Signal, Cauchy IRF

When instrument response function is modeled as normalized cauchy distribution, experimental signal is modeled as convolution of exponentical decay and normalized cauchy distribution.

\[\begin{align*} {Signal}_c(t) &= ({model} * {irf})(t) \\ &= \frac{1}{\pi} \int_0^{\infty} \frac{\gamma \exp(-kx)}{(x-t)^2+\gamma^2} \mathrm{d}x \\ &= \frac{1}{\pi} \Im\left(\int_0^{\infty} \frac{\exp(-kx)}{(x-t)-i\gamma} \mathrm{d}x \right) \end{align*}\]

Assume \(k > 0\), and let \(u=k(x-t)-ik\gamma\), then

\[\begin{align*} {Signal}_c(t) &= \frac{1}{\pi} \exp(-kt) \Im\left(\exp(-ik\gamma) \int_{-k(t+i\gamma)}^{\infty-ik\gamma} \frac{\exp(-u)}{u} \mathrm{d}u \right) \\ &= \frac{1}{\pi} \exp(-kt) \Im(\exp(-ik\gamma)E_1(-k(t+i\gamma)) \end{align*}\]

So, experimental signal could be modeled as

\[\begin{equation*} {Signal_c}(t) = \begin{cases} \frac{1}{2} + \frac{1}{\pi}\arctan\left(\frac{t}{\gamma}\right)& \text{if $k=0$}, \\ \frac{\exp(-kt)}{\pi} \Im(\exp(-ik\gamma)E_1(-k(t+i\gamma)))& \text{if $k>0$}. \end{cases} \end{equation*}\]

\(E_1(z)\) is exponential integral, see dlmf section 6.2.

Implementation Note

At \(|kt| > 700\), \(E_1\) or \(\exp\) term overflows, so, in this region, the following asymptotic expression is used.

\[\begin{equation*} {Signal_c}(t) = -\frac{1}{\pi}\Im\left(\frac{1}{kt+i\gamma}\sum_{i=0}^{10} \frac{i!}{(kt+i\gamma)^i}\right) \end{equation*}\]