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Subject: Re: Please help Explicit integral could not be found
Date: Sun, 21 Nov 2010 03:40:05 +0000 (UTC)
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"rakib hasan" <upol94@gmail.com> wrote in message <ic9nnh$muj$1@fred.mathworks.com>...
> .......
> The value of Z and x is positive and complex. But how could get a solid expression in terms of Z ? Is this impossible to do matlab ?
- - - - - - - - - -
  The warning that "Explicit integral could not be found" means that you cannot use the symbolic toolbox to evaluate this integral, so it must be done numerically, using one of the quadrature routines, 'quad', etc.  You will have to write a function that uses z as input and finds this integral numerically, and z must have a positive real part to evaluate it this way.

  Your integrand, fxy, becomes indeterminate at x = 0.  As x approaches zero fxy must also approach zero, provided the real part of z is positive, as can be proven using L'hospital's rule.  That means that if you define fxy as being zero at x = 0 for such a z, it will constitute a continuous function and consequently be integrable.  However to avoid sending NaNs back to the quadrature routine, it will be necessary to define fxy as a function that is equal to exp((-x^3+2*x-z)/x)/x for x>=e for some suitably small positive quantity e, and equal to the appropriate finite Taylor series for x<e.  If you choose e sufficiently small, you would perhaps need only one or two terms of the series for sufficient accuracy.

  You can show by differentiation of the integrand that this integral has a well-defined derivative with respect to z if z has a positive real part, and it therefore constitutes an analytic function of z at least in the right half complex plane.  I am fairly sure moreover that it can be analytically continued into the left half plane as long as you avoid the essential singularity at z = 0.  However at the moment I have no idea of how it could be computed in this left half in a practical manner.  Certainly for this purpose you cannot use the integral you have defined, since it is not integrable in that left half.   So we'll assume you are content to be restricted to the right half where you can directly compute the integral.

Roger Stafford