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I am considering a Lorentzian integral of the following type: \begin{equation} \int_{-\infty}^\infty\mathrm dx\,\frac{\gamma^2}{(x-x_0)^2+\gamma^2}f(x), \end{equation} where $f(x)$ is a "nice" function (i.e. smooth, rapidly decreasing, etc).

I am trying to find an estimate of such integral for $\gamma\to0$. It is customary in my field to say that the Lorentzian function behaves like a delta "function" for small $\gamma$, \begin{equation} \frac{\gamma^2}{(x-x_0)^2+\gamma^2}\underset{\gamma\to0}\approx\pi\gamma\delta(x-x_0), \end{equation} such that one obtains, \begin{equation} \int_{-\infty}^\infty\mathrm dx\,\frac{\gamma^2}{(x-x_0)^2+\gamma^2}f(x)\underset{\gamma\to0}\approx\pi\gamma f(x_0). \end{equation} To my understanding, this method basically provides us with the leading-order (first-order) behaviour of the integral for small $\gamma$, \begin{equation} \int_{-\infty}^\infty\mathrm dx\,\frac{\gamma^2}{(x-x_0)^2+\gamma^2}f(x)=\pi\gamma f(x_0)+O(\gamma^2). \end{equation}

I am wondering whether there is a systematic way to obtain the next-to-leading order behaviour of the integral. I have run some numerical tests but couldn't identify a pattern for the term of second order in $\gamma$.

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The trick to proving the first fact you have is as follows. Rewrite the integral as

$$\int_{-\infty}^\infty \frac{f(x)}{1+\left(\frac{x-x_0}{\gamma}\right)^2}\:dx$$

which inspires the substitution $u = \frac{x-x_0}{\gamma}$

$$= \gamma \cdot \int_{-\infty}^\infty \frac{f\left(\gamma u + x_0\right)}{1+u^2}du \longrightarrow \gamma \cdot \int_{-\infty}^\infty \frac{f(x_0)}{1+u^2}du = \gamma \pi f(x_0)$$

To find higher order contributions, we're going to use the same trick, but only after subtracting off terms so that we are finding a constant independent of $\gamma$

$$c_2 = \lim_{\gamma\to 0}\frac{\gamma}{\gamma^2} \cdot \int_{-\infty}^\infty \frac{f\left(\gamma u + x_0\right)}{1+u^2}du - \frac{\gamma}{\gamma^2}\pi f(x_0) = \lim_{\gamma\to 0}\frac{1}{\gamma} \cdot \int_{-\infty}^\infty \frac{f\left(\gamma u + x_0\right)-f(x_0)}{1+u^2}du$$

The definition of the derivative gives us the limit

$$\lim_{\gamma\to 0} \int_{-\infty}^\infty \frac{uf'(\gamma u + x_0)}{1+u^2}du$$

and this is where our problems begin. Naively you could try to take the limit and get $0$, but the resulting integral is divergent. You could try to take principal value of the integral then the limit, but this has chaotic numerical properties, to say the least. But even if you were to choose a method to pick this coefficient, the higher order ones definitely don't exist, be it principal value or otherwise.

But this is how you would systematically approach it. Find limits, then keep subtracting off the limits and dividing by the appropriate gamma. Alternatively since $\gamma \to 0$ you could keep taking derivatives before trying the substitution trick, but this should be saved for other functions if you really need good asymptotics.

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  • $\begingroup$ Hi Ninad, sorry for the late reply. Thank you for your answer, it has been very helpful. Your last equation indeed provides a good estimate of the second-order contribution for small gamma. For the cases I have looked at numerically, the limit seems to be well-defined. Ideally, I should now find a way to obtain some expression for this limit, but I doubt that we can find a generic expression that would work for any function f. $\endgroup$
    – Sennin
    Sep 8 '20 at 17:15

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