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I was reading through Lecture of of Prof Patrick Lee on Linear Response Theory. I have found the following relation and could not understand why is it true: $$\Im \left\{\frac{1}{x + i\eta}\right\} = -\pi \delta \left(x\right)$$ where, $\eta \rightarrow 0$. Can someone explain?

Thanks in advance.

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The imaginary part of the function $\frac{1}{x+i\eta}$ is $F_\eta(x)=-\frac{\eta}{x^2+\eta^2}$. It satisfies $\int dx F_\eta(x)=-\pi$, and for any (reasonable) function $f(x)$ it can be proven that $$ \lim_{\eta\rightarrow 0}\int dx f(x)F_\eta(x) = -\pi f(0)$$

Therefore the limit of $F_\eta$ when $\eta\rightarrow 0$ has all the properties one would expect from $-\pi \delta(x)$, and thus can be identified.

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  • $\begingroup$ Presumably you mean that $F_\eta$ is the imaginary part of $1/x+i\eta$, no? $\endgroup$
    – Tobias Fünke
    Commented Sep 17, 2022 at 16:22
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    $\begingroup$ Please, add the details of the evaluation of the integral and its limit, since it contains details that can make the answer more clear, clearly stating the assumptions in this calculation $\endgroup$
    – basics
    Commented Sep 18, 2022 at 8:15

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