(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:
- For each $\epsilon > 0$ the following limit converges: $$ I(\epsilon) = \lim_{R\to\infty} \int_{\epsilon}^{R} \frac{f(x)}{x} \, \mathrm{d}x $$
- There exist constants $m = m(f)$ and $c = c(f)$ such that $$I(\epsilon) = -m\log\epsilon + c + o(1) \qquad \text{as }\epsilon \to 0^+.$$
Remark. Before the actual proof, we make some remarks.
- 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.
- $c(f)$ 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*}
- 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 $$ \int_{0}^{\infty} \frac{J_0(x) - \mathbf{1}_{(0,1)}(x)}{x} \, \mathrm{d}x = \log 2 - \gamma. $$
