Re: Integrate 2 March 1992
- To: mathgroup at yoda.physics.unc.edu
- Subject: Re: Integrate 2 March 1992
- From: David Withoff <withoff>
- Date: Wed, 4 Mar 1992 11:50:44 -0600
> In[2]:= ans = (1 /(4 Pi^2 k^2)) \
> Integrate[(1/(u v)) \
> Exp[-((so + ss)/2) (u^2 + v^2) - u v phi], \
> {u,-Infinity,Infinity},{v,-Infinity,Infinity}]
>
> Infinity::indet:
> Indeterminate expression ComplexInfinity + ComplexInfinity encountered.
>
> Integrate::idiv: Integral does not converge.
>
> Out[2]= Indeterminate
>
> The integral does not diverge. It works out via a sequence of
> transformations that I would like to teach to Mathematica
> (change variable, differentiate, change variable, substitute,
> integrate differential equation) (*)
>
> (1/(2 Pi k^2)) ArcSin(phi/(so + ss)) + c
>
> where c is a constant of integration that happens to
> be 0.
>
> I know this is not an easy sort of integral to do.
> One of the reasons I make the effort to use a
> symbolic math package is so it will save me effort
> evaluating hard integrals. At least, Mathematica
> should say it can't do it, not report false information.
>
> This makes me wonder if it safe to try to each Mathematica
> the steps.
>
> Any clues/suggestions/workarounds ?
>
> Thanks,
>
> Purvis
> purvis at mulab.physiol.upenn.edu
I'm not sure I have any suggestions or workarounds -- maybe just clues.
I can describe what's going on, and it may or may not interfere with
teaching Mathematica the steps you describe. Also, just for fun, I thought
I'd try to summarize what I do when tracking down problems like this.
Definite integrals are normally evaluated (in Mathematica) using one of two
methods -- by evaluating the indefinite integral and taking limits at the
functions.
The first method has the well-known deficiency of ignoring branch cuts
and other singularities. Also, since it uses the Limit code, problems
in Limit will also show up in Integrate. The most common problem of this
type arises from the fact that Limit can't always handle essential
singularities. (Both problems are being worked on, and neither affects
the integral in the present example.)
Various rules are used to decide which method to try first. You can
force Mathematica to use the indefinite integration method by evaluating
the indefinite integral and taking limits at the endpoints yourself.
You can (at least in V2.0) force Mathematica to use the generalized
hypergeometric method by using Integrate`IntegrateG instead of Integrate.
(This is not documented, however, and may change in future versions.)
There is currently no special support in Mathematica for multiple integrals.
Multiple integrals are evaluated as iterated integrals.
Anyway, with that background, I took a look at the integral in the example
above, which is an iterated definite integral of the following function:
In[17]:= f = E^(-(phi*u*v) - ((so + ss)*(u^2 + v^2))/2)/(u*v)
2 2
-(phi u v) - ((so + ss) (u + v ))/2
E
Out[17]= -------------------------------------
u v
This integrand has a pole at v==0, so the corresponding integral has
limited convergence properties (it isn't, for example, uniformly
convergent). It probably has a useful principal value, however. One
method that sometimes works to get the principal value of an integral
with a singularity at, say, x==x0 is to use
Limit[eps -> 0, Integrate[f, {x, a, x0-eps}] +
Integrate[f, {x, x0+eps, b}] ]
The integral in the present example is not elementary and this method
doesn't help.
The first of the iterated integrals is sent to the package that implements
the generalized hypergeometric method, which probably doesn't like
the singularity:
In[20]:= Integrate`IntegrateG[f, {v, -Infinity, Infinity}]
Infinity::indet:
Indeterminate expression ComplexInfinity + ComplexInfinity encountered.
Out[20]= Indeterminate
After this, the second of the iterated integrals is degenerate:
In[21]:= Integrate[Indeterminate, {u, -Infinity, Infinity}]
Integrate::idiv: Integral does not converge.
Out[21]= Indeterminate
The message is arguably a bit misleading, but at least we know where
it comes from.
Like I said -- a few clues, but no workarounds. The implied suggestions
(to provide special support for principal value integrals and multiple
integrals) are certainly worth pursuing.
Dave Withoff
withoff at wri.com