Verifying $x(t)\delta(t)=x(0)\delta(t)$ in Mathematica

Question

How can I verify the property $x(t)\delta(t)=x(0)\delta(t)$ in Mathematica? I tried with:

In[5]:= x[t_] := t

In[6]:= x[t] DiracDelta[t] == x[0] DiracDelta[t]

Out[6]= t DiracDelta[t] == 0

I expect the output True.

Thank you in advance.

Answer 1

As noted by Bob Hanlon in the comments, $\delta(t)$ is not defined outside of integrals, so it doesn't make any sense to ask the question whether $x(t)\delta(t)=x(0)\delta(t)$.

However, what you can do is verify that both expressions behave the same under the integral, i.e. that $\int f(t) g(t)\mathrm{d}t$ is the same for $f(t)=x(t)\delta(t)$ and $f(t)=x(0)\delta(t)$:

In[1]:=  Integrate[x[t] DiracDelta[t] f[t], t] == Integrate[x[0] DiracDelta[t] f[t], t]
Out[1]:= True

Here, you don't need to define anything for x[t], as this is true for all x[t].

Update 2

As noted in the comments, we should only consider definite integrals. This leads to the following definition of GeneralizedEqual:

GeneralizedEqual[f_, g_, t_, opts : OptionsPattern[]] := 
 Integrate[h[t] f, {t, -Infinity, Infinity}, opts]
  == Integrate[h[t] g, {t, -Infinity, Infinity}, opts]

With this definition, we can also prove the generalized version of the original equation (again, see comments), $\delta(t-T)x(t)"="\delta(t-T)x(T)$:

In[1]:= GeneralizedEqual[DiracDelta[t - T] f[t], DiracDelta[t - T] f[T], t]
Out[1]= ConditionalExpression[True, T \[Element] Reals]

Update

To make it a bit nicer to look at, you can introduce a generalized version of ==:

In[1]:=  GeneralizedEqual[f_, g_, t_] := Integrate[h[t] f, t] == Integrate[h[t] g, t]
In[2]:=  GeneralizedEqual[x[t] DiracDelta[t], x[0] DiracDelta[t], t]
Out[2]:= True

Answer 2

You get it directly with FunctionExpand:

 In[1]:= FunctionExpand[DiracDelta[t]*f[t]]

 Out[1]= DiracDelta[t] f[0]

 In[2]:= FunctionExpand[DiracDelta[t-5]*f[t]]

 Out[2]= DiracDelta[-5+t] f[5]

 In[3]:= FunctionExpand[DiracDelta[t+2]*f[t]]

 Out[3]= DiracDelta[2+t] f[-2]