What is $\lim_{h\to0}\int_{-h}^h \delta(i+x)f(x)dx$?

Question

This answer says that

$\int_{-\infty}^\infty \delta(i+x)f(x)dx=f(-i)$ (for well-behaved functions).

But what happens if we make the limits of integration to go to zero? What is $$\lim_{h\to0}\int_{-h}^h \delta(i+x)f(x)dx$$?

Is the answer the same?

Answer 1

When you operate with distributions that are not proper functions, you cannot simply change the limits of integration. In a sense $$ \int_{-\infty}^\infty\delta(x+i) (\cdot) dx$$ is a single symbol and it has a mathematical meaning only as a whole, as the linear map $$ \int_{-\infty}^\infty\delta(x+i) (\cdot) dx : f \mapsto f(-i)$$ defined on a set of functions continuous at $-i$.

Before you ask about the limit $h \to 0$ you need to first properly define what $$ \int_{-h}^h\delta(x+i) (\cdot) dx$$ and it's not obvious how it should be defined. If it was a real number $a$ in place of $i$ one would usually say that $$ \int_{-h}^h\delta(x+a) (\cdot) dx : f \mapsto f(-a){\bf 1}_{[-h,h]}(-a) $$ and is defined on functions $f$ such that $f(x){\bf 1}_{[-h,h]}(x)$ is continuous at $x=-a$. You cannot however use this for $\delta(x+i)$, because ${\bf 1}_{[-h,h]}(x)$ is only defined for real arguments. Therefore, to get an answer you must first define what $\int_{-h}^h\delta(x+i) (\cdot) dx$ means.