Representation of dirac delta function

Question

Can $ \delta (x-x') \cdot e^{k(x-x') } $ equivalent to $ \delta(x-x') $ in the sense of generalised function identity?

As $ \int_{i=-\infty}^\infty \delta (x-x') e^{k(x-x')} G(x) ~dx= \int_{i=-\infty}^\infty \delta(x-x') G(x)~dx = G(x') $ .

Answer 1

As distributions, $d_1=d_2$ iff for all test functions $f\in C_c^{\infty}( \Bbb{R})$, $(d_1,f)=(d_2,f)$.

So you are correct.

Answer 2

This is an example of a general rule for the $\delta$-function. For any smooth function $f$, and any test function $\varphi$: $$\langle f \delta, \varphi, \rangle = \langle \delta, f\varphi \rangle = f(0) \varphi(0) = \langle f(0) \delta , \varphi \rangle.$$

Thus $ \delta f = f(0) \delta$. Likewise, $\delta(x-x') f(x) = f(x') \delta(x-x')$. In your case, since the function $e^{k(x-x')}$ evaluated at $x = x'$ is simply $e^0 = 1$ we find that $\delta (x-x') \cdot e^{k(x-x')} = \delta(x-x')$.