solution of integral

integral { delta / (A1 + (A2 * delta)) } dx

A1,A2 - constant
delta - is delta function http://mathworld.wolfram.com/DeltaFunction.html

Integral writen in TeX : 
\int{\frac{\delta(x-a)}{C_1 + (C_2 * \delta(x-a))} }dx

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Clarification of Question by freemand-ga on 13 Jan 2005 17:50 PST

This is part of more complicated integral. Describe proces where two
physical values has discontinuity in same point, let me say x=a (you
can put a=0), discontinuity described by Heaviside function. Delta
functions are derivations of   these Heaviside functions. In reality
result is not zero (but there is stil posibility of not correct
description of process, so solved integral can be zero).

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Request for Question Clarification by mathtalk-ga on 13 Jan 2005 20:51 PST

Hi, freemand-ga:

I felt sure that there was more to your problem!

I recalled your earlier Question, which was fairly similar except for
the appearance of a Heaviside function (within a square root) in the
denominator:

http://answers.google.com/answers/threadview?id=404752

Did you see my Comment there awhile back?  The delta function and
Heaviside function are a bit of a notational shortcut.  They can be
powerful tools in the hands of someone who appreciates their
limitations well enough, but they can also lead the unwary into traps.

regards, mathtalk-ga

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Clarification of Question by freemand-ga on 13 Jan 2005 22:56 PST

Yes I see your previous answer. Your previos answer was maeby
mathematically correct. But, you have proceses which are described by
same  or similar integrals. You can not say there is many diferent
solution. This is not consistent with real situation/measurements.

New integral, is similar but not same. I try make this easy as
posssible (not involve other things just math, just one thing).
Separate this tricky math things to obtain base as simply as possible
math problem. If it will be something clear I no need opinion of
mathematician.
Thanks for really "helpfull" hands and "whose problem it is" notations.

I'd say zero, unless A2 = 0, in which case I'd say 1/A1.

The delta "function" is not really a function, although it can be
rigorously defined as a "generalized function" in the sense of
distribution theory.

However things are far from being stated in a rigorous framework here.
 The idea is to define a linear functional which "acts" like an
integral with respect to a discontinuous "point mass" distribution,
but which is simply evaluation of a function at a point:

INTEGRAL  delta(x-a) * f(x) dx  =  f(a)

In your case this suggests "evaluating" 1/(A1 + A2*delta) at x = a. 
Since delta is not really a function, it can't rigorously be said to
evaluate to +oo at x = a, but this is sort of the case.  What you can
do is express delta(x-a) as the "weak" limit of a sequence of
honest-to-goodness functions whose densities converge to an infinite
spike at x = a, each having area 1 under the curve.  Using this limit
argument leads, at least for nonzero A2, to "evaluating" a small
result (because the denominator grows large at x = a).

A proper answer to this question requires knowing something more about
the intended use to be made of it.

regards, mathtalk-ga

This is part of more complicated integral. Describe proces where two
physical values has discontinuity in same point, let me say x=a (you
can put a=0), discontinuity described by Heaviside function. Delta
functions are derivations of   these Heaviside functions. In reality
result is not zero (but there is stil posibility of not correct
description of process, so solved integral can be zero).

There is a similarity to your previous question in the sense that with
two limits involved (or two physical processes that share a common
discontinuity), the outcome hinges on how the two limits are taken: 
one after the other or "synchronized" together in some manner.

If you let the delta function in the numerator be the "inner" limit,
this leads to the answer already given, ie. 0 unless A2 = 0, when we
get 1/A1.

If you let the delta function in the denominator be the "inner" limit,
it leads to 1/A1.

My guess is that for the physical outcome/measurements to be
well-defined, some notion of synchronization of the two discontinuous
processes has to be imposed.  A good example of "rigorous"
mathematical treatment of a physical discontinuity is the jump
conditions across a shock in supersonic flow.  Although more than one
flow variable is discontinuous at the shock boundary, there is a
natural way of relating these variables as parts of one single
process.

Perhaps you can devise a similar perspective for your problem, so that
by taking the limits together as a single limiting process, an answer
between the extremes of zero and 1/A1 can be extracted.

regards, mathtalk-ga

> Although more than one flow variable is discontinuous at the shock boundary, 
> there is a natural way of relating these variables as parts of one single
> process.
Can you be more specific in description of this natural way, please?
But for case where is not just discontinuity in flow speed but also
discontinuity in diffusion coefficient.

Just small note. I can accept math solution of integral. Just put it
as answer to give me possibility pay for it.

Second thing is mentioned "good example" where is "natural way of
relating these variables as parts of one single process". I did not
see it there.