       Re: Problem with an integral

• To: mathgroup at smc.vnet.net
• Subject: [mg95611] Re: Problem with an integral
• From: Bill Rowe <readnews at sbcglobal.net>
• Date: Thu, 22 Jan 2009 07:04:46 -0500 (EST)

On 1/21/09 at 7:02 AM, jepessen at gmail.com (Jepessen) wrote:

>I'm working with a little problem in Mathematica 7.0.0. I want to
>integrate this function

>fun = x^(n + 1)*E^(-x + (I*k)/x)

>Assuming that both n and k are greater than zero, I write

>integral = FullSimplify[Assuming[{k > 0, n > 0}, Integrate[fun, {x,
>0, \[Infinity]}]]]

>And I obtain a symbolic result. But, when I want to put some
>specific value, like this

>integral /. {n -> 1, k -> 1}

>I obtain always ComplexInfinity, and/or other errors. So, I've tried
>to evaluate numerically the integral for the same specific values,
>in this way

>NIntegrate[Evaluate[fun /. {n -> 1, k -> 1}], {x, 0, \[Infinity]}]

>And I obtain a finite numeric result. So, there's some error in
>symbolic computation, or I miss something when I try to integrate
>the formula?

Neither. This issue arises due to the way Mathematica does
evaluations. Mathematica first does replacements indicated then
evaluates the result. So,

(x - 2 x +1)/(x - 1)/.x->1

will evaluate as

(1 - 2 + 1)/(1 - 1) = 0/0 and generate an error. That is
Mathematica does not simplify

(x - 2 x +1)/(x - 1) to x - 1 and then do the substitution.

Even using FullSimplify before doing the substitution will not
always resolve the problem particularly for a complex result
returned by Integrate. For a complex integral, FullSimplify may
not be able to simplify the result from Integrate to
successfully remove singularities. Nor does Mathematica do
anything to improve numeric stability of the result. These are
two of the reasons why NIntegrate should always be used in
preference to N[Integrate[... when a numerical result for an
integral is desired.

• Prev by Date: Re: Problem in generating a IF cycle which Mathematica 6.0
• Next by Date: Re: Problem with an integral
• Previous by thread: Re: Problem with an integral
• Next by thread: Re: Problem with an integral