Does this statement make sense or not?

Hi all,

I intuitively had an idea that

f(t) ** g(t) = f'(t) ** int(g(t)) = int(f(t)) ** g'(t)

where " ' " denotes differentiation, int(f(t)) denotes integration, "
** " denotes convolution.

I want to prove it is true but I met with some difficulties for which I
need your help!

Let me first define convolution as integral from -infinity to
+infinity,

if I define int(f(t)) to be integration from -infinity to t,

how about defining int(f(t)) to be integration from 0 to t?

will the above statement work?

------------------

Let me then define convolution as integral from 0 to t,

then how can I define my int(f(t)) and int(g(t)) correspondingly to
make the statement work?

------------------

I think it hard today and tried to prove the statement; but the
interchanging of the order of differentiation and integration inside
the convolution stumbled me... I think I need some condition to
interchange the order of the differentiation and integration
rigorously... But I don't know where to find these conditions ...  Even
worse, it has two integrations ...

So I failed... please help me...

Assume f ,g are are continuously differentiable and vanish a outside a
bounded interval ;convolution is over the whole line and int from -00
to t (0 to t should work also) .Use integration by parts on the
integral from -00 to 00 which defines the convolution.The vanishing
condition should get rid of the end point terms involved in the
integration by parts.smn

"lucy" <losemind@yahoo.com> wrote in message 
news:1128758760.082584.262810@g43g2000cwa.googlegroups.com...
> Hi all,
>
> I intuitively had an idea that
>
> f(t) ** g(t) = f'(t) ** int(g(t)) = int(f(t)) ** g'(t)
>
> where " ' " denotes differentiation, int(f(t)) denotes integration, "
> ** " denotes convolution.
>
> I want to prove it is true but I met with some difficulties for which I
> need your help!
>

other stop snipped.

You may want to look up Fubini's theorem. This will let you interchange the 
idea of double integration with repeated integration. The rest will involve 
the chain rule.

IHTH,

Clay