I am seeking a step-by-step solution of the following integral equation:

u(t) = Integral over tau from lower limit zero to upper limit t of the
expression [J(t-tau) * dP(tau)/dtau)]

where J(t-tau) = [ (1/E) + ((t-tau)/nu) ]
E, nu are constants
and P(tau) = pzero * [H(tau) - H(tau-tzero)]
pzero is a constant
tzero is a known start time within the limits of integration
tau is a psuedo time variable

Any help is greatly appreciated, even just a pointer to a
comprehensive reference with an identical example.

Start by evaluating the derivative dP(tau)/dtau:

     P(tau) = pzero*[H(tau) - H(tau-tzero)]

The derivative of the Heaviside step function is the delta function
(see <http://mathworld.wolfram.com/HeavisideStepFunction.html> and
<http://mathworld.wolfram.com/DeltaFunction.html>)

     dP(tau)/dtau = pzero*[delta(tau) - delta(tau-tzero)]

Plug this into your integral to obtain:

    u(t) = pzero * integral (from 0 to t) of {[J(tau)*delta(tau) -
J(tau)*delta(tau-tzero)]dtau}

(Note that insofar as this integral is concerned, you can write
J(t-tau) as simply J(tau).)

The delta function has the neat property (see above links) that:

     the integral (over any interval that includes a) of
{f(x)*delta(x-a) dx} = f(a).

In other words, the integral over x of a function, f(x), multiplied by
delta(x-a) "picks out" the value of the function at x=a, as long as
the value of a is contained within the limits of integration.

In your case, the first term in the integrand "picks out" the value of
J(tau) at tau=0, and the second term "picks out" the value of J(tau)
at tau = tzero.  Plugging these values of tau into your expression for
J(tau), one obtains:

     u(t) = pzero*[((1/E) + t/nu) - ((1/E) + (t-tzero)/nu))],

which simplifies to:

     u(t) = pzero*tzero/nu, a constant.

Thank you so much for your helpful analysis, hfshaw-ga.   I learned
the functions on Mathworks and was nearly able to repeat your result. 
However, I wonder about the applicability of your statement "the first
term in the integrand "picks out" the value of J(tau) at tau=0".  It
seems inconsistent with the definition of the delta function and also
with your own statement saying "as long as the value of a is contained
within the limits of integration."

Maybe I'm missing something (this is all new math for me), but do my
limits of integration really contain "the value of a".  Isn't the
value of a actually zero for the first term in the integrand?  So my
limits of integration don't bound a; my lower limit of integration is
identically equal to a.

Any suggestions?

Much respect,
elandra-ga

oops...I dashed off that comment pretty quickly and didn't notice the
problem you have pointed out.  I believe you are correct, and that
strictly speaking, the value of your integral is undefined because the
lower limit of intrgration (zero) exactly coincides with the
"location" of one of the delta functions in the integrand.

As you've probably read by now, one of the properties of the delta
function is that the integral from (a-e) to (a+e) of (delta(x-a)*f(x)
dx) = f(a) for all e>0. The hitch with your problem is that the
condition is e>0, not e>=0.

As a physicist, and not a mathematician, I'd be tempted to see if the
answer I gave makes physical sense, and not worry about too much about
the mathematical niceties.  Delta functions are strange beasts <g>. 
Alternatively, is there any reason you cannot change the lower limit
of integration to some value less that zero (say, minus infinity)? 
This would get rid of the problem.

If you want to solve this with mathematical rigor, then you'll need to
use one of the definitions for the delta function, (which all involve
taking the limit of a "peaked" function, e.g., the delta function can
be thought of as a normal distribution in the limit of a zero standard
deviation), crank the definition through the integral and take the
appropriate limits.