Many problems in science and engineering are related to the analytic continuation and in particular infinitesimal analytic continuation to the upper or lower complex plane, i.e., a generic complex function $f(\omega)$ with real $\omega$ is changed to $$F(\omega)=\lim_{\eta\rightarrow0^+}f(\omega\pm i\eta).$$ In many applications, only the part introduced by $\eta$ is necessary, which is $$A(\omega)=\lim_{\eta\rightarrow0^+}[f(\omega+i\eta)-f(\omega-i\eta)].$$

Question: How to symbolically obtain $A(\omega)$?
This can always be calculated by hand; I think MMA is a perfect match for this task.

In such a calculation, as the limit is concerned, only the leading order effect of $\eta$ is relevant to us; hence many higher-order terms of $\eta$ can be dropped.
Also, to finally take the limit and express the result, the following formula is used $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ where $\delta$-function is the effect of $\eta$ and a principal value is formally noted. A more general version $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n+1)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ involves the $(n+1)$-th derivative of $\delta$-function. Note that all these will just appear as convenient symbols, not to be evaluated.

Two examples are given below, where all parameters are real. Therefore the imaginary part purely generated by $\eta$ is taken, which is simply $iA/2$. $$\lim_{\eta\rightarrow0^+}\Im\frac{a+bi\eta+\sqrt{c+2di\eta}}{e}=\lim_{\eta\rightarrow0^+}\frac{(b+d)\eta}{e}=0$$ and $$\lim_{\eta\rightarrow0^+}\Im\frac{a+bi\eta}{c+i\eta}=\lim_{\eta\rightarrow0^+}\Im(-i\pi\delta(c)+\mathcal{P}(\frac{1}{c}))(a+bi\eta)=\lim_{\eta\rightarrow0^+}-a\pi\delta(c)+\eta\mathcal{P}(\frac{1}{c})=-a\pi\delta(c).$$

I thought about this for some time, but due to my very limited MMA skill, I don't see how to realize such a calculation. For instance, I think we need to Series expand with respect to $\eta$ and take the leading order, but it is unclear how to maintain and recognize the pattern of denominators in the formulae.

Many problems in science and engineering are related to the analytic continuation and in particular infinitesimal analytic continuation to the upper or lower complex plane, i.e., a generic complex function $f(\omega)$ with real $\omega$ is changed to $$F(\omega)=\lim_{\eta\rightarrow0^+}f(\omega\pm i\eta).$$ In many applications, only the part introduced by $\eta$ is necessary, which is $$A(\omega)=\lim_{\eta\rightarrow0^+}[f(\omega+i\eta)-f(\omega-i\eta)].$$

Question: How to symbolically obtain $A(\omega)$?
This can always be calculated by hand; I think MMA is a perfect match for this task.

In such a calculation, as the limit is concerned, only the leading order effect of $\eta$ is relevant to us; hence many higher-order terms of $\eta$ can be dropped.
Also, to finally take the limit and express the result, the following formula is used $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ where $\delta$-function is the effect of $\eta$ and a principal value is formally noted. A more general version $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n+1)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ involves the $(n+1)$-th derivative of $\delta$-function. Note that all these will just appear as convenient symbols, not to be evaluated.

Two examples are given below, where all parameters are real. Therefore only the imaginary part is purely generated by $\eta$. For $f(\omega)=\frac{a\omega+\sqrt{c+\omega^2}}{e}$, we have \begin{equation} \begin{split} A(\omega)&=2i\lim_{\eta\rightarrow0^+}\Im f(\omega+i\eta)\\ &=2i\lim_{\eta\rightarrow0^+}\Im\frac{a(\omega+i\eta)+\sqrt{c+(\omega+i\eta)^2}}{e}\\ &=2i\lim_{\eta\rightarrow0^+}\Im\frac{a(\omega+i\eta)+c+\omega^2+i\eta\frac{\omega}{c+\omega^2}}{e}\\ &=2i\lim_{\eta\rightarrow0^+}\frac{ai\eta+i\eta\frac{\omega}{c+\omega^2}}{e}\\ &=0. \end{split} \end{equation} For $f(\omega)=\frac{a+b\omega}{\omega-c}$ we have \begin{equation} \begin{split} A(\omega)&=2i\lim_{\eta\rightarrow0^+}\Im f(\omega+i\eta)\\ &=2i\lim_{\eta\rightarrow0^+}\Im\frac{a+b(\omega+i\eta)}{\omega-c+i\eta}\\ &=2i\lim_{\eta\rightarrow0^+}\Im\,\{[-i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})][a+b(\omega+i\eta)]\}\\ &=2i\lim_{\eta\rightarrow0^+}[-\pi(a+b\omega)\delta(\omega-c)+b\eta\mathcal{P}(\frac{1}{\omega-c})]\\ &=-\pi(a+b\omega)\delta(\omega-c). \end{split} \end{equation}

I thought about this for some time, but due to my very limited MMA skill, I don't see how to realize such a calculation. For instance, I think we need to Series expand with respect to $\eta$ and take the leading order, but it is unclear how to maintain and recognize the pattern of denominators in the formulae.

Many problems in science and engineering are related to the analytic continuation and in particular infinitesimal analytic continuation to the upper or lower complex plane, i.e., a generic complex function $f(\omega)$ with real $\omega$ is changed to $$F(\omega)=\lim_{\eta\rightarrow0^+}f(\omega\pm i\eta).$$ In many applications, only the part introduced by $\eta$ is necessary, which is $$A(\omega)=\lim_{\eta\rightarrow0^+}[f(\omega+i\eta)-f(\omega-i\eta)].$$

Question: How to symbolically obtain $A(\omega)$?
This can always be calculated by hand; I think MMA is a perfect match for this task.

In such a calculation, as the limit is concerned, only the leading order effect of $\eta$ is relevant to us; hence many higher-order terms of $\eta$ can be dropped.
Also, to finally take the limit and express the result, the following formula is used $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ where $\delta$-function is the effect of $\eta$ and a principal value is formally noted. A more general version $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n+1)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ involves the $(n+1)$-th derivative of $\delta$-function. Note that all these will just appear as convenient symbols, not to be evaluated.

Two examples are given below, where all parameters are real. Therefore the imaginary part purely generated by $\eta$ is taken, which is simply $iA/2$. $$\lim_{\eta\rightarrow0^+}\Im\frac{a+bi\eta+\sqrt{c+2di\eta}}{e}=\lim_{\eta\rightarrow0^+}\frac{(b+d)\eta}{e}=0$$ and $$\lim_{\eta\rightarrow0^+}\Im\frac{a+bi\eta}{c+i\eta}=\lim_{\eta\rightarrow0^+}\frac{\eta}{c^2+\eta^2}(-a+bc)=\pi\delta(c)(-a+bc)=-a\pi\delta(c).$$

I thought about this for some time, but due to my very limited MMA skill, I don't see how to realize such a calculation.

Many problems in science and engineering are related to the analytic continuation and in particular infinitesimal analytic continuation to the upper or lower complex plane, i.e., a generic complex function $f(\omega)$ with real $\omega$ is changed to $$F(\omega)=\lim_{\eta\rightarrow0^+}f(\omega\pm i\eta).$$ In many applications, only the part introduced by $\eta$ is necessary, which is $$A(\omega)=\lim_{\eta\rightarrow0^+}[f(\omega+i\eta)-f(\omega-i\eta)].$$

Question: How to symbolically obtain $A(\omega)$?
This can always be calculated by hand; I think MMA is a perfect match for this task.

In such a calculation, as the limit is concerned, only the leading order effect of $\eta$ is relevant to us; hence many higher-order terms of $\eta$ can be dropped.
Also, to finally take the limit and express the result, the following formula is used $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ where $\delta$-function is the effect of $\eta$ and a principal value is formally noted. A more general version $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n+1)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ involves the $(n+1)$-th derivative of $\delta$-function. Note that all these will just appear as convenient symbols, not to be evaluated.

Two examples are given below, where all parameters are real. Therefore the imaginary part purely generated by $\eta$ is taken, which is simply $iA/2$. $$\lim_{\eta\rightarrow0^+}\Im\frac{a+bi\eta+\sqrt{c+2di\eta}}{e}=\lim_{\eta\rightarrow0^+}\frac{(b+d)\eta}{e}=0$$ and $$\lim_{\eta\rightarrow0^+}\Im\frac{a+bi\eta}{c+i\eta}=\lim_{\eta\rightarrow0^+}\Im(-i\pi\delta(c)+\mathcal{P}(\frac{1}{c}))(a+bi\eta)=\lim_{\eta\rightarrow0^+}-a\pi\delta(c)+\eta\mathcal{P}(\frac{1}{c})=-a\pi\delta(c).$$

I thought about this for some time, but due to my very limited MMA skill, I don't see how to realize such a calculation. For instance, I think we need to Series expand with respect to $\eta$ and take the leading order, but it is unclear how to maintain and recognize the pattern of denominators in the formulae.