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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 $A(\omega)$?

A related question here deals with the simplest case (the first formula below), but I am asking about more general cases when symbolic Fourier transform or integral does not work.

Simple Limit does not work, because the Sokhotski–Plemelj formula $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ plays a role in the calculation. Here $\delta$ denotes Dirac delta function (DiracDelta), $\mathcal{P}$ denotes the Cauchy principal value. A more general version of this formula is $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ where $\delta^{(n)}$ denotes the $n$-th derivative of $\delta$-function. But note that all these will just appear as convenient symbols, not to be evaluated anywhere.

Two examples are given below, where all parameters are real. Therefore only the imaginary part is purely generated by $\eta$. For $f_1(\omega)=\frac{a\omega+\sqrt{c+\omega^2}}{e}$, 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}\\ &\approx2i\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 $f_2(\omega)=\frac{a+b\omega}{\omega-c}$ we have \begin{equation} \begin{split} A(\omega)&=\lim_{\eta\rightarrow0^+} f(\omega+i\eta)-f(\omega-i\eta)\\ &=\lim_{\eta\rightarrow0^+}(a + b c)(\frac{1}{\omega-c+i\eta}-\frac{1}{\omega-c-i\eta})\\ &=[-i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]-[+i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]\\ &=-2i\pi(a+bc)\delta(\omega-c). \end{split} \end{equation} Here, simple Limit does not work, which merely gives 0 as shown below.

f[ω_] := (a + b ω)/(ω - c);
Limit[f[ω + I η] - f[ω - I η], η -> 0, Direction -> "FromAbove"]

Edit

We aim at generic functions $f(\omega)$ that can be more complicated than the above minimal examples, but let's assume the denominator, if x is involved, can always be factorized. For instance (use x as the variable),

f[x_] := (a - b x)/(c^3 + 3 c^2 d + 3 c d^2 + d^3) + Sqrt[x + b]/(
   x - e) + (
   b x + Sqrt[x + b^2])/((x - c)^2 (x - Sqrt[d + c^2]) (x^2 + g^2));

In the result, there should be $\delta(x-e),\delta(x-\sqrt{d+c^2}),\delta(x-c),\delta^{(1)}(x-c)$.

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 $A(\omega)$?

A related question here deals with the simplest case (the first formula below), but I am asking about more general cases when symbolic Fourier transform or integral does not work.

Simple Limit does not work, because the Sokhotski–Plemelj formula $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ plays a role in the calculation. Here $\delta$ denotes Dirac delta function (DiracDelta), $\mathcal{P}$ denotes the Cauchy principal value. A more general version of this formula is $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ where $\delta^{(n)}$ denotes the $n$-th derivative of $\delta$-function. But note that all these will just appear as convenient symbols, not to be evaluated anywhere.

Two examples are given below, where all parameters are real. Therefore only the imaginary part is purely generated by $\eta$. For $f_1(\omega)=\frac{a\omega+\sqrt{c+\omega^2}}{e}$, 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}\\ &\approx2i\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 $f_2(\omega)=\frac{a+b\omega}{\omega-c}$ we have \begin{equation} \begin{split} A(\omega)&=\lim_{\eta\rightarrow0^+} f(\omega+i\eta)-f(\omega-i\eta)\\ &=\lim_{\eta\rightarrow0^+}(a + b c)(\frac{1}{\omega-c+i\eta}-\frac{1}{\omega-c-i\eta})\\ &=[-i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]-[+i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]\\ &=-2i\pi(a+bc)\delta(\omega-c). \end{split} \end{equation} Here, simple Limit does not work, which merely gives 0 as shown below.

f[ω_] := (a + b ω)/(ω - c);
Limit[f[ω + I η] - f[ω - I η], η -> 0, Direction -> "FromAbove"]

Edit

We aim at generic functions $f(\omega)$ that can be more complicated than the above minimal examples, but let's assume the denominator, if x is involved, can always be factorized. For instance (use x as the variable),

f[x_] := (a - b x)/(c^3 + 3 c^2 d + 3 c d^2 + d^3) + Sqrt[x + b]/(
   x - e) + (
   b x + Sqrt[x + b^2])/((x - c)^2 (x - Sqrt[d + c^2]) (x^2 + g^2));

In the result, there should be $\delta(x-e),\delta(x-\sqrt{d+c^2}),\delta(x-c),\delta^{(1)}(x-c)$. Also similar

f[x_] := (
       b x + Sqrt[x + b^2])/((x - c)^2 (x - Sqrt[d + c^2]+a b^2) (x^2 + g^2))

which should give $\delta(x-\sqrt{d+c^2}+a b^2),\delta(x-c),\delta^{(1)}(x-c)$.

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 $A(\omega)$?

A related question here deals with the simplest case (the first formula below), but I am asking about more general cases when symbolic Fourier transform or integral does not work.

Simple Limit does not work, because the Sokhotski–Plemelj formula $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ plays a role in the calculation. Here $\delta$ denotes Dirac delta function (DiracDelta), $\mathcal{P}$ denotes the Cauchy principal value. A more general version of this formula is $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ where $\delta^{(n)}$ denotes the $n$-th derivative of $\delta$-function. But note that all these will just appear as convenient symbols, not to be evaluated anywhere.

Two examples are given below, where all parameters are real. Therefore only the imaginary part is purely generated by $\eta$. For $f_1(\omega)=\frac{a\omega+\sqrt{c+\omega^2}}{e}$, 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}\\ &\approx2i\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 $f_2(\omega)=\frac{a+b\omega}{\omega-c}$ we have \begin{equation} \begin{split} A(\omega)&=\lim_{\eta\rightarrow0^+} f(\omega+i\eta)-f(\omega-i\eta)\\ &=\lim_{\eta\rightarrow0^+}(a + b c)(\frac{1}{\omega-c+i\eta}-\frac{1}{\omega-c-i\eta})\\ &=[-i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]-[+i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]\\ &=-2i\pi(a+bc)\delta(\omega-c). \end{split} \end{equation} Here, simple Limit does not work, which merely gives 0 as shown below.

f[ω_] := (a + b ω)/(ω - c);
Limit[f[ω + I η] - f[ω - I η], η -> 0, Direction -> "FromAbove"]

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 $A(\omega)$?

A related question here deals with the simplest case (the first formula below), but I am asking about more general cases when symbolic Fourier transform or integral does not work.

Simple Limit does not work, because the Sokhotski–Plemelj formula $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ plays a role in the calculation. Here $\delta$ denotes Dirac delta function (DiracDelta), $\mathcal{P}$ denotes the Cauchy principal value. A more general version of this formula is $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ where $\delta^{(n)}$ denotes the $n$-th derivative of $\delta$-function. But note that all these will just appear as convenient symbols, not to be evaluated anywhere.

Two examples are given below, where all parameters are real. Therefore only the imaginary part is purely generated by $\eta$. For $f_1(\omega)=\frac{a\omega+\sqrt{c+\omega^2}}{e}$, 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}\\ &\approx2i\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 $f_2(\omega)=\frac{a+b\omega}{\omega-c}$ we have \begin{equation} \begin{split} A(\omega)&=\lim_{\eta\rightarrow0^+} f(\omega+i\eta)-f(\omega-i\eta)\\ &=\lim_{\eta\rightarrow0^+}(a + b c)(\frac{1}{\omega-c+i\eta}-\frac{1}{\omega-c-i\eta})\\ &=[-i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]-[+i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]\\ &=-2i\pi(a+bc)\delta(\omega-c). \end{split} \end{equation} Here, simple Limit does not work, which merely gives 0 as shown below.

f[ω_] := (a + b ω)/(ω - c);
Limit[f[ω + I η] - f[ω - I η], η -> 0, Direction -> "FromAbove"]

Edit

We aim at generic functions $f(\omega)$ that can be more complicated than the above minimal examples, but let's assume the denominator, if x is involved, can always be factorized. For instance (use x as the variable),

f[x_] := (a - b x)/(c^3 + 3 c^2 d + 3 c d^2 + d^3) + Sqrt[x + b]/(
   x - e) + (
   b x + Sqrt[x + b^2])/((x - c)^2 (x - Sqrt[d + c^2]) (x^2 + g^2));

In the result, there should be $\delta(x-e),\delta(x-\sqrt{d+c^2}),\delta(x-c),\delta^{(1)}(x-c)$.

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 $A(\omega)$?

A related question here deals with the simplest case (the first formula below), but I am asking about more general cases when symbolic Fourier transform or integral does not work.

Simple Limit does not work, because the Sokhotski–Plemelj formula $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ plays a role in the calculation. Here $\delta$ denotes Dirac delta function (DiracDelta), $\mathcal{P}$ denotes the Cauchy principal value. A more general version of this formula is $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ where $\delta^{(n)}$ denotes the $n$-th derivative of $\delta$-function. But note that all these will just appear as convenient symbols, not to be evaluated anywhere.

Two examples are given below, where all parameters are real. Therefore only the imaginary part is purely generated by $\eta$. For $f_1(\omega)=\frac{a\omega+\sqrt{c+\omega^2}}{e}$, 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}\\ &\approx2i\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 $f_2(\omega)=\frac{a+b\omega}{\omega-c}$ we have \begin{equation} \begin{split} A(\omega)&=\lim_{\eta\rightarrow0^+} f(\omega+i\eta)-f(\omega-i\eta)\\ &=\lim_{\eta\rightarrow0^+}(a + b c)(\frac{1}{\omega-c+i\eta}-\frac{1}{\omega-c-i\eta})\\ &=[-i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]-[+i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]\\ &=-2i\pi(a+bc)\delta(\omega-c). \end{split} \end{equation} Here, simple Limit does not work, which merely gives 0 as shown below.

f[ω_] := (a + b ω)/(ω - c);
Limit[f[ω + I η] - f[ω - I η], η -> 0, Direction -> "FromAbove"]

Edit

We aim at generic functions $f(\omega)$ that can be more complicated than above minimal examples. For instance, $f_3(\omega)=\frac{\omega}{e}+\frac{\omega^2}{\omega-a}+\frac{a \omega + \sqrt{b + \omega}}{\sqrt{c + \omega^2}}$ or $f_4(\omega)=\sqrt{b + \omega}+\frac{a \omega}{\omega^3+b\omega^2+c\omega+d}$. Therefore, I think expansion with respect to $\eta$ and keeping the leading order is necessary. Note that due to the limit, only the leading order effect is relevant and higher-order terms can be dropped.

Here, let's first take $f_3$ as an example. 1) We find the denominators with $\omega$ involved. 2) Use our formula if possible. The first part $\frac{x}{e}$ does not contribute just as above $f_1$; the second part $\frac{\omega^2}{\omega-a}$ directly uses our formula just as above $f_2$. 3) The third part $\frac{a \omega + \sqrt{b + \omega}}{\sqrt{c + \omega^2}}$ is a bit different. Its denominator can be cast in the form $$\frac{1}{\sqrt{c + (\omega+i\eta)^2}}\approx\frac{\sqrt{|c+\omega^2}}{\omega}\frac{1}{\frac{c+\omega^2}{\omega}+i\eta}$$ by expansion with respect to $\eta$; and the pattern in our formula can be used for $\frac{1}{\frac{c+\omega^2}{\omega}+i\eta}$ to get $\delta(\frac{c+\omega^2}{\omega})$. For $f_4$, it is similar for its second part. We can expand to get the general form $$\frac{a\omega}{\omega^3+b\omega^2+c\omega+d}\big|_{\omega\rightarrow\omega+i\eta}\approx\frac{a(\omega+i\eta)}{X(Y+i\eta)}$$ and then $\delta(Y)$ is obtained with our formula like above $f_2$.

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 $A(\omega)$?

A related question here deals with the simplest case (the first formula below), but I am asking about more general cases when symbolic Fourier transform or integral does not work.

Simple Limit does not work, because the Sokhotski–Plemelj formula $$\lim_{\eta\to0^{+}} \frac{1}{x\pm i\eta}= \mp i\pi\delta(x) + {\mathcal{P}} {\Big(\frac{1}{x}\Big)}$$ plays a role in the calculation. Here $\delta$ denotes Dirac delta function (DiracDelta), $\mathcal{P}$ denotes the Cauchy principal value. A more general version of this formula is $$\lim_{\eta\to0^{+}} \frac{1}{(x\pm i\eta)^{n+1}}= \mp i\pi(-1)^n\frac{\delta^{(n)}(x)}{n!} + {\mathcal{P}} {\Big(\frac{1}{x^{n+1}}\Big)}$$ where $\delta^{(n)}$ denotes the $n$-th derivative of $\delta$-function. But note that all these will just appear as convenient symbols, not to be evaluated anywhere.

Two examples are given below, where all parameters are real. Therefore only the imaginary part is purely generated by $\eta$. For $f_1(\omega)=\frac{a\omega+\sqrt{c+\omega^2}}{e}$, 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}\\ &\approx2i\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 $f_2(\omega)=\frac{a+b\omega}{\omega-c}$ we have \begin{equation} \begin{split} A(\omega)&=\lim_{\eta\rightarrow0^+} f(\omega+i\eta)-f(\omega-i\eta)\\ &=\lim_{\eta\rightarrow0^+}(a + b c)(\frac{1}{\omega-c+i\eta}-\frac{1}{\omega-c-i\eta})\\ &=[-i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]-[+i\pi\delta(\omega-c)+\mathcal{P}(\frac{1}{\omega-c})]\\ &=-2i\pi(a+bc)\delta(\omega-c). \end{split} \end{equation} Here, simple Limit does not work, which merely gives 0 as shown below.

f[ω_] := (a + b ω)/(ω - c);
Limit[f[ω + I η] - f[ω - I η], η -> 0, Direction -> "FromAbove"]

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.