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While solving a problem I encountered $$\int_{a-\epsilon}^{a+\epsilon} dx\,\delta(x-a)\,f(x).$$ What does this evaluate to as $\epsilon \to 0$?

Christoph
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RK1974
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    For every $\epsilon > 0$ this evaluates to $f(a)$, hence the limit is $f(a)$ as well. – Christoph Dec 18 '20 at 22:24
  • Ok thanks. I was for some reason thinking about having the limits as $\pm \infty$. So the result is not affected by the limits as $\delta(x)$ exists only at $x=a$ in this example. – RK1974 Dec 20 '20 at 11:55

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By definition of $\delta_a(x) := \delta(x-a)$, $$ I_{\epsilon} = \int_{a-\epsilon}^{a+\epsilon} f(x)\,\delta_a(\mathrm d x) = \int_{-\infty}^{\infty} \mathbb{1}_{[a-\epsilon,a+\epsilon]}(x)\,f(x)\,\delta_a(\mathrm d x) = \mathbb{1}_{[a-\epsilon,a+\epsilon]}(a)\,f(a) $$ and since $a∈ [a-\epsilon,a+\epsilon]$ for any $\epsilon>0$, $I_{\epsilon} = f(a)$, hence $$ \lim_\limits{\epsilon\to 0^+} I_{\epsilon} = f(a) $$

Remark: However, if $\epsilon<0$, then $I_ϵ = 0$, so $$ \lim_\limits{\epsilon\to 0^-} I_{\epsilon} = 0 $$

LL 3.14
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