Can you hepl me with solution of next integral :

Integral of [ delta(x-a) / sqrt(K1^2 + (K2^2 - K1^2)*H(x-a)) ] dx


So integral of delta divide square root of (K1^2 + (K2^2 - K1^2)*H(x-a)).


K1 = const.
K2 = const.
delta(x-a) is delta function, see http://mathworld.wolfram.com/DeltaFunction.html
H(x-a) is the Heaviside step function, see
http://mathworld.wolfram.com/HeavisideStepFunction.html

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Request for Question Clarification by elmarto-ga on 22 Sep 2004 11:03 PDT

Hi freemand-ga!
What are the limits of this integral? Do you want to find the integral
of this function between minus infinity and plus infinity?

Best regards,
elmarto

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Clarification of Question by freemand-ga on 22 Sep 2004 17:04 PDT

Yes, from minus infinity to plus infinity.

(K1 + K2)/(2K1 K2), or written another way, (1/2)(1/K1 + 1/K2)

Look right?

Yes, from minus infinity to plus infinity.

To : racecar-ga
Please describe how you get result. Thank you.

If look right? No - but I am not sure, but looks not correct. I will
check it in 20 min. more preciselly.

Note :  Integral [ delta(x) / (1+H(x))] dx = ln(x)

To racecar-ga : sorry, your solution looks correct (in my opinion, I
will do now longer numerical test - 2-3hours - and I will tell you
surely if it is correct).

Please, describe how you get it?

Thank's.

method was not rigorous.  Basically, the denominator is a step
function that changes from one level to another (from K1 to K2) at the
point where there's a delta function in the denominator.  If you think
of a delta function as the limit of a gaussian or a boxcar function or
something getting infinitely narrow, it's clear that half the delta
function is over one side of the step function and half is over the
other.  So the integral is just half of the integral of delta(x-a)/K1
plus half the integral of delta(x-a)/K2.

racecar-ga : Looks that (K1 + K2)/(2K1 K2) is correct solution.
Numerical test give me similar solution (I solve something what
contains that integral). Thanks.

But I am still not 100% sure. I need some days for verifying of solution.

The "delta" function is not really a function.  It's a "generalized
function" that can be given a precise definition using the theory of
integration.

Properly speaking the integral is not well-defined, but this has
nothing to do with the limits of integration (provided the point x = a
is within the limits).  We need the factor 1/sqrt(K1^2 + (K2^2 -
K1^2)*H(x-a)) to be continous at x = a for the integral to be defined,
and of course it is precisely at this value that the Heaviside
function introduces a discontinuity.

Racecar's approach is as good as any for interpreting what someone
_might_ mean in writing this expression down.  It's a bit like asking
someone to evaluate 0/0 or 0^0, or any other indeterminate form. 
Depending on how one "approaches" the delta function and the Heaviside
function with continuous approximations, you will obtain various
limits.  Racecar's value represents the case in which the Heaviside
function is used "as is" and a limit is taken with respect to the
delta function using increasingly narrow Gaussian distributions. 
Conversely if you were to use the delta function "as is" but approach
the Heaviside function with a sequence of continuous functions each
passing through (a,1/2), then the limit would be:

  1/sqrt( (1/2)(K1^2 + K2^2) ) = sqrt(2 / (K1^2 + K2^) )

regards, mathtalk-ga