# Distributional expression for $\lim_{\epsilon\to 0^+}\frac{1}{(\epsilon-it)^n}$?

I have computationally verified that the following equation holds

$$\lim_{\epsilon\to 0^+}\int_{-\infty}^{\infty}dt\frac{\phi(t)}{(\epsilon-it)^2}=i\pi\phi'(0)+PV\int_{-\infty}^\infty\frac{\phi(t)-\phi(0)}{t^2}dt$$ for various $$\phi$$, where $$\phi$$ is a test function. I thought this would follow as a consequence of the well known distributional relation

$$\lim_{\epsilon\to 0^+}\frac{1}{\epsilon-it}=\pi\delta(t)+iPV\frac{1}{t}$$

but since the two principal value prescriptions are different from each other, taking a derivative of the last relation doesn't seem to yield the answer immediately. What is a good way to show this?

I have a feeling that complex analysis could work here, and even help with showing the more general conjecture (I verified with Mathematica up to $$n=4$$ as well)

$$i^{n-1}(n-1)!\lim_{\epsilon\to 0^+}\int_{-\infty}^{\infty}dt\frac{\phi(t)}{(\epsilon-it)^n}=(-1)^{n-1}\pi\phi^{(n-1)}(0)+iPV\int_{-\infty}^\infty\frac{\phi(t)-\sum_{k=0}^{n-2}\frac{\phi^{(k)}(0)}{k!}t^k}{t^n}dt$$

• There is a convergence problem in your last integral, please clarify what $\int_{-\infty}^\infty \frac1t dt$ should mean Commented Nov 9, 2022 at 10:43
• Thanks @reuns for pointing that out, in numerical integration I was cutting off the origin anyway, despite the extra factor $\propto t^{n-1}$, so it should be a standard principal value recipe. Commented Nov 9, 2022 at 16:47

Note that we have $$\frac1{\varepsilon -it}=\frac{\varepsilon+it}{\varepsilon^2+t^2}$$. Hence, we can write

\begin{align} \int_{-\infty}^\infty \frac{\phi(t)}{(\varepsilon-it)^2}\,dt&=-i\int_{-\infty}^\infty \phi(t) \frac{d}{dt}\left(\frac1{\varepsilon -it}\right)\,dt\\\\ &=i\int_{-\infty}^\infty \frac{\phi'(t)}{\varepsilon -it}\,dt\\\\ &=i\int_{-\infty}^\infty \frac{\varepsilon+it}{\varepsilon^2+t^2}\phi'(t)\,dt\\\\ &=-\int_{-\infty}^\infty \frac{t}{\varepsilon^2+t^2}\phi'(t)\,dt+i \int_{-\infty}^\infty \frac{\varepsilon}{\varepsilon^2+t^2}\phi'(t)\,dt\tag1 \end{align}

We will now evaluate the limit of the two integrals on the right-hand side of $$(1)$$.

Let $$\delta>0$$ be given and fixed. Then, the limit as $$\varepsilon\to 0^+$$ of the first integral on the right-hand side of $$(1)$$ is

\begin{align} \lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{t}{\varepsilon^2+t^2}\phi'(t)\,dt&=\lim_{\varepsilon\to 0^+}\int_{|t|\le \delta} \frac{t}{\varepsilon^2+t^2}\phi'(t)\,dt+\lim_{\varepsilon\to 0^+}\int_{|t|\ge \delta} \frac{t}{\varepsilon^2+t^2}\phi'(t)\,dt\\\\ &=O(\delta)+\int_{|t|\ge \delta}\frac{\phi'(t)}{t}\,dt\\\\ &=O(\delta)-\int_{|t|\ge\delta}\frac{\phi(t)-\phi(0)}{t^2}\,dt \end{align}

Letting $$\delta \to 0^+$$, we find that

$$\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{t}{\varepsilon^2+t^2}\phi'(t)\,dt=\text{PV}\int_{-\infty}^\infty \frac{\phi(t)-\phi(0)}{t^2}\,dt\tag2$$

The limit of the second integral on the right-hand side of $$(1)$$ is

\begin{align} \lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\varepsilon}{\varepsilon^2+t^2}\phi'(t)\,dt&=\pi \phi'(0)+\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\varepsilon}{\varepsilon^2+t^2}(\phi'(t)-\phi'(0))\,dt\\\\ &=\pi \phi'(0)+\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\phi'(\varepsilon t)-\phi'(0)}{t^2+1}\,dt\\\\ &=\pi \phi'(0)\tag3 \end{align}

Using $$(2)$$ and $$(3)$$ in $$(1)$$ yield the coveted result

$$\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\phi(t)}{(\varepsilon-it)^2}\,dt=-\text{PV}\int_{-\infty}^\infty \frac{\phi(t)-\phi(0)}{t^2}\,dt+i\pi \phi'(0)\tag4$$

from which we deduce that the distributional limit

$$\lim_{\varepsilon\to 0^+}\frac{1}{(\varepsilon-it)^2}=-\text{PV}\left(\frac1{t^2}\right)-i\pi \delta'(t)$$

where the $$PV$$ distribution is defined in $$(4)$$.

For the distributional limit of $$\frac1{(\varepsilon-it)^n}$$, we could follow an analogous procedure. Instead, we proceed as follows. Let $$\psi_\varepsilon(t)=\frac1{\varepsilon-it}$$. Then, $$\psi_\varepsilon^{(n-1)}(t)=\frac{(i)^{n-1}(n-1)!}{(\varepsilon -it)^n}$$ and we can write

\begin{align} \lim_{\varepsilon\to 0^+}\langle (\psi_\varepsilon)^{n} ,\phi \rangle&=\lim_{\varepsilon\to 0^+} \frac{1}{i^{n-1}(n-1)!}\langle \psi_\varepsilon^{(n-1)},\phi \rangle \\\\ &=\lim_{\varepsilon\to 0^+}\frac{i^{n-1}}{ (n-1)!}\langle \psi_\varepsilon ,\phi^{(n-1)}\rangle\\\\ &=\frac{i^{n-1}}{(n-1)!}\left(\pi \phi^{(n-1)}(0)+i\text{PV}\int_{-\infty}^\infty \frac{\phi^{(n-1)}(t)}{t}\,dt\right)\tag5 \end{align}

Integrating by parts $$n-1$$ times the integral on the right-hand side of $$(5)$$ shows that in distribution we have

$$\lim_{\varepsilon\to 0^+}\frac1{(\varepsilon-it)^n} =\frac{(-i)^{n-1}\pi}{(n-1)!} \delta^{(n-1)}(t)+i^n \text{PV}\left(\frac1{t^n}\right)\tag6$$

The distribution $$\text{PV}\left(\frac1{t^n}\right)$$ in $$(6)$$ is defined as its action of a test function and can be written accordingly as

$$\int_{-\infty}^\infty \text{PV}\left(\frac1{t^n}\right)\phi(t)\,dt=\text{PV}\int_{-\infty}^\infty \frac{\phi(t)-\sum_{k=0}^{n-2}\frac{\phi^{(k)}(0)t^k}{k!}}{t^n}\,dt$$