{s^2 + 1}}$$ Using this and $m(J_0) = J_0(0) = 1$, we find that $$\begin{align*} c(J_0) &= \lim_{R \to \infty} \left( \int_{0}^{R} \mathcal{L} \{J_0\}(s) \, \mathrm{d}s - \log R \right) - \gamma \\ &= \lim_{R \to \infty} \left( \operatorname{arsinh}(R) - \log R \right) - \gamma \\ &= \log 2 - \gamma. \end{align*}$$ This also shows $$\int_{0}^{\infty} \frac{J_0(x) - \mathbf{1}_{(0,1)}(x)}{x} \, \mathrm{d}x = \log 2 - \gamma.$$

Fundamental Theorem of Calculus

Theorem. Assume that $f : [a, b] \to \Bbb{R}$ is differentiable on $[a, b]$ and $f'$ is in $L^1$. Then $$\int_{a}^{b} f'(x) \, dx = f(b) - f(a).$$

Remark. This proof is a slightly simplified version of the proof of Theorem 7.21 in Rudin's Real and Complex Analysis, 3rd edition.

Proof. Let $l$ be a lower-semicontinuous function on $[a, b]$ such that $l(x) > f'(x)$ for all $x \in [a, b]$. Define $G : [a, b] \to \Bbb{R}$ by $$G(x) = \int_{a}^{x} l(t) \, dt - [f(x) - f(a)].$$ Then for each $x \in [a, b)$, we have $$\begin{align*} \frac{G(x+h) - G(x)}{h} \geq \left( \inf_{t \in [x,x+h]}l(t) \right) - \frac{f(x+h) - f(x)}{h} \end{align*}$$ and thus $$\begin{align*} \liminf_{h\to 0^+} \frac{G(x+h) - G(x)}{h} &\geq \liminf_{h\to 0^+} l(x+h) - f'(x) \\ &\geq l(x) - f'(x) \\ &> 0. \end{align*}$$ This shows that $G$ is increasing on $[a, b)$ and by continuity, $G(b) \geq G(a) = 0$. From this, we have $$\int_{a}^{b} l(t) \, dt \geq f(b) - f(a).$$ By the Vitali-Caratheodory theorem, $f'$ can be approximated from above by lower-semicontinuous functions in $L^1$. Thus it follows that $$\int_{a}^{b} f'(t) \, dt \geq f(b) - f(a).$$ Replacing $f$ by $-f$ proves the other direction and hence the claim follows. ////