TITLE: delta functions $e^{x}\delta (x)=\delta (x)$
QUESTION [5 upvotes]: How would you prove that;
$$e^{x} \delta (x)= \delta (x)$$
Is it anything to do with the following relationship;
$$ \int_{-\infty}^{\infty} g'(x)h(x)\,dx =  \int_{-\infty}^{\infty} g(x)h'(x)\,dx.$$
Thanks in advance

REPLY [3 votes]: $$
\int_{-\infty}^\infty f(x)\delta(x)\,dx = f(0).
$$
$$
\int_{-\infty}^\infty f(x)\Big(e^x \delta(x)\Big)\,dx = \text{what?}
$$
But look at that last integral this way:
$$
\int_{-\infty}^\infty \Big(f(x)e^x\Big) \delta(x)\,dx.
$$
This is equal to the value of the function $x\mapsto f(x)e^x$ at $x=0$, because the delta function is defined that way.  Letting $g(x)=e^x\delta(x)$, we now have
$$
\int_{-\infty}^\infty f(x)\delta(x)\,dx = \int_{\infty}^\infty f(x)g(x)\,dx \text{ for all suitable functions }f.
$$
The definition of generalized functions is such that that implies that $g=\delta$.  ("Suitable" will mean test functions or Schwarz functions or whatever it is you're using in that role in the context in which you're working.)
So be careful to understand the definition.