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Suppose we have a locally integrable function $f: \mathbb R^+ \to \mathbb R$ and we consider the 'Hilbert transformed' function $$ h(t) := \int_0^\infty \frac{f(\tau)}{t -\tau} \mathrm d \tau. $$ We presume this function is well-defined (possibly in the Cauchy PV sense).

Suppose $f(t) \sim \frac{\beta}{t^\alpha}$ (with $\alpha > 0$) as $t \to \infty$. I would like to show that (or know which extra condition(s) to impose such that) the asymptotics of the Hilbert transform is $$ h(t) \sim_{t \to -\infty} \left\{ \begin{array}{ccl} \frac{\gamma}{|t|^{\min (1,\alpha)}} & & \textrm{if }0 < \alpha \neq 1, \\ \frac{\gamma \ln |t|}{t} & & \textrm{if }\alpha = 1. \end{array} \right. $$

(Note that I am looking at negative $t$.)

The intuitive idea for why I believe this to be true---which I struggle to make rigorous---is the following: I would like to split $h(t)$ up into two parts: $$ h(t) = \underbrace{ \int_0^{t^*} \frac{f(\tau)}{t -\tau} \mathrm d \tau}_{=: h_1(t)} + \underbrace {\int_{t^*}^\infty \frac{f(\tau)}{t -\tau} \mathrm d \tau}_{=: h_2(t)} $$ such that (in a sense to be made precise) in the second integral we can replace $f(\tau)$ by its asymptotic behavior: $$ h_2(t) \approx \beta \int_{t^*}^\infty \frac{1}{\tau^\alpha (t-\tau)}\mathrm d \tau. $$

It is then simple to show that $h_1(t)$ is asymptotically $\sim \frac{\gamma_1}{t}$ (this is the well-known asymptotic behavior of the Hilbert transform of a function on compact support) and $h_2(t)$ has the asymptotics claimed in the yellow box above. To see the latter, simply observe that $\int_{t^*}^\infty \frac{1}{\tau^\alpha (t-\tau)}\mathrm d \tau = -\frac{1}{|t|^\alpha} \int_{t^*/|t|}^\infty \frac{1}{x^\alpha(1+x)} \mathrm d x$ (for negative $t$), and $$\int_\varepsilon^\infty \frac{1}{x^\alpha(1+x)} \mathrm d x \sim_{\varepsilon \to 0^+} \left\{ \begin{array}{lll} \textrm{cst} & & \textrm{if } 0 < \alpha < 1, \\ \textrm{cst} \; \times \; \ln \varepsilon & & \textrm{if } \alpha = 1, \\ \textrm{cst} \; \times \; \varepsilon^{1-\alpha} & & \textrm{if } \alpha \geq 1. \end{array} \right.$$

EDIT: It turns out that all of the above statements are proven in the 1969 article 'Some Asymptotic Behavior of Stieltjes Transforms' by Zimering. (More precisely, case $0 < \alpha < 1$ is 'Theorem 1', case $\alpha = 1$ is 'Theorem 2' and case $\alpha > 1$ is 'Theorem 4'. Also note that the Hilbert/Stieltjes transform considered by the author is related to the above as $-h(-t)$; hence when the author excludes the negative real axis, it corresponds to excluding the positive real axis of $h(t)$, which is fine for our purposes.) I thank Nick Jones for making me aware of this reference.

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General remarks

For the following arguments it is indeed sufficient that $f$ is locally integrable on $[0,\infty)$ . For some $\alpha > 0$, $\beta \in \mathbb{R}\setminus \{0\}$ we have the asymptotic equivalence $f(\tau) \sim \frac{\beta}{\tau^\alpha}$ as $\tau \to \infty$ , which by definition means that $$ \lim_{\tau \to \infty} \left| \frac{f(\tau) - \frac{\beta}{\tau^\alpha}}{\frac{\beta}{\tau^\alpha}} \right| = 0 $$ holds. Equivalently, for all $\varepsilon > 0$ there is a $t_\varepsilon > 0$ such that we have $$ \tag{1} \frac{\beta - \frac{|\beta| \varepsilon}{2}}{\tau^\alpha} \leq f(\tau) \leq \frac{\beta + \frac{|\beta| \varepsilon}{2}}{\tau^\alpha} \, , \, \tau > t_\varepsilon \, . $$

We consider the function $$ h \colon \mathbb{R}^- \to \mathbb{R}\, , \, h(t) = \frac{1}{\pi} \int \limits_0^\infty \frac{f(\tau)}{t - \tau} \, \mathrm{d} \tau \, ,$$ and define $t = -s$ for simplicity. We will also make use of the inequality $$ \tag{2} \left|-\frac{1}{\pi} \int \limits_0^{t_\varepsilon} \frac{f(\tau)}{s + \tau} \, \mathrm{d} \tau \right| \leq \frac{1}{\pi s} \int \limits_0^{t_\varepsilon} |f(\tau)| \, \mathrm{d} \tau \, .$$

The case $\alpha > 1$

For $\alpha > 1$ the local integrability and the asymptotic behaviour imply $f \in L^1 (\mathbb{R})$, so $$ h(-s) = - \frac{1}{\pi s} \int \limits_0^\infty f(\tau) \, \mathrm{d} \tau + \frac{1}{\pi s} \int \limits_0^\infty \frac{f(\tau) \tau}{s + \tau} \, \mathrm{d} \tau \, . $$ We now use equation $(1)$ with $\varepsilon = 2$ to obtain $$\left| \int \limits_0^\infty \frac{f(\tau) \tau}{s + \tau} \, \mathrm{d} \tau \right| \leq \frac{1}{s} \int \limits_0^{t_2} |f(\tau)| \tau \, \mathrm{d} \tau + 2 |\beta| \int \limits_{t_2}^\infty \frac{\tau^{1-\alpha}}{s + \tau} \, \mathrm{d} \tau \stackrel{s \to \infty}{\longrightarrow} 0 $$ by the dominated convergence theorem. Thus we find $$ h (-s) \sim - \frac{1}{\pi s} \int \limits_0^\infty f(\tau) \, \mathrm{d} \tau \, , \, s \to \infty \, . $$

The case $\alpha = 1$

Fix $\varepsilon > 0$ . Then equation $(1)$ yields $$ - \frac{1}{\pi} \int \limits_{t_\varepsilon}^\infty \frac{f(\tau)}{s + \tau} \, \mathrm{d} \tau \leq - \frac{\beta - \frac{|\beta| \varepsilon}{2}}{\pi} \int \limits_{t_\varepsilon}^\infty \frac{1}{\tau(s + \tau)} \, \mathrm{d} \tau = - \frac{\beta - \frac{|\beta| \varepsilon}{2}}{\pi s} \ln \left(1+\frac{s}{t_\varepsilon}\right)$$ and similarly we find $$ - \frac{1}{\pi} \int \limits_{t_\varepsilon}^\infty \frac{f(\tau)}{s + \tau} \, \mathrm{d} \tau \geq - \frac{\beta + \frac{|\beta| \varepsilon}{2}}{\pi s} \ln \left(1+\frac{s}{t_\varepsilon}\right) \, .$$ We can combine these results with equation $(2)$ to obtain $$ \left|\frac{h(-s) + \frac{\beta \ln(s)}{\pi s}}{\frac{\beta \ln(s)}{\pi s}}\right| \leq \frac{1}{\ln(s)} \left[ \frac{1}{\beta} \int \limits_0^{t_\varepsilon} |f(\tau)| \, \mathrm{d} \tau + \left|\ln\left(\frac{1}{s} + \frac{1}{t_\varepsilon}\right)\right| \right] + \frac{\varepsilon}{2} \frac{\left|\ln\left(1+\frac{s}{t_\varepsilon}\right)\right|}{\ln(s)} \, . $$ The first term vanishes as $s \to \infty$, so for $\varepsilon < 1$ we can choose $s$ large enough to make it smaller than $\frac{\varepsilon(1 - \varepsilon)}{2}$ . For sufficiently large $s$ the second term is smaller than $\frac{\varepsilon (1 + \varepsilon)}{2}$ and we get $$ \lim_{s\to\infty} \left|\frac{h(-s) + \frac{\beta \ln(s)}{\pi s}}{\frac{\beta \ln(s)}{\pi s}}\right| \leq \frac{\varepsilon(1 - \varepsilon)}{2} + \frac{\varepsilon(1 + \varepsilon)}{2} = \varepsilon \, . $$ Since $\varepsilon > 0$ was arbitrary, we can let $\varepsilon \searrow 0$ to establish the asymptotic equivalence $$ h_2 (-s) \sim - \frac{\beta \ln(s)}{\pi s} \, , \, s \to \infty \, . $$

The case $\alpha < 1$

We again take $\varepsilon > 0$ and use equation $(1)$ to find \begin{align} - \frac{1}{\pi} \int \limits_{t_\varepsilon}^\infty \frac{f(\tau)}{s + \tau} \, \mathrm{d} \tau &\leq - \frac{\beta - \frac{|\beta| \varepsilon}{2}}{\pi} \int \limits_{t_\varepsilon}^\infty \frac{1}{\tau^\alpha (s + \tau)} \, \mathrm{d} \tau \\ &= - \frac{\beta - \frac{|\beta| \varepsilon}{2}}{\pi} \left[\frac{\pi}{\sin(\pi \alpha) s^\alpha} - \int \limits_0^{t_\varepsilon} \frac{1}{\tau^\alpha (s + \tau)} \, \mathrm{d} \tau \right] \\ &= - \frac{\beta - \frac{|\beta| \varepsilon}{2}}{\sin(\pi \alpha) s^\alpha} \left[1 + \mathcal{O} \left(\frac{1}{s^{1-\alpha}}\right) \right] \end{align} and $$ - \frac{1}{\pi} \int \limits_{t_\varepsilon}^\infty \frac{f(\tau)}{s + \tau} \, \mathrm{d} \tau \geq - \frac{\beta + \frac{|\beta| \varepsilon}{2}}{\sin(\pi \alpha) s^\alpha} \left[1 + \mathcal{O} \left(\frac{1}{s^{1-\alpha}}\right) \right] \, . $$ We can now follow exactly the same steps as in case $\alpha = 1$ to show that $$ h(-s) \sim - \frac{\beta}{\sin(\pi \alpha) s^\alpha} $$ holds as $s \to \infty$ .

Summary

We have $$ h(t) \sim \begin{cases} - \frac{1}{\pi |t|} \int \limits_0^\infty f(\tau)\, \mathrm{d} \tau &, \, \alpha > 1 \\ - \frac{\beta \ln(|t|)}{\pi |t|} &, \, \alpha = 1 \\ - \frac{\beta}{\sin(\pi \alpha) |t|^\alpha} &, \, \alpha < 1 \end{cases} \, , \, t \to - \infty \, ,$$ which agrees with your conjecture.

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  • $\begingroup$ Thank you very much. I had indeed been sloppy in my OP with the case $\alpha = 1$; I fixed the argument accordingly. At the same time as you were writing your post, I was made aware of a paper from the 60's which agrees with all your claims exactly. (I have added the reference to that in the main post.) $\endgroup$ – Ruben Verresen Jul 29 '18 at 19:13
  • $\begingroup$ @RubenVerresen Nice to have my results confirmed, thank you! $\endgroup$ – ComplexYetTrivial Jul 29 '18 at 19:26

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