# Is this case handled by distribution theory?

I want to express a distribution, $f(x)$, that has the following properties: $$\int_{-\infty}^x f(x') \operatorname{d}x' = \left\{\begin{array}{cc} 0 & x \le 0 \\ \frac{1}{x} & x > 0. \end{array}\right. = \frac{\Theta(x)}{x},$$ where $\Theta(x)$ is a unit step function. Formally, I would guess that I could just take the derivative using the standard rules to get: $$f(x) = -\frac{\Theta(x)}{x^2} + \frac{\delta(x)}{x}.$$

The problem I have is that integrating the second equation above to recover the required property runs into an apparent $\infty - \infty$ ambiguity. Is this ambiguity handled in distribution theory in some particular way? Is this, perhaps, related in some way to the Cauchy principle value, for example?

Right now, my best guess for how to write $f$ to make it unambiguously have the needed property is to write it as: $$f(x) = \lim_{s\rightarrow 0}\left(-\frac{1}{x^2} + \left[\frac{1}{x}\right] \frac{\partial}{\partial x} \right) \sigma(x,s),$$ where $\sigma(x,s)$ is a signmoid that satisfies $\lim_{x\rightarrow 0^+} \frac{\sigma(x,s)}{x^2} = 0$, $\sigma(x,s)=0$ if $x \le 0$, $\lim_{x\rightarrow \infty} \sigma(x,s) = 1$, $\lim_{s\rightarrow 0^+} \sigma(x,s) = \Theta(x)$, and the limit is understood, formally, to be delayed until after an integral is taken, the same way many think about the delta function. For example: $$\sigma(x,s) = \frac{1}{2}\operatorname{erfc}\left(-\frac{\ln(x/s)}{s\sqrt{2}}\right) \Theta(x),$$ where $\mathrm{erfc}$ is the the standard complimentary error function.

## 2 Answers

Let $$\ln_+(x) = \begin{cases}\ln x, & x>0 \\ 0, & x\leq 0\end{cases}$$ This is in $L^1(\mathbb R)$ and thus a distribution. The derivative of it obviously satisfies $$\ln_+'(x) = \begin{cases}\frac1x, & x>0 \\ 0, & x\leq 0\end{cases}$$ so you can take $$f = \ln_+''.$$

I understand that you want it expressed in a more explicit form, but that's not always possible.

The ambiguity is not handled by symmetry arguments as in the Cauchy PV. $f$ should be thought of as having an consistently infinite integral against a test function. You could still use distributions, but the result might not be meaningful.

Distributions are defined as a continuous linear functional against a space of test functions. We can suggestively write the evaluation of $T$ against $\phi$ as $\int T(x) \phi(x)dx.$

Let $\beta$ be a bump function with $\beta(0)=1$ and $\beta(x)=\beta(-x)$ and $\beta''(0)=-1.$

In the case of the Cauchy PV, distributions $T$ satisfying $\int T\phi=\int \tfrac 1 x \phi$ wherever the right-hand-side is integrable are determined up to adding a multiple of $\delta_0.$ This is because we can decompose any test function $\phi$ as $\phi(x)=\phi(0)\beta(x)+(\phi(x)-\phi(0)\beta(x)),$ and the value of $T$ against $\phi(x)-\phi(0)\beta(x)$ is uniquely determined. Imposing $\int T \beta=0$ gives a unique solution, and this condition is natural because $1/x$ is odd and $\beta$ is even.

Consider the symmetric problem $\int_{\infty}^x g(x)=|x|^{-1}.$ (To get your $f$ from $\tfrac 1 2 g$, add half the derivative of the Cauchy PV.) By integration by parts we look for a weak solution $T$ satisfying $\int T \phi=-\int |x|^{-1}\phi'(x)$ wherever the right-hand-side is integrable. This determines $T$ uniquely up to a multiple of $\delta'_0.$ But now a symmetry argument doesn't determine the coefficient of $\delta'_0,$ because $T$ should be odd, so symmetry doesn't show that $\int T\beta'$ is necessarily zero.

In fact $\int |x|^{-1}\beta''(x)$ is negative in a neighborhood of zero, and integrable outside that neighborhood. So the integral is unambiguously $-\infty.$ Depending on the application you could still define a distribution which evaluates this integral as a constant $\Lambda,$ and hope to derive some meaningful observable quantities not depending on $\Lambda.$