Finding limit in $D'(R)$

Question

Find a limit when $\varepsilon\rightarrow +0$ of the following distribution:

$$f_\epsilon=\frac{\varepsilon x}{(x^2+\varepsilon^2)^2}$$

I tried to solve this by putting $x=\varepsilon t$, but unsuccessful... Any help is welcome.

Answer 1

Here's one way to do it. By partial integration:

$$\newcommand{\eps}{\varepsilon} \langle f_\eps, \phi \rangle = \int_{-\infty}^\infty \frac{\eps x \phi(x)}{(x^2+\eps^2)^2}\,dx = \frac12 \int_{-\infty}^\infty \frac{\eps}{x^2+\eps^2} \phi'(x)\,dx. $$

Now we can do the change of variables you suggested: $$ \frac12 \int_{-\infty}^\infty \frac{\eps}{x^2+\eps^2} \phi'(x)\,dx = \frac12 \int_{-\infty}^\infty \frac{\phi'(\eps t)}{t^2+1}\,dt \rightarrow \frac12 \int_{-\infty}^\infty \frac{\phi'(0)}{t^2+1}\,dt = \frac{\pi}2 \phi'(0) $$ as $\eps \to 0^+$.

Answer 2

Note that $$\frac{d}{dx} \frac{\epsilon}{x^2+\epsilon^{2}}=-\frac{2\epsilon x}{(x^2+\epsilon^{2})^{2}},$$ integrate by parts and try your substitution again.

Answer 3

Besides the previous answers, one could also directly observe that $$\frac{\varepsilon x}{(x^2+\varepsilon^2)^2}=-\frac{\pi}{2}\mathcal P_{\varepsilon}',$$ where $\mathcal P_\varepsilon$ is the Poisson kernel for the upper half-plane. Thus the desired limit is (kernels being approximate identities) $-\pi\delta'/2$.