An approach to evaluating a Cauchy principal value that yields unexpected extra imaginary term

Question

$\newcommand{\PV}{\operatorname{PV}}$ I have been working on evaluating the first negative moment of a random variable with a piecewise density function by means of the Cauchy principal value, i.e. \begin{equation} \tag{1} PV\!E(X^{-1})=\PV\int_{-\infty}^\infty\frac{f_X(x)}{x}\,\mathrm dx = \lim_{\epsilon\to0}\left(\int_{-\infty}^{-\epsilon}+\int_\epsilon^\infty \right)\frac{f_X(x)}{x}\,\mathrm dx, \end{equation} where $f_X$ is a probability density function. I was unable to get any traction using the definition in $(1)$ but was able to evaluate the moments of $X$ greater than one which yielded \begin{equation} \begin{aligned} E(X^{\epsilon-1}) &=\int_{-\infty}^0 x^{\epsilon-1}f_X(x)\,\mathrm dx% +\int_0^\infty x^{\epsilon-1}f_X(x)\,\mathrm dx,\\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &=\int_0^\infty x^{\epsilon-1}f_X(x)\,\mathrm{d}x% -e^{\mathrm i\pi\epsilon}\int_{-\infty}^0 |x|^{\epsilon-1}f_X(x)\,\mathrm dx,\\ &\propto\Gamma(\epsilon)\left(h_1(\epsilon)-e^{i\pi\epsilon}h_2(\epsilon)\right):=\Gamma(\epsilon)D(\epsilon). \end{aligned} \end{equation} I then tried taking the limit of the moments as $\epsilon\to0$. As $\epsilon\to0$, $D(\epsilon)\to0$ while the gamma term blows up. However, by writing the limit as \begin{equation} \lim_{\epsilon\to0}E(X^{\epsilon-1})\propto\Gamma(\epsilon+1)D(\epsilon)/\epsilon, \end{equation} we see that it is $0/0$ indeterminant. Using L'Hopitals rule and evaluating the limit I got something of the form \begin{equation} \lim_{\epsilon\to0}E(X^{\epsilon-1})= L-f_X(0)\pi\mathrm i. \end{equation} Numerical evaluation revealed that $L=PV\!E(X^{-1})$; thus, what I have found is \begin{equation} \tag{2} \lim_{\epsilon\to0}E(X^{\epsilon-1})= PV\!E(X^{-1})-f_X(0)\pi\mathrm i. \end{equation} I was able to come up with a proof of the result in $(2)$ by means of contour integration but was wondering if there other ways of getting to this result. So with that said, what is going on here? Why does adding the extra power to $t$ result in a residual imaginary term?

Answer 1

What is going on here? Why does the time in which I add the extra power to $t$ matter?

In the first case, for $t < 0$ you multiply the integrand with $e^{i\pi\varepsilon}\cdot \lvert t\rvert^{\varepsilon}$, and in the second case you multiply it just with $\lvert t\rvert^{\varepsilon}$. When we're looking only at the part of the integral for $t < 0$, it is clear by the properties of $f_X$ that

$$I(\varepsilon) := \int_{-\infty}^0 \lvert t\rvert^{\varepsilon}\cdot \frac{f_X(t)}{t}\,dt$$

tends to $-\infty$ as $\varepsilon \searrow 0$. What is not immediately obvious is that the decrease of $I(\varepsilon)$ is essentially proportional to $\varepsilon^{-1}$, and that causes the appearance of the purely imaginary term, since

$$e^{i\pi\varepsilon} = \cos (\pi\varepsilon) + i\sin (\pi\varepsilon) = 1 + i\pi\varepsilon + O(\varepsilon^2)\,.$$

The real part, $\cos (\pi \varepsilon) I(\varepsilon) = I(\varepsilon) + O\bigl(\varepsilon^2 I(\varepsilon)\bigr) = I(\varepsilon) + O(\varepsilon)$, combines with the corresponding integral over $t > 0$ to yield the Cauchy principal value of the integral, and the imaginary part $\sin (\pi\varepsilon)I(\varepsilon)$ produces the $-f_X(0)\pi$.

To see this, split the integral some more. Since $f_X$ is integrable we have

$$\lim_{\varepsilon \to 0} \int_{-\infty}^{-1} \lvert t\rvert^{\varepsilon} \frac{f_X(t)}{t}\,dt = \int_{-\infty}^{-1} \frac{f_X(t)}{t}\,dt$$

by the dominated convergence theorem. We can write the integral from $-1$ to $0$ as

$$\int_{-1}^0 \lvert t\rvert^{\varepsilon} \frac{f_X(t)}{t}\,dt = \int_{-1}^0 \lvert t\rvert^{\varepsilon} \frac{f_X(t) - f_X(0)}{t}\,dt + f_X(0)\int_{-1}^0 \frac{\lvert t\rvert^{\varepsilon}}{t}\,dt\,.$$

If $f_X$ is well-behaved at $0$, the first integral on the right is harmless. Well-behaved means that

$$\int_{-1}^0 \biggl\lvert \frac{f_X(t) - f_X(0)}{t}\biggr\rvert\,dt < +\infty$$

here. For that it is sufficient that $f_X$ is $\alpha$-Hölder continuous for some $\alpha > 0$ (for then the integrand is dominated by a multiple of $\lvert t\rvert^{\alpha - 1}$). This is the case for your $f_X$. Then, again by the dominated convergence theorem, we have

$$\lim_{\varepsilon \to 0} \int_{-1}^0 \lvert t\rvert^{\varepsilon} \frac{f_X(t) - f_X(0)}{t}\,dt = \int_{-1}^0 \frac{f_X(t) - f_X(0)}{t}\,dt\,.$$

And finally we have

$$\int_{-1}^0 \frac{\lvert t\rvert^{\varepsilon}}{t}\,dt = -\int_0^1 u^{\varepsilon - 1}\,du = -\frac{1}{\varepsilon}\,.$$

The analogous splitting of the integral for positive $t$ also produces two parts that tend to a finite limit as $\varepsilon \searrow 0$, plus

$$f_X(0) \int_0^1 t^{\varepsilon - 1}\,dt = \frac{f_X(0)}{\varepsilon}\,.$$

We therefore have

$$\lim_{\varepsilon \searrow 0} \int_{-\infty}^{+\infty} t^{\varepsilon}\frac{f_X(t)}{t}\,dt = \operatorname{PV} \int_{-\infty}^{\infty} \frac{f_X(t)}{t}\,dt + \lim_{\varepsilon \searrow 0}\: f_X(0) \underbrace{\frac{1 - e^{i\pi\varepsilon}}{\varepsilon}}_{\to -\pi i}$$

with

$$\operatorname{PV} \int_{-\infty}^{\infty} \frac{f_X(t)}{t}\,dt = \int_{-1}^1 \frac{f_X(t) - f_X(0)}{t}\,dt + \int_{\lvert t\rvert > 1} \frac{f_X(t)}{t}\,dt\,.$$