delta functions $e^{x}\delta (x)=\delta (x)$ 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
 A: Hint: $\delta(x)$ is a linear functional on some space of suitably nice (in particular, continuous) functions of $\mathbb{R}$. Given a suitably nice function $f$, what does $\delta(x)[f]$ equal? 
When you say $e^x \delta(x) = \delta(x)$, what you're saying is that the left hand side and the right hand side act on continuous functions in the same way. Is that true?
A: Depends on how you define the Dirac delta, if it is defined as a linear functional acting on an $f\in C^{\infty}_0(\mathbb{R})$:
$$
\delta[f](x) = \int^{\infty}_{-\infty} \delta(x)f(x) \,dx = f(0)
$$
Then we can see for any test function $\phi$:
$$
(e^{x}\delta(\cdot) - \delta(\cdot))[\phi](x) = \int^{\infty}_{-\infty} (e^x\delta(x)\phi(x) - \delta(x)\phi(x)) \,dx = e^0\phi(0) - \phi(0) = 0
$$
Hence $e^{x} \delta(x) = \delta(x)$.
A: $$
\int_{-\infty}^\infty f(x)\delta(x)\,dx = f(0).
$$
$$
\int_{-\infty}^\infty f(x)\Big(e^x \delta(x)\Big)\,dx = \text{what?}
$$
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.
A: The $\delta_0$ functional is representable as a measure. You have
$$\delta_0(E) = \cases{1 & if $0 \in E$\cr 0 & otherwise }$$
Then the linear functional here is represented as 
$$f \mapsto f(0) = \int_{R^d} f(\xi)d\delta_0(\xi). $$
It is not hard to show this relation will hold for any continuous function. Your result will follow right away.
